Dilatation of a measurable set is a measurable set.
$begingroup$
Let $E$ Lebesgue measurable set. Let $lambdainmathbb{R},lambdaneq 0$.
Let $lambda E=left{lambda a:ain Eright}$.
Show that $lambda E$ is measurable.
I have this.
Let $f:Eto lambda E$
$f(a)=lambda a$ continuos and bijective.
Now $g:lambda Eto E$ with $g=f^{-1}$. i.e. $g(b)=frac{b}{lambda}$
Then $lambda E=g^{-1}(E)$ and $g$ continuous, therefore $lambda E$ is measurable set.
It is correct?
real-analysis measure-theory lebesgue-measure
asked Dec 3 '18 at 23:18
eraldcoileraldcoil
386211
$endgroup$
add a comment |
$begingroup$
Let $E$ Lebesgue measurable set. Let $lambdainmathbb{R},lambdaneq 0$.
Let $lambda E=left{lambda a:ain Eright}$.
Show that $lambda E$ is measurable.
I have this.
Let $f:Eto lambda E$
$f(a)=lambda a$ continuos and bijective.
Now $g:lambda Eto E$ with $g=f^{-1}$. i.e. $g(b)=frac{b}{lambda}$
Then $lambda E=g^{-1}(E)$ and $g$ continuous, therefore $lambda E$ is measurable set.
It is correct?
real-analysis measure-theory lebesgue-measure
asked Dec 3 '18 at 23:18
eraldcoileraldcoil
386211
$endgroup$
1
$begingroup$
You are mixing up Borel measurability and Lebesgue measurability. Inverse image of a Lebesgue measurable set under a continuous map need not be Lebesgue measurable. In this case $g$ is also Lipschitz and you can use this to complete the argument.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 23:28
$begingroup$
If $f:Eto E'$ Lipschitz function (with same sigma algebra in Domain and Codomain) and $E$ measurable then $E'$ is measurable?
$endgroup$
– eraldcoil
Dec 3 '18 at 23:41
1
$begingroup$
Yes, because Lipschitz functions map null sets to null sets.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 23:43
add a comment |
$begingroup$
Let $E$ Lebesgue measurable set. Let $lambdainmathbb{R},lambdaneq 0$.
Let $lambda E=left{lambda a:ain Eright}$.
Show that $lambda E$ is measurable.
I have this.
Let $f:Eto lambda E$
$f(a)=lambda a$ continuos and bijective.
Now $g:lambda Eto E$ with $g=f^{-1}$. i.e. $g(b)=frac{b}{lambda}$
Then $lambda E=g^{-1}(E)$ and $g$ continuous, therefore $lambda E$ is measurable set.
It is correct?
real-analysis measure-theory lebesgue-measure
asked Dec 3 '18 at 23:18
eraldcoileraldcoil
386211
$endgroup$
Let $E$ Lebesgue measurable set. Let $lambdainmathbb{R},lambdaneq 0$.
Let $lambda E=left{lambda a:ain Eright}$.
Show that $lambda E$ is measurable.
I have this.
Let $f:Eto lambda E$
$f(a)=lambda a$ continuos and bijective.
Now $g:lambda Eto E$ with $g=f^{-1}$. i.e. $g(b)=frac{b}{lambda}$
Then $lambda E=g^{-1}(E)$ and $g$ continuous, therefore $lambda E$ is measurable set.
It is correct?
real-analysis measure-theory lebesgue-measure
real-analysis measure-theory lebesgue-measure
asked Dec 3 '18 at 23:18
eraldcoileraldcoil
386211
asked Dec 3 '18 at 23:18
eraldcoileraldcoil
386211
edited Dec 3 '18 at 23:34
eraldcoil
asked Dec 3 '18 at 23:18
eraldcoileraldcoil
386211
asked Dec 3 '18 at 23:18
eraldcoileraldcoil
386211
asked Dec 3 '18 at 23:18
eraldcoileraldcoil
386211
386211
1
$begingroup$
You are mixing up Borel measurability and Lebesgue measurability. Inverse image of a Lebesgue measurable set under a continuous map need not be Lebesgue measurable. In this case $g$ is also Lipschitz and you can use this to complete the argument.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 23:28
$begingroup$
If $f:Eto E'$ Lipschitz function (with same sigma algebra in Domain and Codomain) and $E$ measurable then $E'$ is measurable?
