Mixing
Lebesgue Integral, an example
$begingroup$
I have been trying to studying the construction of Lebesgue integral for a while now. I am following the Princeton Lectures on Analysis and I am stuck at the part where it defines the integral of non-negative functions. I find the definition to be quite clear but I cannot understand the examples given here.
With the above definition of the integral, there are only two possible cases; the supremum is either finite, or infinite. In the first case, when $int f(x)dx<+infty$, we shall say that $f$ is Lebesgue integrable or simply integrable.
Clearly, if $E$ is any measurable subset of $mathbb R^d$, and $fge0$, then $f_{chi_E}$ is also positive, and we define
$$int f(x)dx=int f(x)chi_E(x)dx.$$
Simple examples of functions on $mathbb R^d$ that are integrable (or non-integrable) are given by
$$f_a(x)=begin{cases}|x|^{-a}&text{ if }|x|le1,\0&text{ if }|x|>1.end{cases}$$
$$F_a(x)=frac1{1+|x|^a}, text{ all }xinmathbb R^d.$$
Then $f_a$ is integrable exactly when $a<d$, while $F_a$ is integrable exactly when $a>d$.
How are the values of "a" here in these two examples making the function integrable or not !? and what does the value of "a" has to do with "d" ?
measure-theory lebesgue-integral lebesgue-measure
edited Jan 13 at 14:15
Lorenzo B.
1,8402520
asked Jan 13 at 9:42
Rpdp_sRpdp_s
234
$endgroup$
add a comment |
$begingroup$
I have been trying to studying the construction of Lebesgue integral for a while now. I am following the Princeton Lectures on Analysis and I am stuck at the part where it defines the integral of non-negative functions. I find the definition to be quite clear but I cannot understand the examples given here.
With the above definition of the integral, there are only two possible cases; the supremum is either finite, or infinite. In the first case, when $int f(x)dx<+infty$, we shall say that $f$ is Lebesgue integrable or simply integrable.
Clearly, if $E$ is any measurable subset of $mathbb R^d$, and $fge0$, then $f_{chi_E}$ is also positive, and we define
$$int f(x)dx=int f(x)chi_E(x)dx.$$
Simple examples of functions on $mathbb R^d$ that are integrable (or non-integrable) are given by
$$f_a(x)=begin{cases}|x|^{-a}&text{ if }|x|le1,\0&text{ if }|x|>1.end{cases}$$
$$F_a(x)=frac1{1+|x|^a}, text{ all }xinmathbb R^d.$$
Then $f_a$ is integrable exactly when $a<d$, while $F_a$ is integrable exactly when $a>d$.
How are the values of "a" here in these two examples making the function integrable or not !? and what does the value of "a" has to do with "d" ?
measure-theory lebesgue-integral lebesgue-measure
edited Jan 13 at 14:15
Lorenzo B.
1,8402520
asked Jan 13 at 9:42
Rpdp_sRpdp_s
234
$endgroup$
$begingroup$
The point discussed in the article is not about measurability. All the maps under discussion are measurable. It is about integrability. And it seems that the integrability of those maps is discussed later on in the article.
$endgroup$
– mathcounterexamples.net
Jan 13 at 9:45
$begingroup$
Im sorry thats what i meant. How does it make the function integrable or not. I am changing the orginal question. And it doesnt discuss anything about these further down.
$endgroup$
– Rpdp_s
Jan 13 at 9:48
1
$begingroup$
It is written see the discussion following Corollary 1.10 and Exercise 10. What are those about?
$endgroup$
– mathcounterexamples.net
Jan 13 at 9:50
$begingroup$
Those are a bit further down after discussing a few more theorems, However, I was worried if I missed something on the way making the example seem hard. Looks like I will have to skip them for now and come back. Thank you for pointing it out.
$endgroup$
– Rpdp_s
Jan 13 at 9:54
add a comment |
$begingroup$
I have been trying to studying the construction of Lebesgue integral for a while now. I am following the Princeton Lectures on Analysis and I am stuck at the part where it defines the integral of non-negative functions. I find the definition to be quite clear but I cannot understand the examples given here.
With the above definition of the integral, there are only two possible cases; the supremum is either finite, or infinite. In the first case, when $int f(x)dx<+infty$, we shall say that $f$ is Lebesgue integrable or simply integrable.
