Monotone Convergence Theorem for Riemann Integrable functions
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I'm having a really hard time proving this statement (this is not homework):
If $f_{n} : [0,1] rightarrow mathbb{R}$ is a Riemann integrable function for all $n in mathbb{N}$, and $0 leq f_{n + 1} leq f_{n}$, and $lim limits_{n rightarrow infty} f_{n} = 0$, I need to prove that $lim limits_{n rightarrow infty} int limits_{0}^{1} f_{n}(x) text{ } mathrm{d}x = 0$.
I'm not allowed to use the Monotone Convergence Theorem for Riemann integrable functions (proving this is actually the first step in proving the MCT).
Now, I know $lim limits_{n rightarrow infty} int limits_{0}^{1} f_{n}(x) text{ } mathrm{d}x$ exists because the sequence $left { int limits_{0}^{1} f_{n}(x) text{ } mathrm{d}x right }_{n =1}^{infty}$ is a monotonically decreasing sequence that is bounded from below by $0$. However, I have no idea how to prove the limit is $0$.
Also, there is a hint to the problem. Assume $lim limits_{n rightarrow infty} int limits_{0}^{1} f_{n}(x) text{ } mathrm{d}x = epsilon > 0$. I must choose a partition $P_{n}$ for $f_{n}$ such that $P_{n + 1}$ is a refinement partition of $P_{n}$ and show that there exists an element in $[0,1]$ such that $f_{n}$ converges to some strictly positive value on that element, which would lead to a contradiction of the hypothesis of pointwise convergence to $0$.
I thought of using a sequence of closed intervals that are nested, because their intersection would be nonempty (since closed subsets of compact spaces are compact), but I can't construct the sequence. If the hint makes the problem harder, is there an easier way to prove this statement? Any help would be greatly appreciated.
real-analysis analysis functional-analysis
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show 6 more comments
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I'm having a really hard time proving this statement (this is not homework):
If $f_{n} : [0,1] rightarrow mathbb{R}$ is a Riemann integrable function for all $n in mathbb{N}$, and $0 leq f_{n + 1} leq f_{n}$, and $lim limits_{n rightarrow infty} f_{n} = 0$, I need to prove that $lim limits_{n rightarrow infty} int limits_{0}^{1} f_{n}(x) text{ } mathrm{d}x = 0$.
I'm not allowed to use the Monotone Convergence Theorem for Riemann integrable functions (proving this is actually the first step in proving the MCT).
Now, I know $lim limits_{n rightarrow infty} int limits_{0}^{1} f_{n}(x) text{ } mathrm{d}x$ exists because the sequence $left { int limits_{0}^{1} f_{n}(x) text{ } mathrm{d}x right }_{n =1}^{infty}$ is a monotonically decreasing sequence that is bounded from below by $0$. However, I have no idea how to prove the limit is $0$.
Also, there is a hint to the problem. Assume $lim limits_{n rightarrow infty} int limits_{0}^{1} f_{n}(x) text{ } mathrm{d}x = epsilon > 0$. I must choose a partition $P_{n}$ for $f_{n}$ such that $P_{n + 1}$ is a refinement partition of $P_{n}$ and show that there exists an element in $[0,1]$ such that $f_{n}$ converges to some strictly positive value on that element, which would lead to a contradiction of the hypothesis of pointwise convergence to $0$.
I thought of using a sequence of closed intervals that are nested, because their intersection would be nonempty (since closed subsets of compact spaces are compact), but I can't construct the sequence. If the hint makes the problem harder, is there an easier way to prove this statement? Any help would be greatly appreciated.
real-analysis analysis functional-analysis
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2
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Here's one proof (not using the hint).
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– David Mitra
Jul 21 '14 at 18:28
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@David Mitra Thanks, I found that on Google when I was searching. This is a hard problem! It doesn't seem like it should be, though.
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– layman
Jul 21 '14 at 20:10
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Hard problem? Yes. Back when I learned about Lebesgue integration, the instructor gave this as an example of the superiority of the Lebesgue theory over the Riemann theory.
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– GEdgar
Mar 3 '15 at 15:50
1
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Here is another proof by C. Niculescu and F. Popovici (2011), theorem 2. It doesn't use the hint but Cousin lemma and Lebesgue criterion of Riemann integrability. Put $A=varnothing$ in the proof.
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– Tony Piccolo
Nov 15 '15 at 8:45
1
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@Tony Piccolo Here is another to add to the collection. It appeared in the Monthly 2010: classicalrealanalysis.info/documents/…
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– B. S. Thomson
Nov 17 '15 at 20:49
|
show 6 more comments
$begingroup$
I'm having a really hard time proving this statement (this is not homework):
If $f_{n} : [0,1] rightarrow mathbb{R}$ is a Riemann integrable function for all $n in mathbb{N}$, and $0 leq f_{n + 1} leq f_{n}$, and $lim limits_{n rightarrow infty} f_{n} = 0$, I need to prove that $lim limits_{n rightarrow infty} int limits_{0}^{1} f_{n}(x) text{ } mathrm{d}x = 0$.
I'm not allowed to use the Monotone Convergence Theorem for Riemann integrable functions (proving this is actually the first step in proving the MCT).
Now, I know $lim limits_{n rightarrow infty} int limits_{0}^{1} f_{n}(x) text{ } mathrm{d}x$ exists because the sequence $left { int limits_{0}^{1} f_{n}(x) text{ } mathrm{d}x right }_{n =1}^{infty}$ is a monotonically decreasing sequence that is bounded from below by $0$. However, I have no idea how to prove the limit is $0$.
Also, there is a hint to the problem. Assume $lim limits_{n rightarrow infty} int limits_{0}^{1} f_{n}(x) text{ } mathrm{d}x = epsilon > 0$. I must choose a partition $P_{n}$ for $f_{n}$ such that $P_{n + 1}$ is a refinement partition of $P_{n}$ and show that there exists an element in $[0,1]$ such that $f_{n}$ converges to some strictly positive value on that element, which would lead to a contradiction of the hypothesis of pointwise convergence to $0$.
I thought of using a sequence of closed intervals that are nested, because their intersection would be nonempty (since closed subsets of compact spaces are compact), but I can't construct the sequence. If the hint makes the problem harder, is there an easier way to prove this statement? Any help would be greatly appreciated.
real-analysis analysis functional-analysis
$endgroup$
I'm having a really hard time proving this statement (this is not homework):
If $f_{n} : [0,1] rightarrow mathbb{R}$ is a Riemann integrable function for all $n in mathbb{N}$, and $0 leq f_{n + 1} leq f_{n}$, and $lim limits_{n rightarrow infty} f_{n} = 0$, I need to prove that $lim limits_{n rightarrow infty} int limits_{0}^{1} f_{n}(x) text{ } mathrm{d}x = 0$.
I'm not allowed to use the Monotone Convergence Theorem for Riemann integrable functions (proving this is actually the first step in proving the MCT).