$endgroup$
– eraldcoil
Dec 3 '18 at 23:41
1
$begingroup$
Yes, because Lipschitz functions map null sets to null sets.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 23:43
add a comment |
1
$begingroup$
You are mixing up Borel measurability and Lebesgue measurability. Inverse image of a Lebesgue measurable set under a continuous map need not be Lebesgue measurable. In this case $g$ is also Lipschitz and you can use this to complete the argument.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 23:28
$begingroup$
If $f:Eto E'$ Lipschitz function (with same sigma algebra in Domain and Codomain) and $E$ measurable then $E'$ is measurable?
$endgroup$
– eraldcoil
Dec 3 '18 at 23:41
1
$begingroup$
Yes, because Lipschitz functions map null sets to null sets.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 23:43
1
1
$begingroup$
You are mixing up Borel measurability and Lebesgue measurability. Inverse image of a Lebesgue measurable set under a continuous map need not be Lebesgue measurable. In this case $g$ is also Lipschitz and you can use this to complete the argument.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 23:28
$begingroup$
You are mixing up Borel measurability and Lebesgue measurability. Inverse image of a Lebesgue measurable set under a continuous map need not be Lebesgue measurable. In this case $g$ is also Lipschitz and you can use this to complete the argument.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 23:28
$begingroup$
If $f:Eto E'$ Lipschitz function (with same sigma algebra in Domain and Codomain) and $E$ measurable then $E'$ is measurable?
$endgroup$
– eraldcoil
Dec 3 '18 at 23:41
$begingroup$
If $f:Eto E'$ Lipschitz function (with same sigma algebra in Domain and Codomain) and $E$ measurable then $E'$ is measurable?
$endgroup$
– eraldcoil
Dec 3 '18 at 23:41
1
1
$begingroup$
Yes, because Lipschitz functions map null sets to null sets.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 23:43
$begingroup$
Yes, because Lipschitz functions map null sets to null sets.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 23:43
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
First we show that if $A$ has measure zero, then $lambda A$ has measure zero.
Let $A subset mathbb{R}$ have measure zero. Let $epsilon > 0$ be given. Then there exists a countable collection of intervals ${(a_i, b_i)}_{i = 1}^{n}$ such that $A subset bigcup_{i = 1}^{infty}(a_i, b_i)$ and $sum_{i = 1}^{infty}(b_i - a_i) < frac{epsilon}{|lambda|}$. Let $a in A$. Then $a in (a_i, b_i)$ for some $i in mathbb{N}$. Now consider 2 cases:
Case 1 ($lambda < 0$): If $lambda < 0$ then $a in (lambda b_i, lambda a_i)$. Thus $lambda A subset bigcup_{i = 1}^{infty}(lambda b_i, lambda a_i)$. We see
$$sum_{i = 1}^{infty}lambda a_i - lambda b_i = -lambda sum_{i = 1}^{n}left(b_i - a_iright) < epsilon.$$
Case 2 ($lambda > 0$): If $lambda > 0$ then $a in (lambda a_i, lambda b_i)$. Thus $lambda A subset bigcup_{i = 1}^{infty}(lambda a_i, lambda b_i)$. We see
$$sum_{i = 1}^{infty}lambda b_i - lambda a_i = lambda sum_{i = 1}^{n}left(b_i - a_iright) < epsilon.$$
Let $m^{ast}$ denote the Lebesgue outer-measure on $mathbb{R}$. We see
$$ m^{ast}(lambda A) = infleft{sum_{i = 1}^{infty} b_i - a_i : A subset bigcup_{i = 1}^{infty}(a_i, b_i)right} < epsilon.$$
Thus $m^{ast}(lambda A) = 0$. Since Lebesgue measure is complete, $lambda A$ is Lebesgue measurable.