Clearly, if $E$ is any measurable subset of $mathbb R^d$, and $fge0$, then $f_{chi_E}$ is also positive, and we define
$$int f(x)dx=int f(x)chi_E(x)dx.$$
Simple examples of functions on $mathbb R^d$ that are integrable (or non-integrable) are given by
$$f_a(x)=begin{cases}|x|^{-a}&text{ if }|x|le1,\0&text{ if }|x|>1.end{cases}$$
$$F_a(x)=frac1{1+|x|^a}, text{ all }xinmathbb R^d.$$
Then $f_a$ is integrable exactly when $a<d$, while $F_a$ is integrable exactly when $a>d$.
How are the values of "a" here in these two examples making the function integrable or not !? and what does the value of "a" has to do with "d" ?
measure-theory lebesgue-integral lebesgue-measure
edited Jan 13 at 14:15
Lorenzo B.
1,8402520
asked Jan 13 at 9:42
Rpdp_sRpdp_s
234
$endgroup$
I have been trying to studying the construction of Lebesgue integral for a while now. I am following the Princeton Lectures on Analysis and I am stuck at the part where it defines the integral of non-negative functions. I find the definition to be quite clear but I cannot understand the examples given here.
With the above definition of the integral, there are only two possible cases; the supremum is either finite, or infinite. In the first case, when $int f(x)dx<+infty$, we shall say that $f$ is Lebesgue integrable or simply integrable.
Clearly, if $E$ is any measurable subset of $mathbb R^d$, and $fge0$, then $f_{chi_E}$ is also positive, and we define
$$int f(x)dx=int f(x)chi_E(x)dx.$$
Simple examples of functions on $mathbb R^d$ that are integrable (or non-integrable) are given by
$$f_a(x)=begin{cases}|x|^{-a}&text{ if }|x|le1,\0&text{ if }|x|>1.end{cases}$$
$$F_a(x)=frac1{1+|x|^a}, text{ all }xinmathbb R^d.$$
Then $f_a$ is integrable exactly when $a<d$, while $F_a$ is integrable exactly when $a>d$.
How are the values of "a" here in these two examples making the function integrable or not !? and what does the value of "a" has to do with "d" ?
measure-theory lebesgue-integral lebesgue-measure
measure-theory lebesgue-integral lebesgue-measure
edited Jan 13 at 14:15
Lorenzo B.
1,8402520
asked Jan 13 at 9:42
Rpdp_sRpdp_s
234
edited Jan 13 at 14:15
Lorenzo B.
1,8402520
asked Jan 13 at 9:42
Rpdp_sRpdp_s
234
edited Jan 13 at 14:15
Lorenzo B.
1,8402520
edited Jan 13 at 14:15
Lorenzo B.
1,8402520
edited Jan 13 at 14:15
Lorenzo B.
1,8402520
1,8402520
asked Jan 13 at 9:42
Rpdp_sRpdp_s
234
asked Jan 13 at 9:42
Rpdp_sRpdp_s
234
asked Jan 13 at 9:42
Rpdp_sRpdp_s
234
234
$begingroup$
The point discussed in the article is not about measurability. All the maps under discussion are measurable. It is about integrability. And it seems that the integrability of those maps is discussed later on in the article.
$endgroup$
– mathcounterexamples.net
Jan 13 at 9:45
$begingroup$
Im sorry thats what i meant. How does it make the function integrable or not. I am changing the orginal question. And it doesnt discuss anything about these further down.
$endgroup$
– Rpdp_s
Jan 13 at 9:48
1
$begingroup$
It is written see the discussion following Corollary 1.10 and Exercise 10. What are those about?
$endgroup$
– mathcounterexamples.net
Jan 13 at 9:50
$begingroup$
Those are a bit further down after discussing a few more theorems, However, I was worried if I missed something on the way making the example seem hard. Looks like I will have to skip them for now and come back. Thank you for pointing it out.
$endgroup$
– Rpdp_s
Jan 13 at 9:54
add a comment |
$begingroup$
The point discussed in the article is not about measurability. All the maps under discussion are measurable. It is about integrability. And it seems that the integrability of those maps is discussed later on in the article.
$endgroup$
– mathcounterexamples.net
Jan 13 at 9:45
$begingroup$
Im sorry thats what i meant. How does it make the function integrable or not. I am changing the orginal question. And it doesnt discuss anything about these further down.
$endgroup$
– Rpdp_s
Jan 13 at 9:48
1
$begingroup$
It is written see the discussion following Corollary 1.10 and Exercise 10. What are those about?
$endgroup$
– mathcounterexamples.net
Jan 13 at 9:50
$begingroup$
Those are a bit further down after discussing a few more theorems, However, I was worried if I missed something on the way making the example seem hard. Looks like I will have to skip them for now and come back. Thank you for pointing it out.