Now, I know $lim limits_{n rightarrow infty} int limits_{0}^{1} f_{n}(x) text{ } mathrm{d}x$ exists because the sequence $left { int limits_{0}^{1} f_{n}(x) text{ } mathrm{d}x right }_{n =1}^{infty}$ is a monotonically decreasing sequence that is bounded from below by $0$. However, I have no idea how to prove the limit is $0$.
Also, there is a hint to the problem. Assume $lim limits_{n rightarrow infty} int limits_{0}^{1} f_{n}(x) text{ } mathrm{d}x = epsilon > 0$. I must choose a partition $P_{n}$ for $f_{n}$ such that $P_{n + 1}$ is a refinement partition of $P_{n}$ and show that there exists an element in $[0,1]$ such that $f_{n}$ converges to some strictly positive value on that element, which would lead to a contradiction of the hypothesis of pointwise convergence to $0$.
I thought of using a sequence of closed intervals that are nested, because their intersection would be nonempty (since closed subsets of compact spaces are compact), but I can't construct the sequence. If the hint makes the problem harder, is there an easier way to prove this statement? Any help would be greatly appreciated.
real-analysis analysis functional-analysis
real-analysis analysis functional-analysis
edited Jul 22 '14 at 17:15
layman
asked Jul 21 '14 at 17:33
laymanlayman
15.1k22156
15.1k22156
2
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Here's one proof (not using the hint).
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– David Mitra
Jul 21 '14 at 18:28
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@David Mitra Thanks, I found that on Google when I was searching. This is a hard problem! It doesn't seem like it should be, though.
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– layman
Jul 21 '14 at 20:10
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Hard problem? Yes. Back when I learned about Lebesgue integration, the instructor gave this as an example of the superiority of the Lebesgue theory over the Riemann theory.
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– GEdgar
Mar 3 '15 at 15:50
1
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Here is another proof by C. Niculescu and F. Popovici (2011), theorem 2. It doesn't use the hint but Cousin lemma and Lebesgue criterion of Riemann integrability. Put $A=varnothing$ in the proof.
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– Tony Piccolo
Nov 15 '15 at 8:45
1
$begingroup$
@Tony Piccolo Here is another to add to the collection. It appeared in the Monthly 2010: classicalrealanalysis.info/documents/…
$endgroup$
– B. S. Thomson
Nov 17 '15 at 20:49
|
show 6 more comments
2
$begingroup$
Here's one proof (not using the hint).
$endgroup$
– David Mitra
Jul 21 '14 at 18:28
$begingroup$
@David Mitra Thanks, I found that on Google when I was searching. This is a hard problem! It doesn't seem like it should be, though.
$endgroup$
– layman
Jul 21 '14 at 20:10
$begingroup$
Hard problem? Yes. Back when I learned about Lebesgue integration, the instructor gave this as an example of the superiority of the Lebesgue theory over the Riemann theory.
$endgroup$
– GEdgar
Mar 3 '15 at 15:50
1
$begingroup$
Here is another proof by C. Niculescu and F. Popovici (2011), theorem 2. It doesn't use the hint but Cousin lemma and Lebesgue criterion of Riemann integrability. Put $A=varnothing$ in the proof.
$endgroup$
– Tony Piccolo
Nov 15 '15 at 8:45
1
$begingroup$
@Tony Piccolo Here is another to add to the collection. It appeared in the Monthly 2010: classicalrealanalysis.info/documents/…
$endgroup$
– B. S. Thomson
Nov 17 '15 at 20:49
2
2
$begingroup$
Here's one proof (not using the hint).
$endgroup$
– David Mitra
Jul 21 '14 at 18:28
$begingroup$
Here's one proof (not using the hint).
$endgroup$
– David Mitra
Jul 21 '14 at 18:28
$begingroup$
@David Mitra Thanks, I found that on Google when I was searching. This is a hard problem! It doesn't seem like it should be, though.
$endgroup$
– layman
Jul 21 '14 at 20:10
$begingroup$
@David Mitra Thanks, I found that on Google when I was searching. This is a hard problem! It doesn't seem like it should be, though.
$endgroup$
– layman
Jul 21 '14 at 20:10
$begingroup$
Hard problem? Yes. Back when I learned about Lebesgue integration, the instructor gave this as an example of the superiority of the Lebesgue theory over the Riemann theory.
$endgroup$
– GEdgar
Mar 3 '15 at 15:50
$begingroup$
Hard problem? Yes. Back when I learned about Lebesgue integration, the instructor gave this as an example of the superiority of the Lebesgue theory over the Riemann theory.
$endgroup$
– GEdgar
Mar 3 '15 at 15:50
1
1
$begingroup$
Here is another proof by C. Niculescu and F. Popovici (2011), theorem 2. It doesn't use the hint but Cousin lemma and Lebesgue criterion of Riemann integrability. Put $A=varnothing$ in the proof.
$endgroup$
– Tony Piccolo
Nov 15 '15 at 8:45
$begingroup$
Here is another proof by C. Niculescu and F. Popovici (2011), theorem 2. It doesn't use the hint but Cousin lemma and Lebesgue criterion of Riemann integrability. Put $A=varnothing$ in the proof.
$endgroup$
– Tony Piccolo
Nov 15 '15 at 8:45
1
1
$begingroup$
@Tony Piccolo Here is another to add to the collection. It appeared in the Monthly 2010: classicalrealanalysis.info/documents/…
$endgroup$
– B. S. Thomson
Nov 17 '15 at 20:49
$begingroup$
@Tony Piccolo Here is another to add to the collection. It appeared in the Monthly 2010: classicalrealanalysis.info/documents/…
$endgroup$
– B. S. Thomson
Nov 17 '15 at 20:49
|
show 6 more comments
4 Answers
4
active
oldest
votes
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We need to prove the following lemma:
Lemma: If $f$ is non-negative and Riemann integrable on $[a, b]$ with $$int_{a}^{b}f(x),dx > 0$$ then there is a sub-interval of $[a, b]$ on which $f$ is positive.
The lemma is proved easily by using the definition of Riemann integral. Let $$I = int_{a}^{b}f(x),dx > 0$$ and let $0 leq f(x) < M$ for all $x in [a, b]$. By definition of Riemann integral there is a partition $$P = {x_{0}, x_{1}, x_{2}, dots, x_{n}}$$ of $[a, b]$ such that the Riemann sum $$S(f, P) = sum_{i = 1}^{n}f(t_{i})(x_{i} - x_{i - 1}) > frac{I}{2}$$ Let $$epsilon = frac{I}{2(M + b - a)}$$ and consider the set $J = {x: x in [a,b], f(x) geq epsilon}$. Now we split the Riemann sum $S(f, P)$ as $$S(f, P) = sum_{i in A}f(t_{i})(x_{i}- x_{i - 1}) + sum_{i in B}f(t_{i})(x_{i}- x_{i - 1})$$ where $$A = {i: [x_{i - 1}, x_{i}] subseteq J}, B = {i: i notin A}$$ Now for $i in A$ we can see that $f(t_{i}) leq M$ and for $i in B$ we can choose $t_{i}$ such that $f(t_{i}) < epsilon$. Then we can see that $$epsilon(M + b - a) = frac{I}{2} < S(f, P) < Msum_{i in A}(x_{i} - x_{i - 1}) + epsilon(b - a)$$ It now follows by simple algebra that $$sum_{iin A}(x_{i} - x_{i - 1}) > epsilon$$ and therefore there exist intervals of type $[x_{i - 1}, x_{i}]$ where $i in A$ where $f(x) geq epsilon$. The above proof is taken from my favorite book Mathematical Analysis by Tom M. Apostol.
