if ${x| f(x) neq g(x))}$ has zero measure then $int f = int g$.
$begingroup$
Let $f,g$ be integrable functions in a closed box $B subset mathbb R^n$.
Suppose that the set $A = {x in B| f(x) neq g(x)}$ is negligible. Show that $int_Bf =int_B g$
What I tried:
Notice that $B = A cup (A^c cap B)$ and that $A cap (A^c cap B) = emptyset$ obviously.
Then $int_Bf - int_B g = int_B f-g = int_{A}f-g+int_{A^c cap B}f-g = int_{A}f-g$
Now see that $0 = v(A)inf_{x in A}(f-g)(x) leq int_{A}f-g leq v(A) sup_{x in A}(f-g)(x) =0$ because $v(A) = 0$ since it's negligible, and $f,g$ are integrable on $B$ so we know they are bounded.
Where is my issue:
I'm unsure about
$int_B f-g = int_{A}f-g+int_{A^c cap B}f-g$, I realize it's true when $A,A^c cap B$ are boxes. If I split a big box into two smaller non intersecting boxes, then the integral on the big box is equal to the sum of the integrals of the two smaller boxes, but is it also true for general shapes?
Is there a way to prove this without resorting to integrals over non boxes?
calculus measure-theory riemann-integration
asked Dec 13 '18 at 20:58
Oria GruberOria Gruber
6,48032460
$endgroup$
add a comment |
$begingroup$
Let $f,g$ be integrable functions in a closed box $B subset mathbb R^n$.
Suppose that the set $A = {x in B| f(x) neq g(x)}$ is negligible. Show that $int_Bf =int_B g$
What I tried:
Notice that $B = A cup (A^c cap B)$ and that $A cap (A^c cap B) = emptyset$ obviously.
Then $int_Bf - int_B g = int_B f-g = int_{A}f-g+int_{A^c cap B}f-g = int_{A}f-g$
Now see that $0 = v(A)inf_{x in A}(f-g)(x) leq int_{A}f-g leq v(A) sup_{x in A}(f-g)(x) =0$ because $v(A) = 0$ since it's negligible, and $f,g$ are integrable on $B$ so we know they are bounded.
Where is my issue:
I'm unsure about
$int_B f-g = int_{A}f-g+int_{A^c cap B}f-g$, I realize it's true when $A,A^c cap B$ are boxes. If I split a big box into two smaller non intersecting boxes, then the integral on the big box is equal to the sum of the integrals of the two smaller boxes, but is it also true for general shapes?
Is there a way to prove this without resorting to integrals over non boxes?
calculus measure-theory riemann-integration
asked Dec 13 '18 at 20:58
Oria GruberOria Gruber
6,48032460
$endgroup$
$begingroup$
Not sure about your tags... I would not tag calculus and I would tag Lebesgue integration instead of Riemann Integration.
$endgroup$
– Zuriel
Dec 13 '18 at 21:23
add a comment |
$begingroup$
Let $f,g$ be integrable functions in a closed box $B subset mathbb R^n$.
Suppose that the set $A = {x in B| f(x) neq g(x)}$ is negligible. Show that $int_Bf =int_B g$
What I tried:
Notice that $B = A cup (A^c cap B)$ and that $A cap (A^c cap B) = emptyset$ obviously.
Then $int_Bf - int_B g = int_B f-g = int_{A}f-g+int_{A^c cap B}f-g = int_{A}f-g$
Now see that $0 = v(A)inf_{x in A}(f-g)(x) leq int_{A}f-g leq v(A) sup_{x in A}(f-g)(x) =0$ because $v(A) = 0$ since it's negligible, and $f,g$ are integrable on $B$ so we know they are bounded.
Where is my issue:
I'm unsure about
$int_B f-g = int_{A}f-g+int_{A^c cap B}f-g$, I realize it's true when $A,A^c cap B$ are boxes. If I split a big box into two smaller non intersecting boxes, then the integral on the big box is equal to the sum of the integrals of the two smaller boxes, but is it also true for general shapes?
Is there a way to prove this without resorting to integrals over non boxes?
calculus measure-theory riemann-integration
asked Dec 13 '18 at 20:58
Oria GruberOria Gruber
6,48032460
$endgroup$
Let $f,g$ be integrable functions in a closed box $B subset mathbb R^n$.
Suppose that the set $A = {x in B| f(x) neq g(x)}$ is negligible. Show that $int_Bf =int_B g$
What I tried:
Notice that $B = A cup (A^c cap B)$ and that $A cap (A^c cap B) = emptyset$ obviously.