$endgroup$
– Rpdp_s
Jan 13 at 9:54
$begingroup$
The point discussed in the article is not about measurability. All the maps under discussion are measurable. It is about integrability. And it seems that the integrability of those maps is discussed later on in the article.
$endgroup$
– mathcounterexamples.net
Jan 13 at 9:45
$begingroup$
The point discussed in the article is not about measurability. All the maps under discussion are measurable. It is about integrability. And it seems that the integrability of those maps is discussed later on in the article.
$endgroup$
– mathcounterexamples.net
Jan 13 at 9:45
$begingroup$
Im sorry thats what i meant. How does it make the function integrable or not. I am changing the orginal question. And it doesnt discuss anything about these further down.
$endgroup$
– Rpdp_s
Jan 13 at 9:48
$begingroup$
Im sorry thats what i meant. How does it make the function integrable or not. I am changing the orginal question. And it doesnt discuss anything about these further down.
$endgroup$
– Rpdp_s
Jan 13 at 9:48
1
1
$begingroup$
It is written see the discussion following Corollary 1.10 and Exercise 10. What are those about?
$endgroup$
– mathcounterexamples.net
Jan 13 at 9:50
$begingroup$
It is written see the discussion following Corollary 1.10 and Exercise 10. What are those about?
$endgroup$
– mathcounterexamples.net
Jan 13 at 9:50
$begingroup$
Those are a bit further down after discussing a few more theorems, However, I was worried if I missed something on the way making the example seem hard. Looks like I will have to skip them for now and come back. Thank you for pointing it out.
$endgroup$
– Rpdp_s
Jan 13 at 9:54
$begingroup$
Those are a bit further down after discussing a few more theorems, However, I was worried if I missed something on the way making the example seem hard. Looks like I will have to skip them for now and come back. Thank you for pointing it out.
$endgroup$
– Rpdp_s
Jan 13 at 9:54
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider a map $f : mathbb R^d to mathbb R$ which is only dependent on the radius $R$ in the $n$-sphere Spherical coordinates. I.e. $f(x)=g(vert xvert)$ where $g$ is a real map.
You then have
$$int_{mathbb R^d} f(x) dx = K_d int_0^infty g(R) R^{d-1} dR$$ where $K_d$ is a constant that only depends on $d$. This uses an integral by substitution in dimension $d$.
You can use that to look at the integrability of the maps of your original question.
answered Jan 13 at 10:05
mathcounterexamples.netmathcounterexamples.net
26.8k22157
$endgroup$
add a comment |
$begingroup$
Likely this is explained further down in your text. Anyway...
The Lebesgue integral of $f_a$ over $mathbb R^d$ is:
$$int_{mathbb R^d} f_a,dmu = int_0^infty mubig({xin B_d(1):f_a(x)>y}big),dy
= int_0^infty mubig({xin B_d(1):|x|^{-a}>y}big),dy$$
where $B_d(r)$ is the unit ball in $d$ dimensions with radius $r$.
If $ale 0$ there is no singularity nor infinity involved, and the result is finite.
If $a>0$ then the measure at $y$ is the volume of the $d$-ball $B_d(r)$ with radius $r=y^{-frac 1a}le 1$. That also means that $yge 1$. The volume of $B_d(r)$ is $C_d r^d$ for some constant $C_d$ that depends only on $d$ (for instance $C_2=pi$ and $C_3=frac 43pi$). So the integral becomes:
$$int_{mathbb R^d} f_a,dmu
= int_1^infty C_dcdot(y^{-frac 1a})^d,dy
= frac{C_d}{-frac da+1} y^{-frac da+1}Bigg|_1^infty
$$
This is finite iff $-frac da+1<0 iff a<dquad$ (for the case $a>0$).
Therefore $f_a$ is Lebesgue integrable over $mathbb R^d$ iff $a<d$.
answered Jan 13 at 11:18
I like SerenaI like Serena
4,2071722
$endgroup$
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%2f3071840%2flebesgue-integral-an-example%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
$begingroup$
Consider a map $f : mathbb R^d to mathbb R$ which is only dependent on the radius $R$ in the $n$-sphere Spherical coordinates. I.e. $f(x)=g(vert xvert)$ where $g$ is a real map.
You then have
$$int_{mathbb R^d} f(x) dx = K_d int_0^infty g(R) R^{d-1} dR$$ where $K_d$ is a constant that only depends on $d$. This uses an integral by substitution in dimension $d$.