Now we need to make use of this lemma for solving the current problem. Let's assume that $$lim_{n to infty}int_{0}^{1}f_{n}(x),dx = lim_{n to infty}I_{n} = c > 0$$ Then it follows that after a certain value of $n$ we have $I_{n} > c/2$. And by our lemma it follows that $f_{n}(x) geq d > 0$ where $d$ is some constant dependent only on $c$ and not on $n$. Moreover we see that there is a partition $P_{n}$ of $[0, 1]$ such that $f_{n}(x)geq d$ in some of the sub-intervals made by $P_{n}$ (whose total length also exceeds $d$) and $f_{n}(x) < d$ for some point in remaining sub-intervals. If necessary we can replace $P_{n + 1}$ by $P_{n} cup P_{n + 1}$ and hence it is OK to assume that $P_{n + 1}$ is a refinement of $P_{n}$.
Since $f_{n}$ is monotonically decreasing it follows that the sub-intervals of partition $P_{n + 1}$ where $f_{n + 1}(x) geq d$ are contained in the corresponding sub-intervals of $P_{n}$ and again these sub-intervals of $P_{n + 1}$ have a total length greater than $d$. Thus it follows that there is a sub-interval of $[0, 1]$ where $f_{n}(x) geq d > 0$ for all $n$ after a certain value. This contradicts the fact that $f_{n}$ converges pointwise to $0$.
The above proof is wrong and the wrong inferences are striked out. The approach however can be salvaged with some more effort. An alternative approach is given in this answer.
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add a comment |
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The following proof uses the concept of uniform convergence.
First note that $;f_n to 0$ on $[0,1]$, which is a compact set, the limit function is continuous and $;f_n$ decreases monotonically. Then $;f_n$ converges uniformly to $0$ by Dini's theorem. Thus
begin{equation}
lim_{n to infty} int_0^1 f_n(x) operatorname{d}!x = int_0^1lim_{n to infty} f_n(x) operatorname{d}!x = int_0^1 0 operatorname{d}!x = 0.
end{equation}
(The interchange of limits is legitimate due to uniform convergence of $f_n$ and the fact that the interval $[0,1]$ is compact.)
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add a comment |
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Since $int_0^1 f_n $ is monotone decreasing in $n$, argue by contradiction, we assume that
$$ int_0^1 f_n geq 2 epsilon quad forall n.$$
By the definition of Riemann integral $int_0^1 f_n = sup_{gleq f_n, text{ piecewise constant }} int_0^1 g$, there exists a piecewise constant function $g_n$ for each $f_n $ such that
$$int_0^1 g_n geq epsilonquad forall n.$$
Now, $g_n leq f_1$ for each $n$, and because $f_1$ is Riemann integrable, $f_1$ is bounded above by a constant $C$ , thus we have $g_nleq C$.
$$epsilon leq int_0^1 g_n = sum a^n_i |I^n_i| leq C sum |I^n_i|$$
which means that
$$frac{epsilon}{C}leq sum |I^n_i| quad forall n.$$
The above inequality implies that the sum of the length of the intervals which each $g_n$ takes a positive value is bounded below by $frac{epsilon}{C}$ for each $n$. And such $g_nleq f_n$ exists for all $f_n$ , this contradicts the fact that $f_n rightarrow 0$ pointwise.
Edit: If you are allowed to use results from measure theory and Lebesgue integration, see Bounded Convergence theorem or Dominated Convergence theorem.
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1
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I don't get the end. These intervals might not be nested as $n$ increases.
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– Ted Shifrin
Jul 21 '14 at 23:07
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The bound $frac{epsilon}{C}$ does not depend on $n$. So for each $f_n$ there exists a function $gleq f_n$ such that $m({x: g(x) > 0}) geq frac{epsilon}{C}$. And because the domain is bounded, the mass can not escape like the function $chi_{[n,n+1]}$ define on $mathbb{R}$. So you will not get $f_n rightarrow 0$ pointwise.
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– Xiao
Jul 22 '14 at 7:47
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Please explain: you say "there exists a piecewise constant function $g$ for each $f_n$ such that ..." So the choice of $g$ depends on $f_n$ in which case there should be a label $g_n$ that depends on $n$. And then how does the proof go?
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– B. S. Thomson
Nov 17 '15 at 22:26
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I didn't notice the date so Mr Xiao is probably not around to answer, so I leave it as a fatal flaw in an attempted proof. It is not quite that easy, although maybe someone can find easier proofs than those published.
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– B. S. Thomson
Nov 18 '15 at 0:32
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@B.S.Thomson I kinda just did a quick edit of the answer as you said. I will take a closer look of this.
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– Xiao
Nov 18 '15 at 0:41
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show 6 more comments
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The following proof uses the countable additivity of Jordan content on its domain (i.e. if one knows from the start that the disjoint union has Jordan content).
Let $A_i={(x,y) in [0,1] times mathbb R : 0<y<f_i(x)}$.
By the monotone convergence of ${f_i}$ to zero, $A_{i+1} subseteq A_i$ and $,bigcap_{i=1}^ infty A_i = varnothing$.
So, if $,B_i=A_i setminus A_{i+1}$, then ${B_i}$ is a disjoint sequence and $,bigcup_{i=1}^ infty B_i = A_1$.
By the R-integrability of $f_i$, the set $,B_i,$ has Jordan content (in $mathbb R^2$) and , by the countable additivity of Jordan content on its domain, the series $sum_{i=1}^ infty m(B_i)$ is convergent, so the sequence of its remainders ${sum_{i=n}^ infty m(B_i)}$ is null.
But $,sum_{i=n}^ infty m(B_i) = m(A_n)= int_0^1 f_n ,dots$
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add a comment |
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4 Answers
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4 Answers
4
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$begingroup$
We need to prove the following lemma:
Lemma: If $f$ is non-negative and Riemann integrable on $[a, b]$ with $$int_{a}^{b}f(x),dx > 0$$ then there is a sub-interval of $[a, b]$ on which $f$ is positive.