Let $E subset mathbb{R}$ be Lebesgue measurable. There is a Borel set $G$ such that $E subset G$ and $G backslash E$ has measure zero. Define $g: mathbb{R} to mathbb{R}$ by $g(x) = frac{1}{lambda}x$. Clearly $g$ is continuous. Since continuous functions are Borel measurable, $lambda G = g^{-1}(G)$ is measurable. Since $G backslash E$ has measure zero, $lambda Gbackslash E$ is Lebesgue measurable by the argument above. Thus $lambda E = lambda G backslash (lambda G backslash E)$ is Lebesgue measurable.
answered Dec 4 '18 at 6:11
Sean HaightSean Haight
675519
$endgroup$
1
$begingroup$
Why $lambda E=g^{-1}(G)$?
$endgroup$
– eraldcoil
Dec 7 '18 at 2:24
1
$begingroup$
Must be $lambda G=g^{-1}(G)$
$endgroup$
– eraldcoil
Dec 7 '18 at 4:25
$begingroup$
Yes, you're absolutely right!
$endgroup$
– Sean Haight
Dec 7 '18 at 5:58
add a comment |
$begingroup$
If $E$ measurable then for any $epsilon>0$ exists $G$ open set such that $Esubset G$ and $m(G-E)<epsilon$.
Now, $Esubset G$ then $lambda Esubsetlambda G$ and $m(lambda G-lambda E)=m(lambda (G-E))=|lambda|m(G-E)<|lambda|epsilon$
Therefore $lambda E$ is measurable.
It is correct?
answered Dec 6 '18 at 22:28
eraldcoileraldcoil
386211
$endgroup$
$begingroup$
Why is $lambda G - lambda E$ measurable?
$endgroup$
– Sean Haight
Dec 6 '18 at 22:35
$begingroup$
Oh. Sorry. Should be $m^{ast}$ outer measure
$endgroup$
– eraldcoil
Dec 6 '18 at 22:37
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First we show that if $A$ has measure zero, then $lambda A$ has measure zero.
Let $A subset mathbb{R}$ have measure zero. Let $epsilon > 0$ be given. Then there exists a countable collection of intervals ${(a_i, b_i)}_{i = 1}^{n}$ such that $A subset bigcup_{i = 1}^{infty}(a_i, b_i)$ and $sum_{i = 1}^{infty}(b_i - a_i) < frac{epsilon}{|lambda|}$. Let $a in A$. Then $a in (a_i, b_i)$ for some $i in mathbb{N}$. Now consider 2 cases:
Case 1 ($lambda < 0$): If $lambda < 0$ then $a in (lambda b_i, lambda a_i)$. Thus $lambda A subset bigcup_{i = 1}^{infty}(lambda b_i, lambda a_i)$. We see
$$sum_{i = 1}^{infty}lambda a_i - lambda b_i = -lambda sum_{i = 1}^{n}left(b_i - a_iright) < epsilon.$$
Case 2 ($lambda > 0$): If $lambda > 0$ then $a in (lambda a_i, lambda b_i)$. Thus $lambda A subset bigcup_{i = 1}^{infty}(lambda a_i, lambda b_i)$. We see
$$sum_{i = 1}^{infty}lambda b_i - lambda a_i = lambda sum_{i = 1}^{n}left(b_i - a_iright) < epsilon.$$
Let $m^{ast}$ denote the Lebesgue outer-measure on $mathbb{R}$. We see
$$ m^{ast}(lambda A) = infleft{sum_{i = 1}^{infty} b_i - a_i : A subset bigcup_{i = 1}^{infty}(a_i, b_i)right} < epsilon.$$
Thus $m^{ast}(lambda A) = 0$. Since Lebesgue measure is complete, $lambda A$ is Lebesgue measurable.