Then $int_Bf - int_B g = int_B f-g = int_{A}f-g+int_{A^c cap B}f-g = int_{A}f-g$
Now see that $0 = v(A)inf_{x in A}(f-g)(x) leq int_{A}f-g leq v(A) sup_{x in A}(f-g)(x) =0$ because $v(A) = 0$ since it's negligible, and $f,g$ are integrable on $B$ so we know they are bounded.
Where is my issue:
I'm unsure about
$int_B f-g = int_{A}f-g+int_{A^c cap B}f-g$, I realize it's true when $A,A^c cap B$ are boxes. If I split a big box into two smaller non intersecting boxes, then the integral on the big box is equal to the sum of the integrals of the two smaller boxes, but is it also true for general shapes?
Is there a way to prove this without resorting to integrals over non boxes?
calculus measure-theory riemann-integration
calculus measure-theory riemann-integration
asked Dec 13 '18 at 20:58
Oria GruberOria Gruber
6,48032460
asked Dec 13 '18 at 20:58
Oria GruberOria Gruber
6,48032460
asked Dec 13 '18 at 20:58
Oria GruberOria Gruber
6,48032460
asked Dec 13 '18 at 20:58
Oria GruberOria Gruber
6,48032460
asked Dec 13 '18 at 20:58
Oria GruberOria Gruber
6,48032460
6,48032460
$begingroup$
Not sure about your tags... I would not tag calculus and I would tag Lebesgue integration instead of Riemann Integration.
$endgroup$
– Zuriel
Dec 13 '18 at 21:23
add a comment |
$begingroup$
Not sure about your tags... I would not tag calculus and I would tag Lebesgue integration instead of Riemann Integration.
$endgroup$
– Zuriel
Dec 13 '18 at 21:23
$begingroup$
Not sure about your tags... I would not tag calculus and I would tag Lebesgue integration instead of Riemann Integration.
$endgroup$
– Zuriel
Dec 13 '18 at 21:23
$begingroup$
Not sure about your tags... I would not tag calculus and I would tag Lebesgue integration instead of Riemann Integration.
$endgroup$
– Zuriel
Dec 13 '18 at 21:23
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We can also prove this in the framework of Riemann integration.
Take $h = f- g$ and
$$A = {x in B ,|, f(x) neq g(x) } = { x in B ,|, h(x) neq 0}$$ where $A$
has zero measure.
If $P$ is any partition of $B$ into sub-rectangles (boxes) then any such sub-rectangle $R$ has non-zero content and so must include points $y$ not in A such that $h(y) = 0$.
Thus, $inf_{x in R}h(x) leqslant 0 leqslant sup_{x in R} h(x)$ for each $R in P$ and the lower and upper Darboux sums satisfy
$$L(P,h) leqslant 0 leqslant U(P,h), \ implies L(P,h) leqslant sup_P L(P,h) leqslant 0 leqslant inf_P ,U(P,h) leqslant U(P,h)$$
However, we are given that $f$ and $g$, and , hence, $h$ are integrable.
Thus,
$$int_B h = sup_P L(P,h) geqslant 0 , quad int_B h = inf_P ,L(P,h) leqslant 0,$$
which implies
$$int_Bf - int_B g = int_Bh = 0$$
answered Dec 13 '18 at 22:03
RRLRRL
51.4k42573
$endgroup$
$begingroup$
Why must any such sub rectangle include points not from $A$? How did you use the fact that $A$ has zero measure?
$endgroup$
– Oria Gruber
Dec 13 '18 at 22:27
$begingroup$
In Riemann integration over boxes in $mathbb{R}^n$ we always use partitions where the sub-boxes are of the form $R = [a_1,b_1] times [a_2,b_2] times ldots times [a_n,b_n]$ and each interval is non-empty. Otherwise $vol(R) = 0$ and it contributes nothing to a Darboux or Rieman sum. It is impossible for $R subset A$ since the measure of $R$ is non-zero. Every partition element has to include points where $h$ is $0$. This is the easiest proof of this theorem requiring the least amount of measure theory.
$endgroup$
– RRL
Dec 13 '18 at 22:45
add a comment |
$begingroup$
Since $f$ and $g$ are integrable, so is $f-g$, so $A=(f-g)^{-1}(mathbb{R}setminus{0})$ is a measurable set. Hence also $A^ccap B$ is measurable.
Then use the fact that for any integrable function $f$ and partitioning measurable sets $A$ and $B$ we have $f=chi_Af+chi_Bf$ and hence $int f=intchi_Af+chi_Bf=intchi_Af+intchi_Bf=int_Af+int_Bf$.