The lemma is proved easily by using the definition of Riemann integral. Let $$I = int_{a}^{b}f(x),dx > 0$$ and let $0 leq f(x) < M$ for all $x in [a, b]$. By definition of Riemann integral there is a partition $$P = {x_{0}, x_{1}, x_{2}, dots, x_{n}}$$ of $[a, b]$ such that the Riemann sum $$S(f, P) = sum_{i = 1}^{n}f(t_{i})(x_{i} - x_{i - 1}) > frac{I}{2}$$ Let $$epsilon = frac{I}{2(M + b - a)}$$ and consider the set $J = {x: x in [a,b], f(x) geq epsilon}$. Now we split the Riemann sum $S(f, P)$ as $$S(f, P) = sum_{i in A}f(t_{i})(x_{i}- x_{i - 1}) + sum_{i in B}f(t_{i})(x_{i}- x_{i - 1})$$ where $$A = {i: [x_{i - 1}, x_{i}] subseteq J}, B = {i: i notin A}$$ Now for $i in A$ we can see that $f(t_{i}) leq M$ and for $i in B$ we can choose $t_{i}$ such that $f(t_{i}) < epsilon$. Then we can see that $$epsilon(M + b - a) = frac{I}{2} < S(f, P) < Msum_{i in A}(x_{i} - x_{i - 1}) + epsilon(b - a)$$ It now follows by simple algebra that $$sum_{iin A}(x_{i} - x_{i - 1}) > epsilon$$ and therefore there exist intervals of type $[x_{i - 1}, x_{i}]$ where $i in A$ where $f(x) geq epsilon$. The above proof is taken from my favorite book Mathematical Analysis by Tom M. Apostol.
Now we need to make use of this lemma for solving the current problem. Let's assume that $$lim_{n to infty}int_{0}^{1}f_{n}(x),dx = lim_{n to infty}I_{n} = c > 0$$ Then it follows that after a certain value of $n$ we have $I_{n} > c/2$. And by our lemma it follows that $f_{n}(x) geq d > 0$ where $d$ is some constant dependent only on $c$ and not on $n$. Moreover we see that there is a partition $P_{n}$ of $[0, 1]$ such that $f_{n}(x)geq d$ in some of the sub-intervals made by $P_{n}$ (whose total length also exceeds $d$) and $f_{n}(x) < d$ for some point in remaining sub-intervals. If necessary we can replace $P_{n + 1}$ by $P_{n} cup P_{n + 1}$ and hence it is OK to assume that $P_{n + 1}$ is a refinement of $P_{n}$.
Since $f_{n}$ is monotonically decreasing it follows that the sub-intervals of partition $P_{n + 1}$ where $f_{n + 1}(x) geq d$ are contained in the corresponding sub-intervals of $P_{n}$ and again these sub-intervals of $P_{n + 1}$ have a total length greater than $d$. Thus it follows that there is a sub-interval of $[0, 1]$ where $f_{n}(x) geq d > 0$ for all $n$ after a certain value. This contradicts the fact that $f_{n}$ converges pointwise to $0$.
The above proof is wrong and the wrong inferences are striked out. The approach however can be salvaged with some more effort. An alternative approach is given in this answer.
$endgroup$
add a comment |
$begingroup$
We need to prove the following lemma:
Lemma: If $f$ is non-negative and Riemann integrable on $[a, b]$ with $$int_{a}^{b}f(x),dx > 0$$ then there is a sub-interval of $[a, b]$ on which $f$ is positive.
The lemma is proved easily by using the definition of Riemann integral. Let $$I = int_{a}^{b}f(x),dx > 0$$ and let $0 leq f(x) < M$ for all $x in [a, b]$. By definition of Riemann integral there is a partition $$P = {x_{0}, x_{1}, x_{2}, dots, x_{n}}$$ of $[a, b]$ such that the Riemann sum $$S(f, P) = sum_{i = 1}^{n}f(t_{i})(x_{i} - x_{i - 1}) > frac{I}{2}$$ Let $$epsilon = frac{I}{2(M + b - a)}$$ and consider the set $J = {x: x in [a,b], f(x) geq epsilon}$. Now we split the Riemann sum $S(f, P)$ as $$S(f, P) = sum_{i in A}f(t_{i})(x_{i}- x_{i - 1}) + sum_{i in B}f(t_{i})(x_{i}- x_{i - 1})$$ where $$A = {i: [x_{i - 1}, x_{i}] subseteq J}, B = {i: i notin A}$$ Now for $i in A$ we can see that $f(t_{i}) leq M$ and for $i in B$ we can choose $t_{i}$ such that $f(t_{i}) < epsilon$. Then we can see that $$epsilon(M + b - a) = frac{I}{2} < S(f, P) < Msum_{i in A}(x_{i} - x_{i - 1}) + epsilon(b - a)$$ It now follows by simple algebra that $$sum_{iin A}(x_{i} - x_{i - 1}) > epsilon$$ and therefore there exist intervals of type $[x_{i - 1}, x_{i}]$ where $i in A$ where $f(x) geq epsilon$. The above proof is taken from my favorite book Mathematical Analysis by Tom M. Apostol.
Now we need to make use of this lemma for solving the current problem. Let's assume that $$lim_{n to infty}int_{0}^{1}f_{n}(x),dx = lim_{n to infty}I_{n} = c > 0$$ Then it follows that after a certain value of $n$ we have $I_{n} > c/2$. And by our lemma it follows that $f_{n}(x) geq d > 0$ where $d$ is some constant dependent only on $c$ and not on $n$. Moreover we see that there is a partition $P_{n}$ of $[0, 1]$ such that $f_{n}(x)geq d$ in some of the sub-intervals made by $P_{n}$ (whose total length also exceeds $d$) and $f_{n}(x) < d$ for some point in remaining sub-intervals. If necessary we can replace $P_{n + 1}$ by $P_{n} cup P_{n + 1}$ and hence it is OK to assume that $P_{n + 1}$ is a refinement of $P_{n}$.
Since $f_{n}$ is monotonically decreasing it follows that the sub-intervals of partition $P_{n + 1}$ where $f_{n + 1}(x) geq d$ are contained in the corresponding sub-intervals of $P_{n}$ and again these sub-intervals of $P_{n + 1}$ have a total length greater than $d$. Thus it follows that there is a sub-interval of $[0, 1]$ where $f_{n}(x) geq d > 0$ for all $n$ after a certain value. This contradicts the fact that $f_{n}$ converges pointwise to $0$.
The above proof is wrong and the wrong inferences are striked out. The approach however can be salvaged with some more effort. An alternative approach is given in this answer.
$endgroup$
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$begingroup$
We need to prove the following lemma:
Lemma: If $f$ is non-negative and Riemann integrable on $[a, b]$ with $$int_{a}^{b}f(x),dx > 0$$ then there is a sub-interval of $[a, b]$ on which $f$ is positive.