Let $E subset mathbb{R}$ be Lebesgue measurable. There is a Borel set $G$ such that $E subset G$ and $G backslash E$ has measure zero. Define $g: mathbb{R} to mathbb{R}$ by $g(x) = frac{1}{lambda}x$. Clearly $g$ is continuous. Since continuous functions are Borel measurable, $lambda G = g^{-1}(G)$ is measurable. Since $G backslash E$ has measure zero, $lambda Gbackslash E$ is Lebesgue measurable by the argument above. Thus $lambda E = lambda G backslash (lambda G backslash E)$ is Lebesgue measurable.
answered Dec 4 '18 at 6:11
Sean HaightSean Haight
675519
$endgroup$
1
$begingroup$
Why $lambda E=g^{-1}(G)$?
$endgroup$
– eraldcoil
Dec 7 '18 at 2:24
1
$begingroup$
Must be $lambda G=g^{-1}(G)$
$endgroup$
– eraldcoil
Dec 7 '18 at 4:25
$begingroup$
Yes, you're absolutely right!
$endgroup$
– Sean Haight
Dec 7 '18 at 5:58
add a comment |
$begingroup$
First we show that if $A$ has measure zero, then $lambda A$ has measure zero.
Let $A subset mathbb{R}$ have measure zero. Let $epsilon > 0$ be given. Then there exists a countable collection of intervals ${(a_i, b_i)}_{i = 1}^{n}$ such that $A subset bigcup_{i = 1}^{infty}(a_i, b_i)$ and $sum_{i = 1}^{infty}(b_i - a_i) < frac{epsilon}{|lambda|}$. Let $a in A$. Then $a in (a_i, b_i)$ for some $i in mathbb{N}$. Now consider 2 cases:
Case 1 ($lambda < 0$): If $lambda < 0$ then $a in (lambda b_i, lambda a_i)$. Thus $lambda A subset bigcup_{i = 1}^{infty}(lambda b_i, lambda a_i)$. We see
$$sum_{i = 1}^{infty}lambda a_i - lambda b_i = -lambda sum_{i = 1}^{n}left(b_i - a_iright) < epsilon.$$
Case 2 ($lambda > 0$): If $lambda > 0$ then $a in (lambda a_i, lambda b_i)$. Thus $lambda A subset bigcup_{i = 1}^{infty}(lambda a_i, lambda b_i)$. We see
$$sum_{i = 1}^{infty}lambda b_i - lambda a_i = lambda sum_{i = 1}^{n}left(b_i - a_iright) < epsilon.$$
Let $m^{ast}$ denote the Lebesgue outer-measure on $mathbb{R}$. We see
$$ m^{ast}(lambda A) = infleft{sum_{i = 1}^{infty} b_i - a_i : A subset bigcup_{i = 1}^{infty}(a_i, b_i)right} < epsilon.$$
Thus $m^{ast}(lambda A) = 0$. Since Lebesgue measure is complete, $lambda A$ is Lebesgue measurable.
Let $E subset mathbb{R}$ be Lebesgue measurable. There is a Borel set $G$ such that $E subset G$ and $G backslash E$ has measure zero. Define $g: mathbb{R} to mathbb{R}$ by $g(x) = frac{1}{lambda}x$. Clearly $g$ is continuous. Since continuous functions are Borel measurable, $lambda G = g^{-1}(G)$ is measurable. Since $G backslash E$ has measure zero, $lambda Gbackslash E$ is Lebesgue measurable by the argument above. Thus $lambda E = lambda G backslash (lambda G backslash E)$ is Lebesgue measurable.
answered Dec 4 '18 at 6:11
Sean HaightSean Haight
675519
$endgroup$
1
$begingroup$
Why $lambda E=g^{-1}(G)$?
$endgroup$
– eraldcoil
Dec 7 '18 at 2:24
1
$begingroup$
Must be $lambda G=g^{-1}(G)$
$endgroup$
– eraldcoil
Dec 7 '18 at 4:25
$begingroup$
Yes, you're absolutely right!