Here $chi_A$ denotes the indicator function, which is $1$ in $A$ and $0$ elsewhere.
answered Dec 13 '18 at 21:20
SmileyCraftSmileyCraft
3,591517
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038561%2fif-x-fx-neq-gx-has-zero-measure-then-int-f-int-g%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We can also prove this in the framework of Riemann integration.
Take $h = f- g$ and
$$A = {x in B ,|, f(x) neq g(x) } = { x in B ,|, h(x) neq 0}$$ where $A$
has zero measure.
If $P$ is any partition of $B$ into sub-rectangles (boxes) then any such sub-rectangle $R$ has non-zero content and so must include points $y$ not in A such that $h(y) = 0$.
Thus, $inf_{x in R}h(x) leqslant 0 leqslant sup_{x in R} h(x)$ for each $R in P$ and the lower and upper Darboux sums satisfy
$$L(P,h) leqslant 0 leqslant U(P,h), \ implies L(P,h) leqslant sup_P L(P,h) leqslant 0 leqslant inf_P ,U(P,h) leqslant U(P,h)$$
However, we are given that $f$ and $g$, and , hence, $h$ are integrable.
Thus,
$$int_B h = sup_P L(P,h) geqslant 0 , quad int_B h = inf_P ,L(P,h) leqslant 0,$$
which implies
$$int_Bf - int_B g = int_Bh = 0$$
answered Dec 13 '18 at 22:03
RRLRRL
51.4k42573
$endgroup$
$begingroup$
Why must any such sub rectangle include points not from $A$? How did you use the fact that $A$ has zero measure?
$endgroup$
– Oria Gruber
Dec 13 '18 at 22:27
$begingroup$
In Riemann integration over boxes in $mathbb{R}^n$ we always use partitions where the sub-boxes are of the form $R = [a_1,b_1] times [a_2,b_2] times ldots times [a_n,b_n]$ and each interval is non-empty. Otherwise $vol(R) = 0$ and it contributes nothing to a Darboux or Rieman sum. It is impossible for $R subset A$ since the measure of $R$ is non-zero. Every partition element has to include points where $h$ is $0$. This is the easiest proof of this theorem requiring the least amount of measure theory.
$endgroup$
– RRL
Dec 13 '18 at 22:45
add a comment |
$begingroup$
We can also prove this in the framework of Riemann integration.
Take $h = f- g$ and
$$A = {x in B ,|, f(x) neq g(x) } = { x in B ,|, h(x) neq 0}$$ where $A$
has zero measure.
If $P$ is any partition of $B$ into sub-rectangles (boxes) then any such sub-rectangle $R$ has non-zero content and so must include points $y$ not in A such that $h(y) = 0$.
Thus, $inf_{x in R}h(x) leqslant 0 leqslant sup_{x in R} h(x)$ for each $R in P$ and the lower and upper Darboux sums satisfy
$$L(P,h) leqslant 0 leqslant U(P,h), \ implies L(P,h) leqslant sup_P L(P,h) leqslant 0 leqslant inf_P ,U(P,h) leqslant U(P,h)$$
However, we are given that $f$ and $g$, and , hence, $h$ are integrable.
Thus,
$$int_B h = sup_P L(P,h) geqslant 0 , quad int_B h = inf_P ,L(P,h) leqslant 0,$$
which implies
$$int_Bf - int_B g = int_Bh = 0$$
answered Dec 13 '18 at 22:03
RRLRRL
51.4k42573
$endgroup$
$begingroup$
Why must any such sub rectangle include points not from $A$? How did you use the fact that $A$ has zero measure?
$endgroup$
– Oria Gruber
Dec 13 '18 at 22:27
$begingroup$
In Riemann integration over boxes in $mathbb{R}^n$ we always use partitions where the sub-boxes are of the form $R = [a_1,b_1] times [a_2,b_2] times ldots times [a_n,b_n]$ and each interval is non-empty. Otherwise $vol(R) = 0$ and it contributes nothing to a Darboux or Rieman sum. It is impossible for $R subset A$ since the measure of $R$ is non-zero. Every partition element has to include points where $h$ is $0$. This is the easiest proof of this theorem requiring the least amount of measure theory.
$endgroup$
– RRL
Dec 13 '18 at 22:45
add a comment |
$begingroup$
We can also prove this in the framework of Riemann integration.