The lemma is proved easily by using the definition of Riemann integral. Let $$I = int_{a}^{b}f(x),dx > 0$$ and let $0 leq f(x) < M$ for all $x in [a, b]$. By definition of Riemann integral there is a partition $$P = {x_{0}, x_{1}, x_{2}, dots, x_{n}}$$ of $[a, b]$ such that the Riemann sum $$S(f, P) = sum_{i = 1}^{n}f(t_{i})(x_{i} - x_{i - 1}) > frac{I}{2}$$ Let $$epsilon = frac{I}{2(M + b - a)}$$ and consider the set $J = {x: x in [a,b], f(x) geq epsilon}$. Now we split the Riemann sum $S(f, P)$ as $$S(f, P) = sum_{i in A}f(t_{i})(x_{i}- x_{i - 1}) + sum_{i in B}f(t_{i})(x_{i}- x_{i - 1})$$ where $$A = {i: [x_{i - 1}, x_{i}] subseteq J}, B = {i: i notin A}$$ Now for $i in A$ we can see that $f(t_{i}) leq M$ and for $i in B$ we can choose $t_{i}$ such that $f(t_{i}) < epsilon$. Then we can see that $$epsilon(M + b - a) = frac{I}{2} < S(f, P) < Msum_{i in A}(x_{i} - x_{i - 1}) + epsilon(b - a)$$ It now follows by simple algebra that $$sum_{iin A}(x_{i} - x_{i - 1}) > epsilon$$ and therefore there exist intervals of type $[x_{i - 1}, x_{i}]$ where $i in A$ where $f(x) geq epsilon$. The above proof is taken from my favorite book Mathematical Analysis by Tom M. Apostol.
Now we need to make use of this lemma for solving the current problem. Let's assume that $$lim_{n to infty}int_{0}^{1}f_{n}(x),dx = lim_{n to infty}I_{n} = c > 0$$ Then it follows that after a certain value of $n$ we have $I_{n} > c/2$. And by our lemma it follows that $f_{n}(x) geq d > 0$ where $d$ is some constant dependent only on $c$ and not on $n$. Moreover we see that there is a partition $P_{n}$ of $[0, 1]$ such that $f_{n}(x)geq d$ in some of the sub-intervals made by $P_{n}$ (whose total length also exceeds $d$) and $f_{n}(x) < d$ for some point in remaining sub-intervals. If necessary we can replace $P_{n + 1}$ by $P_{n} cup P_{n + 1}$ and hence it is OK to assume that $P_{n + 1}$ is a refinement of $P_{n}$.
Since $f_{n}$ is monotonically decreasing it follows that the sub-intervals of partition $P_{n + 1}$ where $f_{n + 1}(x) geq d$ are contained in the corresponding sub-intervals of $P_{n}$ and again these sub-intervals of $P_{n + 1}$ have a total length greater than $d$. Thus it follows that there is a sub-interval of $[0, 1]$ where $f_{n}(x) geq d > 0$ for all $n$ after a certain value. This contradicts the fact that $f_{n}$ converges pointwise to $0$.
The above proof is wrong and the wrong inferences are striked out. The approach however can be salvaged with some more effort. An alternative approach is given in this answer.
$endgroup$
We need to prove the following lemma:
Lemma: If $f$ is non-negative and Riemann integrable on $[a, b]$ with $$int_{a}^{b}f(x),dx > 0$$ then there is a sub-interval of $[a, b]$ on which $f$ is positive.
The lemma is proved easily by using the definition of Riemann integral. Let $$I = int_{a}^{b}f(x),dx > 0$$ and let $0 leq f(x) < M$ for all $x in [a, b]$. By definition of Riemann integral there is a partition $$P = {x_{0}, x_{1}, x_{2}, dots, x_{n}}$$ of $[a, b]$ such that the Riemann sum $$S(f, P) = sum_{i = 1}^{n}f(t_{i})(x_{i} - x_{i - 1}) > frac{I}{2}$$ Let $$epsilon = frac{I}{2(M + b - a)}$$ and consider the set $J = {x: x in [a,b], f(x) geq epsilon}$. Now we split the Riemann sum $S(f, P)$ as $$S(f, P) = sum_{i in A}f(t_{i})(x_{i}- x_{i - 1}) + sum_{i in B}f(t_{i})(x_{i}- x_{i - 1})$$ where $$A = {i: [x_{i - 1}, x_{i}] subseteq J}, B = {i: i notin A}$$ Now for $i in A$ we can see that $f(t_{i}) leq M$ and for $i in B$ we can choose $t_{i}$ such that $f(t_{i}) < epsilon$. Then we can see that $$epsilon(M + b - a) = frac{I}{2} < S(f, P) < Msum_{i in A}(x_{i} - x_{i - 1}) + epsilon(b - a)$$ It now follows by simple algebra that $$sum_{iin A}(x_{i} - x_{i - 1}) > epsilon$$ and therefore there exist intervals of type $[x_{i - 1}, x_{i}]$ where $i in A$ where $f(x) geq epsilon$. The above proof is taken from my favorite book Mathematical Analysis by Tom M. Apostol.
Now we need to make use of this lemma for solving the current problem. Let's assume that $$lim_{n to infty}int_{0}^{1}f_{n}(x),dx = lim_{n to infty}I_{n} = c > 0$$ Then it follows that after a certain value of $n$ we have $I_{n} > c/2$. And by our lemma it follows that $f_{n}(x) geq d > 0$ where $d$ is some constant dependent only on $c$ and not on $n$. Moreover we see that there is a partition $P_{n}$ of $[0, 1]$ such that $f_{n}(x)geq d$ in some of the sub-intervals made by $P_{n}$ (whose total length also exceeds $d$) and $f_{n}(x) < d$ for some point in remaining sub-intervals. If necessary we can replace $P_{n + 1}$ by $P_{n} cup P_{n + 1}$ and hence it is OK to assume that $P_{n + 1}$ is a refinement of $P_{n}$.
Since $f_{n}$ is monotonically decreasing it follows that the sub-intervals of partition $P_{n + 1}$ where $f_{n + 1}(x) geq d$ are contained in the corresponding sub-intervals of $P_{n}$ and again these sub-intervals of $P_{n + 1}$ have a total length greater than $d$. Thus it follows that there is a sub-interval of $[0, 1]$ where $f_{n}(x) geq d > 0$ for all $n$ after a certain value. This contradicts the fact that $f_{n}$ converges pointwise to $0$.
The above proof is wrong and the wrong inferences are striked out. The approach however can be salvaged with some more effort. An alternative approach is given in this answer.
edited Dec 14 '18 at 8:59
answered Feb 17 '17 at 9:54
Paramanand SinghParamanand Singh
50.1k556163
50.1k556163
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The following proof uses the concept of uniform convergence.
First note that $;f_n to 0$ on $[0,1]$, which is a compact set, the limit function is continuous and $;f_n$ decreases monotonically. Then $;f_n$ converges uniformly to $0$ by Dini's theorem. Thus
begin{equation}
lim_{n to infty} int_0^1 f_n(x) operatorname{d}!x = int_0^1lim_{n to infty} f_n(x) operatorname{d}!x = int_0^1 0 operatorname{d}!x = 0.
end{equation}
(The interchange of limits is legitimate due to uniform convergence of $f_n$ and the fact that the interval $[0,1]$ is compact.)