$endgroup$
– Sean Haight
Dec 7 '18 at 5:58
add a comment |
$begingroup$
First we show that if $A$ has measure zero, then $lambda A$ has measure zero.
Let $A subset mathbb{R}$ have measure zero. Let $epsilon > 0$ be given. Then there exists a countable collection of intervals ${(a_i, b_i)}_{i = 1}^{n}$ such that $A subset bigcup_{i = 1}^{infty}(a_i, b_i)$ and $sum_{i = 1}^{infty}(b_i - a_i) < frac{epsilon}{|lambda|}$. Let $a in A$. Then $a in (a_i, b_i)$ for some $i in mathbb{N}$. Now consider 2 cases:
Case 1 ($lambda < 0$): If $lambda < 0$ then $a in (lambda b_i, lambda a_i)$. Thus $lambda A subset bigcup_{i = 1}^{infty}(lambda b_i, lambda a_i)$. We see
$$sum_{i = 1}^{infty}lambda a_i - lambda b_i = -lambda sum_{i = 1}^{n}left(b_i - a_iright) < epsilon.$$
Case 2 ($lambda > 0$): If $lambda > 0$ then $a in (lambda a_i, lambda b_i)$. Thus $lambda A subset bigcup_{i = 1}^{infty}(lambda a_i, lambda b_i)$. We see
$$sum_{i = 1}^{infty}lambda b_i - lambda a_i = lambda sum_{i = 1}^{n}left(b_i - a_iright) < epsilon.$$
Let $m^{ast}$ denote the Lebesgue outer-measure on $mathbb{R}$. We see
$$ m^{ast}(lambda A) = infleft{sum_{i = 1}^{infty} b_i - a_i : A subset bigcup_{i = 1}^{infty}(a_i, b_i)right} < epsilon.$$
Thus $m^{ast}(lambda A) = 0$. Since Lebesgue measure is complete, $lambda A$ is Lebesgue measurable.
Let $E subset mathbb{R}$ be Lebesgue measurable. There is a Borel set $G$ such that $E subset G$ and $G backslash E$ has measure zero. Define $g: mathbb{R} to mathbb{R}$ by $g(x) = frac{1}{lambda}x$. Clearly $g$ is continuous. Since continuous functions are Borel measurable, $lambda G = g^{-1}(G)$ is measurable. Since $G backslash E$ has measure zero, $lambda Gbackslash E$ is Lebesgue measurable by the argument above. Thus $lambda E = lambda G backslash (lambda G backslash E)$ is Lebesgue measurable.
answered Dec 4 '18 at 6:11
Sean HaightSean Haight
675519
$endgroup$
First we show that if $A$ has measure zero, then $lambda A$ has measure zero.
Let $A subset mathbb{R}$ have measure zero. Let $epsilon > 0$ be given. Then there exists a countable collection of intervals ${(a_i, b_i)}_{i = 1}^{n}$ such that $A subset bigcup_{i = 1}^{infty}(a_i, b_i)$ and $sum_{i = 1}^{infty}(b_i - a_i) < frac{epsilon}{|lambda|}$. Let $a in A$. Then $a in (a_i, b_i)$ for some $i in mathbb{N}$. Now consider 2 cases:
Case 1 ($lambda < 0$): If $lambda < 0$ then $a in (lambda b_i, lambda a_i)$. Thus $lambda A subset bigcup_{i = 1}^{infty}(lambda b_i, lambda a_i)$. We see
$$sum_{i = 1}^{infty}lambda a_i - lambda b_i = -lambda sum_{i = 1}^{n}left(b_i - a_iright) < epsilon.$$
Case 2 ($lambda > 0$): If $lambda > 0$ then $a in (lambda a_i, lambda b_i)$. Thus $lambda A subset bigcup_{i = 1}^{infty}(lambda a_i, lambda b_i)$. We see
$$sum_{i = 1}^{infty}lambda b_i - lambda a_i = lambda sum_{i = 1}^{n}left(b_i - a_iright) < epsilon.$$
Let $m^{ast}$ denote the Lebesgue outer-measure on $mathbb{R}$. We see
$$ m^{ast}(lambda A) = infleft{sum_{i = 1}^{infty} b_i - a_i : A subset bigcup_{i = 1}^{infty}(a_i, b_i)right} < epsilon.$$
Thus $m^{ast}(lambda A) = 0$. Since Lebesgue measure is complete, $lambda A$ is Lebesgue measurable.