Take $h = f- g$ and
$$A = {x in B ,|, f(x) neq g(x) } = { x in B ,|, h(x) neq 0}$$ where $A$
has zero measure.
If $P$ is any partition of $B$ into sub-rectangles (boxes) then any such sub-rectangle $R$ has non-zero content and so must include points $y$ not in A such that $h(y) = 0$.
Thus, $inf_{x in R}h(x) leqslant 0 leqslant sup_{x in R} h(x)$ for each $R in P$ and the lower and upper Darboux sums satisfy
$$L(P,h) leqslant 0 leqslant U(P,h), \ implies L(P,h) leqslant sup_P L(P,h) leqslant 0 leqslant inf_P ,U(P,h) leqslant U(P,h)$$
However, we are given that $f$ and $g$, and , hence, $h$ are integrable.
Thus,
$$int_B h = sup_P L(P,h) geqslant 0 , quad int_B h = inf_P ,L(P,h) leqslant 0,$$
which implies
$$int_Bf - int_B g = int_Bh = 0$$
answered Dec 13 '18 at 22:03
RRLRRL
51.4k42573
$endgroup$
We can also prove this in the framework of Riemann integration.
Take $h = f- g$ and
$$A = {x in B ,|, f(x) neq g(x) } = { x in B ,|, h(x) neq 0}$$ where $A$
has zero measure.
If $P$ is any partition of $B$ into sub-rectangles (boxes) then any such sub-rectangle $R$ has non-zero content and so must include points $y$ not in A such that $h(y) = 0$.
Thus, $inf_{x in R}h(x) leqslant 0 leqslant sup_{x in R} h(x)$ for each $R in P$ and the lower and upper Darboux sums satisfy
$$L(P,h) leqslant 0 leqslant U(P,h), \ implies L(P,h) leqslant sup_P L(P,h) leqslant 0 leqslant inf_P ,U(P,h) leqslant U(P,h)$$
However, we are given that $f$ and $g$, and , hence, $h$ are integrable.
Thus,
$$int_B h = sup_P L(P,h) geqslant 0 , quad int_B h = inf_P ,L(P,h) leqslant 0,$$
which implies
$$int_Bf - int_B g = int_Bh = 0$$
answered Dec 13 '18 at 22:03
RRLRRL
51.4k42573
edited Dec 13 '18 at 22:12
answered Dec 13 '18 at 22:03
RRLRRL
51.4k42573
answered Dec 13 '18 at 22:03
RRLRRL
51.4k42573
answered Dec 13 '18 at 22:03
RRLRRL
51.4k42573
51.4k42573
$begingroup$
Why must any such sub rectangle include points not from $A$? How did you use the fact that $A$ has zero measure?
$endgroup$
– Oria Gruber
Dec 13 '18 at 22:27
$begingroup$
In Riemann integration over boxes in $mathbb{R}^n$ we always use partitions where the sub-boxes are of the form $R = [a_1,b_1] times [a_2,b_2] times ldots times [a_n,b_n]$ and each interval is non-empty. Otherwise $vol(R) = 0$ and it contributes nothing to a Darboux or Rieman sum. It is impossible for $R subset A$ since the measure of $R$ is non-zero. Every partition element has to include points where $h$ is $0$. This is the easiest proof of this theorem requiring the least amount of measure theory.
$endgroup$
– RRL
Dec 13 '18 at 22:45
add a comment |
$begingroup$
Why must any such sub rectangle include points not from $A$? How did you use the fact that $A$ has zero measure?
$endgroup$
– Oria Gruber
Dec 13 '18 at 22:27
$begingroup$
In Riemann integration over boxes in $mathbb{R}^n$ we always use partitions where the sub-boxes are of the form $R = [a_1,b_1] times [a_2,b_2] times ldots times [a_n,b_n]$ and each interval is non-empty. Otherwise $vol(R) = 0$ and it contributes nothing to a Darboux or Rieman sum. It is impossible for $R subset A$ since the measure of $R$ is non-zero. Every partition element has to include points where $h$ is $0$. This is the easiest proof of this theorem requiring the least amount of measure theory.
$endgroup$
– RRL
Dec 13 '18 at 22:45
$begingroup$
Why must any such sub rectangle include points not from $A$? How did you use the fact that $A$ has zero measure?
$endgroup$
– Oria Gruber
Dec 13 '18 at 22:27
$begingroup$
Why must any such sub rectangle include points not from $A$? How did you use the fact that $A$ has zero measure?