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add a comment |
$begingroup$
The following proof uses the concept of uniform convergence.
First note that $;f_n to 0$ on $[0,1]$, which is a compact set, the limit function is continuous and $;f_n$ decreases monotonically. Then $;f_n$ converges uniformly to $0$ by Dini's theorem. Thus
begin{equation}
lim_{n to infty} int_0^1 f_n(x) operatorname{d}!x = int_0^1lim_{n to infty} f_n(x) operatorname{d}!x = int_0^1 0 operatorname{d}!x = 0.
end{equation}
(The interchange of limits is legitimate due to uniform convergence of $f_n$ and the fact that the interval $[0,1]$ is compact.)
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add a comment |
$begingroup$
The following proof uses the concept of uniform convergence.
First note that $;f_n to 0$ on $[0,1]$, which is a compact set, the limit function is continuous and $;f_n$ decreases monotonically. Then $;f_n$ converges uniformly to $0$ by Dini's theorem. Thus
begin{equation}
lim_{n to infty} int_0^1 f_n(x) operatorname{d}!x = int_0^1lim_{n to infty} f_n(x) operatorname{d}!x = int_0^1 0 operatorname{d}!x = 0.
end{equation}
(The interchange of limits is legitimate due to uniform convergence of $f_n$ and the fact that the interval $[0,1]$ is compact.)
$endgroup$
The following proof uses the concept of uniform convergence.
First note that $;f_n to 0$ on $[0,1]$, which is a compact set, the limit function is continuous and $;f_n$ decreases monotonically. Then $;f_n$ converges uniformly to $0$ by Dini's theorem. Thus
begin{equation}
lim_{n to infty} int_0^1 f_n(x) operatorname{d}!x = int_0^1lim_{n to infty} f_n(x) operatorname{d}!x = int_0^1 0 operatorname{d}!x = 0.
end{equation}
(The interchange of limits is legitimate due to uniform convergence of $f_n$ and the fact that the interval $[0,1]$ is compact.)
answered Dec 14 '18 at 10:04
Ken HungKen Hung
613
613
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Since $int_0^1 f_n $ is monotone decreasing in $n$, argue by contradiction, we assume that
$$ int_0^1 f_n geq 2 epsilon quad forall n.$$
By the definition of Riemann integral $int_0^1 f_n = sup_{gleq f_n, text{ piecewise constant }} int_0^1 g$, there exists a piecewise constant function $g_n$ for each $f_n $ such that
$$int_0^1 g_n geq epsilonquad forall n.$$
Now, $g_n leq f_1$ for each $n$, and because $f_1$ is Riemann integrable, $f_1$ is bounded above by a constant $C$ , thus we have $g_nleq C$.
$$epsilon leq int_0^1 g_n = sum a^n_i |I^n_i| leq C sum |I^n_i|$$
which means that
$$frac{epsilon}{C}leq sum |I^n_i| quad forall n.$$
The above inequality implies that the sum of the length of the intervals which each $g_n$ takes a positive value is bounded below by $frac{epsilon}{C}$ for each $n$. And such $g_nleq f_n$ exists for all $f_n$ , this contradicts the fact that $f_n rightarrow 0$ pointwise.
Edit: If you are allowed to use results from measure theory and Lebesgue integration, see Bounded Convergence theorem or Dominated Convergence theorem.
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1
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I don't get the end. These intervals might not be nested as $n$ increases.
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– Ted Shifrin
Jul 21 '14 at 23:07
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The bound $frac{epsilon}{C}$ does not depend on $n$. So for each $f_n$ there exists a function $gleq f_n$ such that $m({x: g(x) > 0}) geq frac{epsilon}{C}$. And because the domain is bounded, the mass can not escape like the function $chi_{[n,n+1]}$ define on $mathbb{R}$. So you will not get $f_n rightarrow 0$ pointwise.
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– Xiao
Jul 22 '14 at 7:47
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Please explain: you say "there exists a piecewise constant function $g$ for each $f_n$ such that ..." So the choice of $g$ depends on $f_n$ in which case there should be a label $g_n$ that depends on $n$. And then how does the proof go?
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– B. S. Thomson
Nov 17 '15 at 22:26
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I didn't notice the date so Mr Xiao is probably not around to answer, so I leave it as a fatal flaw in an attempted proof. It is not quite that easy, although maybe someone can find easier proofs than those published.
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– B. S. Thomson
Nov 18 '15 at 0:32
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@B.S.Thomson I kinda just did a quick edit of the answer as you said. I will take a closer look of this.
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– Xiao
Nov 18 '15 at 0:41
|
show 6 more comments
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Since $int_0^1 f_n $ is monotone decreasing in $n$, argue by contradiction, we assume that
$$ int_0^1 f_n geq 2 epsilon quad forall n.$$
By the definition of Riemann integral $int_0^1 f_n = sup_{gleq f_n, text{ piecewise constant }} int_0^1 g$, there exists a piecewise constant function $g_n$ for each $f_n $ such that
$$int_0^1 g_n geq epsilonquad forall n.$$
Now, $g_n leq f_1$ for each $n$, and because $f_1$ is Riemann integrable, $f_1$ is bounded above by a constant $C$ , thus we have $g_nleq C$.
$$epsilon leq int_0^1 g_n = sum a^n_i |I^n_i| leq C sum |I^n_i|$$
which means that
$$frac{epsilon}{C}leq sum |I^n_i| quad forall n.$$
The above inequality implies that the sum of the length of the intervals which each $g_n$ takes a positive value is bounded below by $frac{epsilon}{C}$ for each $n$. And such $g_nleq f_n$ exists for all $f_n$ , this contradicts the fact that $f_n rightarrow 0$ pointwise.
Edit: If you are allowed to use results from measure theory and Lebesgue integration, see Bounded Convergence theorem or Dominated Convergence theorem.
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1
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I don't get the end. These intervals might not be nested as $n$ increases.
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– Ted Shifrin
Jul 21 '14 at 23:07
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The bound $frac{epsilon}{C}$ does not depend on $n$. So for each $f_n$ there exists a function $gleq f_n$ such that $m({x: g(x) > 0}) geq frac{epsilon}{C}$. And because the domain is bounded, the mass can not escape like the function $chi_{[n,n+1]}$ define on $mathbb{R}$. So you will not get $f_n rightarrow 0$ pointwise.
$endgroup$
– Xiao
Jul 22 '14 at 7:47
$begingroup$
Please explain: you say "there exists a piecewise constant function $g$ for each $f_n$ such that ..." So the choice of $g$ depends on $f_n$ in which case there should be a label $g_n$ that depends on $n$. And then how does the proof go?