Let $E subset mathbb{R}$ be Lebesgue measurable. There is a Borel set $G$ such that $E subset G$ and $G backslash E$ has measure zero. Define $g: mathbb{R} to mathbb{R}$ by $g(x) = frac{1}{lambda}x$. Clearly $g$ is continuous. Since continuous functions are Borel measurable, $lambda G = g^{-1}(G)$ is measurable. Since $G backslash E$ has measure zero, $lambda Gbackslash E$ is Lebesgue measurable by the argument above. Thus $lambda E = lambda G backslash (lambda G backslash E)$ is Lebesgue measurable.
answered Dec 4 '18 at 6:11
Sean HaightSean Haight
675519
edited Dec 7 '18 at 5:58
answered Dec 4 '18 at 6:11
Sean HaightSean Haight
675519
answered Dec 4 '18 at 6:11
Sean HaightSean Haight
675519
answered Dec 4 '18 at 6:11
Sean HaightSean Haight
675519
675519
1
$begingroup$
Why $lambda E=g^{-1}(G)$?
$endgroup$
– eraldcoil
Dec 7 '18 at 2:24
1
$begingroup$
Must be $lambda G=g^{-1}(G)$
$endgroup$
– eraldcoil
Dec 7 '18 at 4:25
$begingroup$
Yes, you're absolutely right!
$endgroup$
– Sean Haight
Dec 7 '18 at 5:58
add a comment |
1
$begingroup$
Why $lambda E=g^{-1}(G)$?
$endgroup$
– eraldcoil
Dec 7 '18 at 2:24
1
$begingroup$
Must be $lambda G=g^{-1}(G)$
$endgroup$
– eraldcoil
Dec 7 '18 at 4:25
$begingroup$
Yes, you're absolutely right!
$endgroup$
– Sean Haight
Dec 7 '18 at 5:58
1
1
$begingroup$
Why $lambda E=g^{-1}(G)$?
$endgroup$
– eraldcoil
Dec 7 '18 at 2:24
$begingroup$
Why $lambda E=g^{-1}(G)$?
$endgroup$
– eraldcoil
Dec 7 '18 at 2:24
1
1
$begingroup$
Must be $lambda G=g^{-1}(G)$
$endgroup$
– eraldcoil
Dec 7 '18 at 4:25
$begingroup$
Must be $lambda G=g^{-1}(G)$
$endgroup$
– eraldcoil
Dec 7 '18 at 4:25
$begingroup$
Yes, you're absolutely right!
$endgroup$
– Sean Haight
Dec 7 '18 at 5:58
$begingroup$
Yes, you're absolutely right!
$endgroup$
– Sean Haight
Dec 7 '18 at 5:58
add a comment |
$begingroup$
If $E$ measurable then for any $epsilon>0$ exists $G$ open set such that $Esubset G$ and $m(G-E)<epsilon$.
Now, $Esubset G$ then $lambda Esubsetlambda G$ and $m(lambda G-lambda E)=m(lambda (G-E))=|lambda|m(G-E)<|lambda|epsilon$
Therefore $lambda E$ is measurable.
It is correct?
answered Dec 6 '18 at 22:28
eraldcoileraldcoil
386211
$endgroup$
$begingroup$
Why is $lambda G - lambda E$ measurable?