$endgroup$
– Oria Gruber
Dec 13 '18 at 22:27
$begingroup$
In Riemann integration over boxes in $mathbb{R}^n$ we always use partitions where the sub-boxes are of the form $R = [a_1,b_1] times [a_2,b_2] times ldots times [a_n,b_n]$ and each interval is non-empty. Otherwise $vol(R) = 0$ and it contributes nothing to a Darboux or Rieman sum. It is impossible for $R subset A$ since the measure of $R$ is non-zero. Every partition element has to include points where $h$ is $0$. This is the easiest proof of this theorem requiring the least amount of measure theory.
$endgroup$
– RRL
Dec 13 '18 at 22:45
$begingroup$
In Riemann integration over boxes in $mathbb{R}^n$ we always use partitions where the sub-boxes are of the form $R = [a_1,b_1] times [a_2,b_2] times ldots times [a_n,b_n]$ and each interval is non-empty. Otherwise $vol(R) = 0$ and it contributes nothing to a Darboux or Rieman sum. It is impossible for $R subset A$ since the measure of $R$ is non-zero. Every partition element has to include points where $h$ is $0$. This is the easiest proof of this theorem requiring the least amount of measure theory.
$endgroup$
– RRL
Dec 13 '18 at 22:45
add a comment |
$begingroup$
Since $f$ and $g$ are integrable, so is $f-g$, so $A=(f-g)^{-1}(mathbb{R}setminus{0})$ is a measurable set. Hence also $A^ccap B$ is measurable.
Then use the fact that for any integrable function $f$ and partitioning measurable sets $A$ and $B$ we have $f=chi_Af+chi_Bf$ and hence $int f=intchi_Af+chi_Bf=intchi_Af+intchi_Bf=int_Af+int_Bf$.
Here $chi_A$ denotes the indicator function, which is $1$ in $A$ and $0$ elsewhere.
answered Dec 13 '18 at 21:20
SmileyCraftSmileyCraft
3,591517
$endgroup$
add a comment |
$begingroup$
Since $f$ and $g$ are integrable, so is $f-g$, so $A=(f-g)^{-1}(mathbb{R}setminus{0})$ is a measurable set. Hence also $A^ccap B$ is measurable.
Then use the fact that for any integrable function $f$ and partitioning measurable sets $A$ and $B$ we have $f=chi_Af+chi_Bf$ and hence $int f=intchi_Af+chi_Bf=intchi_Af+intchi_Bf=int_Af+int_Bf$.
Here $chi_A$ denotes the indicator function, which is $1$ in $A$ and $0$ elsewhere.
answered Dec 13 '18 at 21:20
SmileyCraftSmileyCraft
3,591517
$endgroup$
add a comment |
$begingroup$
Since $f$ and $g$ are integrable, so is $f-g$, so $A=(f-g)^{-1}(mathbb{R}setminus{0})$ is a measurable set. Hence also $A^ccap B$ is measurable.
Then use the fact that for any integrable function $f$ and partitioning measurable sets $A$ and $B$ we have $f=chi_Af+chi_Bf$ and hence $int f=intchi_Af+chi_Bf=intchi_Af+intchi_Bf=int_Af+int_Bf$.
Here $chi_A$ denotes the indicator function, which is $1$ in $A$ and $0$ elsewhere.
answered Dec 13 '18 at 21:20
SmileyCraftSmileyCraft
3,591517
$endgroup$
Since $f$ and $g$ are integrable, so is $f-g$, so $A=(f-g)^{-1}(mathbb{R}setminus{0})$ is a measurable set. Hence also $A^ccap B$ is measurable.
Then use the fact that for any integrable function $f$ and partitioning measurable sets $A$ and $B$ we have $f=chi_Af+chi_Bf$ and hence $int f=intchi_Af+chi_Bf=intchi_Af+intchi_Bf=int_Af+int_Bf$.
Here $chi_A$ denotes the indicator function, which is $1$ in $A$ and $0$ elsewhere.
answered Dec 13 '18 at 21:20
SmileyCraftSmileyCraft
3,591517
edited Dec 13 '18 at 21:31
answered Dec 13 '18 at 21:20
SmileyCraftSmileyCraft
3,591517
answered Dec 13 '18 at 21:20
SmileyCraftSmileyCraft
3,591517
answered Dec 13 '18 at 21:20
SmileyCraftSmileyCraft
3,591517
3,591517
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038561%2fif-x-fx-neq-gx-has-zero-measure-then-int-f-int-g%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Not sure about your tags... I would not tag calculus and I would tag Lebesgue integration instead of Riemann Integration.
$endgroup$
– Zuriel
Dec 13 '18 at 21:23