$endgroup$
– B. S. Thomson
Nov 17 '15 at 22:26
$begingroup$
I didn't notice the date so Mr Xiao is probably not around to answer, so I leave it as a fatal flaw in an attempted proof. It is not quite that easy, although maybe someone can find easier proofs than those published.
$endgroup$
– B. S. Thomson
Nov 18 '15 at 0:32
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@B.S.Thomson I kinda just did a quick edit of the answer as you said. I will take a closer look of this.
$endgroup$
– Xiao
Nov 18 '15 at 0:41
|
show 6 more comments
$begingroup$
Since $int_0^1 f_n $ is monotone decreasing in $n$, argue by contradiction, we assume that
$$ int_0^1 f_n geq 2 epsilon quad forall n.$$
By the definition of Riemann integral $int_0^1 f_n = sup_{gleq f_n, text{ piecewise constant }} int_0^1 g$, there exists a piecewise constant function $g_n$ for each $f_n $ such that
$$int_0^1 g_n geq epsilonquad forall n.$$
Now, $g_n leq f_1$ for each $n$, and because $f_1$ is Riemann integrable, $f_1$ is bounded above by a constant $C$ , thus we have $g_nleq C$.
$$epsilon leq int_0^1 g_n = sum a^n_i |I^n_i| leq C sum |I^n_i|$$
which means that
$$frac{epsilon}{C}leq sum |I^n_i| quad forall n.$$
The above inequality implies that the sum of the length of the intervals which each $g_n$ takes a positive value is bounded below by $frac{epsilon}{C}$ for each $n$. And such $g_nleq f_n$ exists for all $f_n$ , this contradicts the fact that $f_n rightarrow 0$ pointwise.
Edit: If you are allowed to use results from measure theory and Lebesgue integration, see Bounded Convergence theorem or Dominated Convergence theorem.
$endgroup$
Since $int_0^1 f_n $ is monotone decreasing in $n$, argue by contradiction, we assume that
$$ int_0^1 f_n geq 2 epsilon quad forall n.$$
By the definition of Riemann integral $int_0^1 f_n = sup_{gleq f_n, text{ piecewise constant }} int_0^1 g$, there exists a piecewise constant function $g_n$ for each $f_n $ such that
$$int_0^1 g_n geq epsilonquad forall n.$$
Now, $g_n leq f_1$ for each $n$, and because $f_1$ is Riemann integrable, $f_1$ is bounded above by a constant $C$ , thus we have $g_nleq C$.
$$epsilon leq int_0^1 g_n = sum a^n_i |I^n_i| leq C sum |I^n_i|$$
which means that
$$frac{epsilon}{C}leq sum |I^n_i| quad forall n.$$
The above inequality implies that the sum of the length of the intervals which each $g_n$ takes a positive value is bounded below by $frac{epsilon}{C}$ for each $n$. And such $g_nleq f_n$ exists for all $f_n$ , this contradicts the fact that $f_n rightarrow 0$ pointwise.
Edit: If you are allowed to use results from measure theory and Lebesgue integration, see Bounded Convergence theorem or Dominated Convergence theorem.
edited Nov 18 '15 at 0:38
answered Jul 21 '14 at 22:53
XiaoXiao
4,81611435
4,81611435
1
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I don't get the end. These intervals might not be nested as $n$ increases.
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– Ted Shifrin
Jul 21 '14 at 23:07
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The bound $frac{epsilon}{C}$ does not depend on $n$. So for each $f_n$ there exists a function $gleq f_n$ such that $m({x: g(x) > 0}) geq frac{epsilon}{C}$. And because the domain is bounded, the mass can not escape like the function $chi_{[n,n+1]}$ define on $mathbb{R}$. So you will not get $f_n rightarrow 0$ pointwise.
$endgroup$
– Xiao
Jul 22 '14 at 7:47
$begingroup$
Please explain: you say "there exists a piecewise constant function $g$ for each $f_n$ such that ..." So the choice of $g$ depends on $f_n$ in which case there should be a label $g_n$ that depends on $n$. And then how does the proof go?
$endgroup$
– B. S. Thomson
Nov 17 '15 at 22:26
$begingroup$
I didn't notice the date so Mr Xiao is probably not around to answer, so I leave it as a fatal flaw in an attempted proof. It is not quite that easy, although maybe someone can find easier proofs than those published.
$endgroup$
– B. S. Thomson
Nov 18 '15 at 0:32
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@B.S.Thomson I kinda just did a quick edit of the answer as you said. I will take a closer look of this.
$endgroup$
– Xiao
Nov 18 '15 at 0:41
|
show 6 more comments
1
$begingroup$
I don't get the end. These intervals might not be nested as $n$ increases.
$endgroup$
– Ted Shifrin
Jul 21 '14 at 23:07
$begingroup$
The bound $frac{epsilon}{C}$ does not depend on $n$. So for each $f_n$ there exists a function $gleq f_n$ such that $m({x: g(x) > 0}) geq frac{epsilon}{C}$. And because the domain is bounded, the mass can not escape like the function $chi_{[n,n+1]}$ define on $mathbb{R}$. So you will not get $f_n rightarrow 0$ pointwise.
$endgroup$
– Xiao
Jul 22 '14 at 7:47
$begingroup$
Please explain: you say "there exists a piecewise constant function $g$ for each $f_n$ such that ..." So the choice of $g$ depends on $f_n$ in which case there should be a label $g_n$ that depends on $n$. And then how does the proof go?
$endgroup$
– B. S. Thomson
Nov 17 '15 at 22:26
$begingroup$
I didn't notice the date so Mr Xiao is probably not around to answer, so I leave it as a fatal flaw in an attempted proof. It is not quite that easy, although maybe someone can find easier proofs than those published.
$endgroup$
– B. S. Thomson
Nov 18 '15 at 0:32
$begingroup$
@B.S.Thomson I kinda just did a quick edit of the answer as you said. I will take a closer look of this.
$endgroup$
– Xiao
Nov 18 '15 at 0:41
1
1
$begingroup$
I don't get the end. These intervals might not be nested as $n$ increases.
$endgroup$
– Ted Shifrin
Jul 21 '14 at 23:07
$begingroup$
I don't get the end. These intervals might not be nested as $n$ increases.
$endgroup$
– Ted Shifrin
Jul 21 '14 at 23:07
$begingroup$
The bound $frac{epsilon}{C}$ does not depend on $n$. So for each $f_n$ there exists a function $gleq f_n$ such that $m({x: g(x) > 0}) geq frac{epsilon}{C}$. And because the domain is bounded, the mass can not escape like the function $chi_{[n,n+1]}$ define on $mathbb{R}$. So you will not get $f_n rightarrow 0$ pointwise.
$endgroup$
– Xiao
Jul 22 '14 at 7:47
$begingroup$
The bound $frac{epsilon}{C}$ does not depend on $n$. So for each $f_n$ there exists a function $gleq f_n$ such that $m({x: g(x) > 0}) geq frac{epsilon}{C}$. And because the domain is bounded, the mass can not escape like the function $chi_{[n,n+1]}$ define on $mathbb{R}$. So you will not get $f_n rightarrow 0$ pointwise.