$endgroup$
– Sean Haight
Dec 6 '18 at 22:35
$begingroup$
Oh. Sorry. Should be $m^{ast}$ outer measure
$endgroup$
– eraldcoil
Dec 6 '18 at 22:37
add a comment |
$begingroup$
If $E$ measurable then for any $epsilon>0$ exists $G$ open set such that $Esubset G$ and $m(G-E)<epsilon$.
Now, $Esubset G$ then $lambda Esubsetlambda G$ and $m(lambda G-lambda E)=m(lambda (G-E))=|lambda|m(G-E)<|lambda|epsilon$
Therefore $lambda E$ is measurable.
It is correct?
answered Dec 6 '18 at 22:28
eraldcoileraldcoil
386211
$endgroup$
$begingroup$
Why is $lambda G - lambda E$ measurable?
$endgroup$
– Sean Haight
Dec 6 '18 at 22:35
$begingroup$
Oh. Sorry. Should be $m^{ast}$ outer measure
$endgroup$
– eraldcoil
Dec 6 '18 at 22:37
add a comment |
$begingroup$
If $E$ measurable then for any $epsilon>0$ exists $G$ open set such that $Esubset G$ and $m(G-E)<epsilon$.
Now, $Esubset G$ then $lambda Esubsetlambda G$ and $m(lambda G-lambda E)=m(lambda (G-E))=|lambda|m(G-E)<|lambda|epsilon$
Therefore $lambda E$ is measurable.
It is correct?
answered Dec 6 '18 at 22:28
eraldcoileraldcoil
386211
$endgroup$
If $E$ measurable then for any $epsilon>0$ exists $G$ open set such that $Esubset G$ and $m(G-E)<epsilon$.
Now, $Esubset G$ then $lambda Esubsetlambda G$ and $m(lambda G-lambda E)=m(lambda (G-E))=|lambda|m(G-E)<|lambda|epsilon$
Therefore $lambda E$ is measurable.
It is correct?
answered Dec 6 '18 at 22:28
eraldcoileraldcoil
386211
answered Dec 6 '18 at 22:28
eraldcoileraldcoil
386211
answered Dec 6 '18 at 22:28
eraldcoileraldcoil
386211
answered Dec 6 '18 at 22:28
eraldcoileraldcoil
386211
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Why is $lambda G - lambda E$ measurable?
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– Sean Haight
Dec 6 '18 at 22:35
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Oh. Sorry. Should be $m^{ast}$ outer measure
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– eraldcoil
Dec 6 '18 at 22:37
add a comment |
$begingroup$
Why is $lambda G - lambda E$ measurable?
$endgroup$
– Sean Haight
Dec 6 '18 at 22:35
$begingroup$
Oh. Sorry. Should be $m^{ast}$ outer measure
$endgroup$
– eraldcoil
Dec 6 '18 at 22:37
$begingroup$
Why is $lambda G - lambda E$ measurable?
$endgroup$
– Sean Haight
Dec 6 '18 at 22:35
$begingroup$
Why is $lambda G - lambda E$ measurable?
$endgroup$
– Sean Haight
Dec 6 '18 at 22:35
$begingroup$
Oh. Sorry. Should be $m^{ast}$ outer measure
$endgroup$
– eraldcoil
Dec 6 '18 at 22:37
$begingroup$
Oh. Sorry. Should be $m^{ast}$ outer measure
$endgroup$
– eraldcoil
Dec 6 '18 at 22:37
add a comment |
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You are mixing up Borel measurability and Lebesgue measurability. Inverse image of a Lebesgue measurable set under a continuous map need not be Lebesgue measurable. In this case $g$ is also Lipschitz and you can use this to complete the argument.
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– Kavi Rama Murthy
Dec 3 '18 at 23:28
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If $f:Eto E'$ Lipschitz function (with same sigma algebra in Domain and Codomain) and $E$ measurable then $E'$ is measurable?
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– eraldcoil
Dec 3 '18 at 23:41
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Yes, because Lipschitz functions map null sets to null sets.
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– Kavi Rama Murthy
Dec 3 '18 at 23:43