You can use that to look at the integrability of the maps of your original question.
answered Jan 13 at 10:05
mathcounterexamples.netmathcounterexamples.net
26.8k22157
$endgroup$
add a comment |
$begingroup$
Consider a map $f : mathbb R^d to mathbb R$ which is only dependent on the radius $R$ in the $n$-sphere Spherical coordinates. I.e. $f(x)=g(vert xvert)$ where $g$ is a real map.
You then have
$$int_{mathbb R^d} f(x) dx = K_d int_0^infty g(R) R^{d-1} dR$$ where $K_d$ is a constant that only depends on $d$. This uses an integral by substitution in dimension $d$.
You can use that to look at the integrability of the maps of your original question.
answered Jan 13 at 10:05
mathcounterexamples.netmathcounterexamples.net
26.8k22157
$endgroup$
add a comment |
$begingroup$
Consider a map $f : mathbb R^d to mathbb R$ which is only dependent on the radius $R$ in the $n$-sphere Spherical coordinates. I.e. $f(x)=g(vert xvert)$ where $g$ is a real map.
You then have
$$int_{mathbb R^d} f(x) dx = K_d int_0^infty g(R) R^{d-1} dR$$ where $K_d$ is a constant that only depends on $d$. This uses an integral by substitution in dimension $d$.
You can use that to look at the integrability of the maps of your original question.
answered Jan 13 at 10:05
mathcounterexamples.netmathcounterexamples.net
26.8k22157
$endgroup$
Consider a map $f : mathbb R^d to mathbb R$ which is only dependent on the radius $R$ in the $n$-sphere Spherical coordinates. I.e. $f(x)=g(vert xvert)$ where $g$ is a real map.
You then have
$$int_{mathbb R^d} f(x) dx = K_d int_0^infty g(R) R^{d-1} dR$$ where $K_d$ is a constant that only depends on $d$. This uses an integral by substitution in dimension $d$.
You can use that to look at the integrability of the maps of your original question.
answered Jan 13 at 10:05
mathcounterexamples.netmathcounterexamples.net
26.8k22157
edited Jan 13 at 13:28
answered Jan 13 at 10:05
mathcounterexamples.netmathcounterexamples.net
26.8k22157
answered Jan 13 at 10:05
mathcounterexamples.netmathcounterexamples.net
26.8k22157
answered Jan 13 at 10:05
mathcounterexamples.netmathcounterexamples.net
26.8k22157
26.8k22157
add a comment |
add a comment |
$begingroup$
Likely this is explained further down in your text. Anyway...
The Lebesgue integral of $f_a$ over $mathbb R^d$ is:
$$int_{mathbb R^d} f_a,dmu = int_0^infty mubig({xin B_d(1):f_a(x)>y}big),dy
= int_0^infty mubig({xin B_d(1):|x|^{-a}>y}big),dy$$
where $B_d(r)$ is the unit ball in $d$ dimensions with radius $r$.
If $ale 0$ there is no singularity nor infinity involved, and the result is finite.
If $a>0$ then the measure at $y$ is the volume of the $d$-ball $B_d(r)$ with radius $r=y^{-frac 1a}le 1$. That also means that $yge 1$. The volume of $B_d(r)$ is $C_d r^d$ for some constant $C_d$ that depends only on $d$ (for instance $C_2=pi$ and $C_3=frac 43pi$). So the integral becomes:
$$int_{mathbb R^d} f_a,dmu
= int_1^infty C_dcdot(y^{-frac 1a})^d,dy
= frac{C_d}{-frac da+1} y^{-frac da+1}Bigg|_1^infty
$$
This is finite iff $-frac da+1<0 iff a<dquad$ (for the case $a>0$).
Therefore $f_a$ is Lebesgue integrable over $mathbb R^d$ iff $a<d$.
answered Jan 13 at 11:18
I like SerenaI like Serena
4,2071722
$endgroup$
add a comment |
$begingroup$
Likely this is explained further down in your text. Anyway...
The Lebesgue integral of $f_a$ over $mathbb R^d$ is:
$$int_{mathbb R^d} f_a,dmu = int_0^infty mubig({xin B_d(1):f_a(x)>y}big),dy
= int_0^infty mubig({xin B_d(1):|x|^{-a}>y}big),dy$$
where $B_d(r)$ is the unit ball in $d$ dimensions with radius $r$.
If $ale 0$ there is no singularity nor infinity involved, and the result is finite.