$endgroup$
– Xiao
Jul 22 '14 at 7:47
$begingroup$
Please explain: you say "there exists a piecewise constant function $g$ for each $f_n$ such that ..." So the choice of $g$ depends on $f_n$ in which case there should be a label $g_n$ that depends on $n$. And then how does the proof go?
$endgroup$
– B. S. Thomson
Nov 17 '15 at 22:26
$begingroup$
Please explain: you say "there exists a piecewise constant function $g$ for each $f_n$ such that ..." So the choice of $g$ depends on $f_n$ in which case there should be a label $g_n$ that depends on $n$. And then how does the proof go?
$endgroup$
– B. S. Thomson
Nov 17 '15 at 22:26
$begingroup$
I didn't notice the date so Mr Xiao is probably not around to answer, so I leave it as a fatal flaw in an attempted proof. It is not quite that easy, although maybe someone can find easier proofs than those published.
$endgroup$
– B. S. Thomson
Nov 18 '15 at 0:32
$begingroup$
I didn't notice the date so Mr Xiao is probably not around to answer, so I leave it as a fatal flaw in an attempted proof. It is not quite that easy, although maybe someone can find easier proofs than those published.
$endgroup$
– B. S. Thomson
Nov 18 '15 at 0:32
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@B.S.Thomson I kinda just did a quick edit of the answer as you said. I will take a closer look of this.
$endgroup$
– Xiao
Nov 18 '15 at 0:41
$begingroup$
@B.S.Thomson I kinda just did a quick edit of the answer as you said. I will take a closer look of this.
$endgroup$
– Xiao
Nov 18 '15 at 0:41
|
show 6 more comments
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The following proof uses the countable additivity of Jordan content on its domain (i.e. if one knows from the start that the disjoint union has Jordan content).
Let $A_i={(x,y) in [0,1] times mathbb R : 0<y<f_i(x)}$.
By the monotone convergence of ${f_i}$ to zero, $A_{i+1} subseteq A_i$ and $,bigcap_{i=1}^ infty A_i = varnothing$.
So, if $,B_i=A_i setminus A_{i+1}$, then ${B_i}$ is a disjoint sequence and $,bigcup_{i=1}^ infty B_i = A_1$.
By the R-integrability of $f_i$, the set $,B_i,$ has Jordan content (in $mathbb R^2$) and , by the countable additivity of Jordan content on its domain, the series $sum_{i=1}^ infty m(B_i)$ is convergent, so the sequence of its remainders ${sum_{i=n}^ infty m(B_i)}$ is null.
But $,sum_{i=n}^ infty m(B_i) = m(A_n)= int_0^1 f_n ,dots$
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add a comment |
$begingroup$
The following proof uses the countable additivity of Jordan content on its domain (i.e. if one knows from the start that the disjoint union has Jordan content).
Let $A_i={(x,y) in [0,1] times mathbb R : 0<y<f_i(x)}$.
By the monotone convergence of ${f_i}$ to zero, $A_{i+1} subseteq A_i$ and $,bigcap_{i=1}^ infty A_i = varnothing$.
So, if $,B_i=A_i setminus A_{i+1}$, then ${B_i}$ is a disjoint sequence and $,bigcup_{i=1}^ infty B_i = A_1$.
By the R-integrability of $f_i$, the set $,B_i,$ has Jordan content (in $mathbb R^2$) and , by the countable additivity of Jordan content on its domain, the series $sum_{i=1}^ infty m(B_i)$ is convergent, so the sequence of its remainders ${sum_{i=n}^ infty m(B_i)}$ is null.
But $,sum_{i=n}^ infty m(B_i) = m(A_n)= int_0^1 f_n ,dots$
$endgroup$
add a comment |
$begingroup$
The following proof uses the countable additivity of Jordan content on its domain (i.e. if one knows from the start that the disjoint union has Jordan content).
Let $A_i={(x,y) in [0,1] times mathbb R : 0<y<f_i(x)}$.
By the monotone convergence of ${f_i}$ to zero, $A_{i+1} subseteq A_i$ and $,bigcap_{i=1}^ infty A_i = varnothing$.
So, if $,B_i=A_i setminus A_{i+1}$, then ${B_i}$ is a disjoint sequence and $,bigcup_{i=1}^ infty B_i = A_1$.
By the R-integrability of $f_i$, the set $,B_i,$ has Jordan content (in $mathbb R^2$) and , by the countable additivity of Jordan content on its domain, the series $sum_{i=1}^ infty m(B_i)$ is convergent, so the sequence of its remainders ${sum_{i=n}^ infty m(B_i)}$ is null.
But $,sum_{i=n}^ infty m(B_i) = m(A_n)= int_0^1 f_n ,dots$
$endgroup$
The following proof uses the countable additivity of Jordan content on its domain (i.e. if one knows from the start that the disjoint union has Jordan content).
Let $A_i={(x,y) in [0,1] times mathbb R : 0<y<f_i(x)}$.
By the monotone convergence of ${f_i}$ to zero, $A_{i+1} subseteq A_i$ and $,bigcap_{i=1}^ infty A_i = varnothing$.
So, if $,B_i=A_i setminus A_{i+1}$, then ${B_i}$ is a disjoint sequence and $,bigcup_{i=1}^ infty B_i = A_1$.
By the R-integrability of $f_i$, the set $,B_i,$ has Jordan content (in $mathbb R^2$) and , by the countable additivity of Jordan content on its domain, the series $sum_{i=1}^ infty m(B_i)$ is convergent, so the sequence of its remainders ${sum_{i=n}^ infty m(B_i)}$ is null.
But $,sum_{i=n}^ infty m(B_i) = m(A_n)= int_0^1 f_n ,dots$
edited Nov 18 '15 at 5:58
answered Nov 17 '15 at 18:33
Tony PiccoloTony Piccolo
3,2152719
3,2152719
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2
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Here's one proof (not using the hint).
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– David Mitra
Jul 21 '14 at 18:28
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@David Mitra Thanks, I found that on Google when I was searching. This is a hard problem! It doesn't seem like it should be, though.
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– layman
Jul 21 '14 at 20:10
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Hard problem? Yes. Back when I learned about Lebesgue integration, the instructor gave this as an example of the superiority of the Lebesgue theory over the Riemann theory.
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– GEdgar
Mar 3 '15 at 15:50
1
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Here is another proof by C. Niculescu and F. Popovici (2011), theorem 2. It doesn't use the hint but Cousin lemma and Lebesgue criterion of Riemann integrability. Put $A=varnothing$ in the proof.
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– Tony Piccolo
Nov 15 '15 at 8:45
1
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@Tony Piccolo Here is another to add to the collection. It appeared in the Monthly 2010: classicalrealanalysis.info/documents/…
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– B. S. Thomson
Nov 17 '15 at 20:49