Mixing
function is not Lebesgue integrable
How to prove that $f(x)=2xsin(frac{1}{x^2})-frac{2}{x}cos(frac{1}{x^2})$ on if $xneq0$ and $f(x)=0$ if $x=0$ is not Lebesgue-integrable on $[-1,1]$? I think the reason is, that $frac{2}{x}$ is part of the function Showing that $1/x$ is NOT Lebesgue Integrable on $(0,1]$.
I tried to construct step functions using the step function for $1/x$ on $(0,1]$: Let $lfloor wrfloor$ be the greatest integer less than or equal to $w$. Then the function
$$xmapsto begin{cases} lfloor 2/xrfloor & text{if } lfloor 1/xrfloorle n \[8pt]
n & text{otherwise} end{cases}$$ is simple. It is $le 2/x$ and its integral over $(0,1]$ approaches $infty$ as $ntoinfty$.
Now with the upper gaussian brackets and using that $-1le sin(frac{1}{x^2}),cos(frac{1}{x^2})le 1$. But I don't get it
analysis measure-theory lebesgue-integral
edited Nov 21 '18 at 15:32
Tianlalu
3,09621038
asked Nov 21 '18 at 15:31
Gero
23319
add a comment |
How to prove that $f(x)=2xsin(frac{1}{x^2})-frac{2}{x}cos(frac{1}{x^2})$ on if $xneq0$ and $f(x)=0$ if $x=0$ is not Lebesgue-integrable on $[-1,1]$? I think the reason is, that $frac{2}{x}$ is part of the function Showing that $1/x$ is NOT Lebesgue Integrable on $(0,1]$.
I tried to construct step functions using the step function for $1/x$ on $(0,1]$: Let $lfloor wrfloor$ be the greatest integer less than or equal to $w$. Then the function
$$xmapsto begin{cases} lfloor 2/xrfloor & text{if } lfloor 1/xrfloorle n \[8pt]
n & text{otherwise} end{cases}$$ is simple. It is $le 2/x$ and its integral over $(0,1]$ approaches $infty$ as $ntoinfty$.
Now with the upper gaussian brackets and using that $-1le sin(frac{1}{x^2}),cos(frac{1}{x^2})le 1$. But I don't get it
analysis measure-theory lebesgue-integral
edited Nov 21 '18 at 15:32
Tianlalu
3,09621038
asked Nov 21 '18 at 15:31
Gero
23319
It's worth commenting that this thing integrates to $x^2 sin x^{-2}$, and is Riemann integrable ...
– Richard Martin
Nov 21 '18 at 15:43
The problem with the Lebesgue integral is the wild oscillation of $f'$ near zero.
– Richard Martin
Nov 21 '18 at 15:45
add a comment |
How to prove that $f(x)=2xsin(frac{1}{x^2})-frac{2}{x}cos(frac{1}{x^2})$ on if $xneq0$ and $f(x)=0$ if $x=0$ is not Lebesgue-integrable on $[-1,1]$? I think the reason is, that $frac{2}{x}$ is part of the function Showing that $1/x$ is NOT Lebesgue Integrable on $(0,1]$.
I tried to construct step functions using the step function for $1/x$ on $(0,1]$: Let $lfloor wrfloor$ be the greatest integer less than or equal to $w$. Then the function
$$xmapsto begin{cases} lfloor 2/xrfloor & text{if } lfloor 1/xrfloorle n \[8pt]
n & text{otherwise} end{cases}$$ is simple. It is $le 2/x$ and its integral over $(0,1]$ approaches $infty$ as $ntoinfty$.
Now with the upper gaussian brackets and using that $-1le sin(frac{1}{x^2}),cos(frac{1}{x^2})le 1$. But I don't get it
analysis measure-theory lebesgue-integral
edited Nov 21 '18 at 15:32
Tianlalu
3,09621038
asked Nov 21 '18 at 15:31
Gero
23319
How to prove that $f(x)=2xsin(frac{1}{x^2})-frac{2}{x}cos(frac{1}{x^2})$ on if $xneq0$ and $f(x)=0$ if $x=0$ is not Lebesgue-integrable on $[-1,1]$? I think the reason is, that $frac{2}{x}$ is part of the function Showing that $1/x$ is NOT Lebesgue Integrable on $(0,1]$.
I tried to construct step functions using the step function for $1/x$ on $(0,1]$: Let $lfloor wrfloor$ be the greatest integer less than or equal to $w$. Then the function
$$xmapsto begin{cases} lfloor 2/xrfloor & text{if } lfloor 1/xrfloorle n \[8pt]
n & text{otherwise} end{cases}$$ is simple. It is $le 2/x$ and its integral over $(0,1]$ approaches $infty$ as $ntoinfty$.