If $a>0$ then the measure at $y$ is the volume of the $d$-ball $B_d(r)$ with radius $r=y^{-frac 1a}le 1$. That also means that $yge 1$. The volume of $B_d(r)$ is $C_d r^d$ for some constant $C_d$ that depends only on $d$ (for instance $C_2=pi$ and $C_3=frac 43pi$). So the integral becomes:
$$int_{mathbb R^d} f_a,dmu
= int_1^infty C_dcdot(y^{-frac 1a})^d,dy
= frac{C_d}{-frac da+1} y^{-frac da+1}Bigg|_1^infty
$$
This is finite iff $-frac da+1<0 iff a<dquad$ (for the case $a>0$).
Therefore $f_a$ is Lebesgue integrable over $mathbb R^d$ iff $a<d$.
answered Jan 13 at 11:18
I like SerenaI like Serena
4,2071722
$endgroup$
add a comment |
$begingroup$
Likely this is explained further down in your text. Anyway...
The Lebesgue integral of $f_a$ over $mathbb R^d$ is:
$$int_{mathbb R^d} f_a,dmu = int_0^infty mubig({xin B_d(1):f_a(x)>y}big),dy
= int_0^infty mubig({xin B_d(1):|x|^{-a}>y}big),dy$$
where $B_d(r)$ is the unit ball in $d$ dimensions with radius $r$.
If $ale 0$ there is no singularity nor infinity involved, and the result is finite.
If $a>0$ then the measure at $y$ is the volume of the $d$-ball $B_d(r)$ with radius $r=y^{-frac 1a}le 1$. That also means that $yge 1$. The volume of $B_d(r)$ is $C_d r^d$ for some constant $C_d$ that depends only on $d$ (for instance $C_2=pi$ and $C_3=frac 43pi$). So the integral becomes:
$$int_{mathbb R^d} f_a,dmu
= int_1^infty C_dcdot(y^{-frac 1a})^d,dy
= frac{C_d}{-frac da+1} y^{-frac da+1}Bigg|_1^infty
$$
This is finite iff $-frac da+1<0 iff a<dquad$ (for the case $a>0$).
Therefore $f_a$ is Lebesgue integrable over $mathbb R^d$ iff $a<d$.
answered Jan 13 at 11:18
I like SerenaI like Serena
4,2071722
$endgroup$
Likely this is explained further down in your text. Anyway...
The Lebesgue integral of $f_a$ over $mathbb R^d$ is:
$$int_{mathbb R^d} f_a,dmu = int_0^infty mubig({xin B_d(1):f_a(x)>y}big),dy
= int_0^infty mubig({xin B_d(1):|x|^{-a}>y}big),dy$$
where $B_d(r)$ is the unit ball in $d$ dimensions with radius $r$.
If $ale 0$ there is no singularity nor infinity involved, and the result is finite.
If $a>0$ then the measure at $y$ is the volume of the $d$-ball $B_d(r)$ with radius $r=y^{-frac 1a}le 1$. That also means that $yge 1$. The volume of $B_d(r)$ is $C_d r^d$ for some constant $C_d$ that depends only on $d$ (for instance $C_2=pi$ and $C_3=frac 43pi$). So the integral becomes:
$$int_{mathbb R^d} f_a,dmu
= int_1^infty C_dcdot(y^{-frac 1a})^d,dy
= frac{C_d}{-frac da+1} y^{-frac da+1}Bigg|_1^infty
$$
This is finite iff $-frac da+1<0 iff a<dquad$ (for the case $a>0$).
Therefore $f_a$ is Lebesgue integrable over $mathbb R^d$ iff $a<d$.
answered Jan 13 at 11:18
I like SerenaI like Serena
4,2071722
answered Jan 13 at 11:18
I like SerenaI like Serena
4,2071722
answered Jan 13 at 11:18
I like SerenaI like Serena
4,2071722
answered Jan 13 at 11:18
I like SerenaI like Serena
4,2071722
4,2071722
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%2f3071840%2flebesgue-integral-an-example%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
$begingroup$
The point discussed in the article is not about measurability. All the maps under discussion are measurable. It is about integrability. And it seems that the integrability of those maps is discussed later on in the article.
$endgroup$
– mathcounterexamples.net
Jan 13 at 9:45
$begingroup$
Im sorry thats what i meant. How does it make the function integrable or not. I am changing the orginal question. And it doesnt discuss anything about these further down.
$endgroup$
– Rpdp_s
Jan 13 at 9:48
1
$begingroup$
It is written see the discussion following Corollary 1.10 and Exercise 10. What are those about?
$endgroup$
– mathcounterexamples.net
Jan 13 at 9:50
$begingroup$
Those are a bit further down after discussing a few more theorems, However, I was worried if I missed something on the way making the example seem hard. Looks like I will have to skip them for now and come back. Thank you for pointing it out.
$endgroup$
– Rpdp_s
Jan 13 at 9:54