Now with the upper gaussian brackets and using that $-1le sin(frac{1}{x^2}),cos(frac{1}{x^2})le 1$. But I don't get it
analysis measure-theory lebesgue-integral
analysis measure-theory lebesgue-integral
edited Nov 21 '18 at 15:32
Tianlalu
3,09621038
asked Nov 21 '18 at 15:31
Gero
23319
edited Nov 21 '18 at 15:32
Tianlalu
3,09621038
asked Nov 21 '18 at 15:31
Gero
23319
edited Nov 21 '18 at 15:32
Tianlalu
3,09621038
edited Nov 21 '18 at 15:32
Tianlalu
3,09621038
edited Nov 21 '18 at 15:32
Tianlalu
3,09621038
3,09621038
asked Nov 21 '18 at 15:31
Gero
23319
asked Nov 21 '18 at 15:31
Gero
23319
asked Nov 21 '18 at 15:31
Gero
23319
23319
It's worth commenting that this thing integrates to $x^2 sin x^{-2}$, and is Riemann integrable ...
– Richard Martin
Nov 21 '18 at 15:43
The problem with the Lebesgue integral is the wild oscillation of $f'$ near zero.
– Richard Martin
Nov 21 '18 at 15:45
add a comment |
It's worth commenting that this thing integrates to $x^2 sin x^{-2}$, and is Riemann integrable ...
– Richard Martin
Nov 21 '18 at 15:43
The problem with the Lebesgue integral is the wild oscillation of $f'$ near zero.
– Richard Martin
Nov 21 '18 at 15:45
It's worth commenting that this thing integrates to $x^2 sin x^{-2}$, and is Riemann integrable ...
– Richard Martin
Nov 21 '18 at 15:43
It's worth commenting that this thing integrates to $x^2 sin x^{-2}$, and is Riemann integrable ...
– Richard Martin
Nov 21 '18 at 15:43
The problem with the Lebesgue integral is the wild oscillation of $f'$ near zero.
– Richard Martin
Nov 21 '18 at 15:45
The problem with the Lebesgue integral is the wild oscillation of $f'$ near zero.
– Richard Martin
Nov 21 '18 at 15:45
add a comment |
2 Answers
2
active
oldest
votes
The precise meaning that a function $f$ is Lebesgue integrable on some measurable set $E$ is that both positive and negative parts, $f^+(x) = max(f(x),0)$ and $f^-(x) = max(-f(x),0)$, must have finite-valued integrals
$$int_E f^+ < +infty, ,, int_E f^- < +infty$$
Since $f^+$ and $f^-$ are nonnegative and $|f| = f^+ +f^-$, Lebesgue integrability requires that $int_E|f| < +infty$.
In this case there is no problem with $x mapsto 2xsinfrac{1}{x^2}$ as this can be continuously extended at $0$ and so is both Riemann and Lebesgue integrable on $[0,1]$ and $[-1,0]$. Recall that a bounded function that is Riemann integrable on a bounded interval is always Lebesgue integrable.
On the other hand, $x mapsto frac{2}{x} cos frac{1}{x^2}$ is not Lebesgue integrable on $[0,1]$. If it were we could apply the change-of-variables $x to frac{1}{sqrt{t}}$ and have
$$int_1^infty frac{|cos t|}{t} , dt = int_0^1 frac{2}{x} left|cos frac{1}{x^2}right| , dx < +infty$$
However,
$$int_1^infty frac{|cos t|}{t} , dt > sum_{k=1}^inftyint_{kpi}^{kpi + pi} frac{|cos t|}{t} , dt > sum_{k=1}^infty frac{1}{kpi +pi}int_{kpi}^{kpi + pi} |cos t| , dt = frac{2}{pi}sum_{k=1}^{infty} frac{1}{k+1} \ = +infty $$
since the harmonic series on the RHS is divergent.
Note that the improper (Riemann) integral, without the absolute value present,
$$int_0^1 frac{2}{x} cos frac{1}{x^2} , dx $$
is convergent, but this does not enforce Lebesgue integrability.
answered Nov 22 '18 at 1:40
RRL
49.3k42573
add a comment |
The derivative of $F(x)= x^2 sin (1/x^2)$ if $f$, except at $x=0$, where the former is zero and the latter is not defined.
Now if $f$ were integrable, then $F$ would be absolutely continuous.
We show that this is not the case.
For $j=1,2,dots$ let $a_j= frac{1}{sqrt{pi j}}$ and $b_j = frac{1}{sqrt{pi( j + 1/2)}}$.
Then $F(a_j)=0$ and $F(b_j)=pi (j+1/2)$.
To conclude, for any $j_1$:
$sum_{j>j_1}| a_j - b_j| < a_{j_1}to 0$ as $j_1toinfty$, while
$sum_{j>j_1} |F(a_j)-F(b_j)| = sum_{j>j_1} pi (j+1/2)= infty$.
answered Nov 22 '18 at 2:11
Fnacool
4,976511
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%2f3007873%2ffunction-is-not-lebesgue-integrable%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
The precise meaning that a function $f$ is Lebesgue integrable on some measurable set $E$ is that both positive and negative parts, $f^+(x) = max(f(x),0)$ and $f^-(x) = max(-f(x),0)$, must have finite-valued integrals
$$int_E f^+ < +infty, ,, int_E f^- < +infty$$
Since $f^+$ and $f^-$ are nonnegative and $|f| = f^+ +f^-$, Lebesgue integrability requires that $int_E|f| < +infty$.
In this case there is no problem with $x mapsto 2xsinfrac{1}{x^2}$ as this can be continuously extended at $0$ and so is both Riemann and Lebesgue integrable on $[0,1]$ and $[-1,0]$. Recall that a bounded function that is Riemann integrable on a bounded interval is always Lebesgue integrable.
On the other hand, $x mapsto frac{2}{x} cos frac{1}{x^2}$ is not Lebesgue integrable on $[0,1]$. If it were we could apply the change-of-variables $x to frac{1}{sqrt{t}}$ and have
$$int_1^infty frac{|cos t|}{t} , dt = int_0^1 frac{2}{x} left|cos frac{1}{x^2}right| , dx < +infty$$
However,
$$int_1^infty frac{|cos t|}{t} , dt > sum_{k=1}^inftyint_{kpi}^{kpi + pi} frac{|cos t|}{t} , dt > sum_{k=1}^infty frac{1}{kpi +pi}int_{kpi}^{kpi + pi} |cos t| , dt = frac{2}{pi}sum_{k=1}^{infty} frac{1}{k+1} \ = +infty $$
since the harmonic series on the RHS is divergent.
Note that the improper (Riemann) integral, without the absolute value present,
$$int_0^1 frac{2}{x} cos frac{1}{x^2} , dx $$
is convergent, but this does not enforce Lebesgue integrability.
answered Nov 22 '18 at 1:40
RRL
49.3k42573
add a comment |
The precise meaning that a function $f$ is Lebesgue integrable on some measurable set $E$ is that both positive and negative parts, $f^+(x) = max(f(x),0)$ and $f^-(x) = max(-f(x),0)$, must have finite-valued integrals
$$int_E f^+ < +infty, ,, int_E f^- < +infty$$
Since $f^+$ and $f^-$ are nonnegative and $|f| = f^+ +f^-$, Lebesgue integrability requires that $int_E|f| < +infty$.
In this case there is no problem with $x mapsto 2xsinfrac{1}{x^2}$ as this can be continuously extended at $0$ and so is both Riemann and Lebesgue integrable on $[0,1]$ and $[-1,0]$. Recall that a bounded function that is Riemann integrable on a bounded interval is always Lebesgue integrable.
On the other hand, $x mapsto frac{2}{x} cos frac{1}{x^2}$ is not Lebesgue integrable on $[0,1]$. If it were we could apply the change-of-variables $x to frac{1}{sqrt{t}}$ and have
$$int_1^infty frac{|cos t|}{t} , dt = int_0^1 frac{2}{x} left|cos frac{1}{x^2}right| , dx < +infty$$
However,
$$int_1^infty frac{|cos t|}{t} , dt > sum_{k=1}^inftyint_{kpi}^{kpi + pi} frac{|cos t|}{t} , dt > sum_{k=1}^infty frac{1}{kpi +pi}int_{kpi}^{kpi + pi} |cos t| , dt = frac{2}{pi}sum_{k=1}^{infty} frac{1}{k+1} \ = +infty $$
since the harmonic series on the RHS is divergent.
Note that the improper (Riemann) integral, without the absolute value present,
$$int_0^1 frac{2}{x} cos frac{1}{x^2} , dx $$
is convergent, but this does not enforce Lebesgue integrability.
answered Nov 22 '18 at 1:40
RRL
49.3k42573
add a comment |
The precise meaning that a function $f$ is Lebesgue integrable on some measurable set $E$ is that both positive and negative parts, $f^+(x) = max(f(x),0)$ and $f^-(x) = max(-f(x),0)$, must have finite-valued integrals
$$int_E f^+ < +infty, ,, int_E f^- < +infty$$
Since $f^+$ and $f^-$ are nonnegative and $|f| = f^+ +f^-$, Lebesgue integrability requires that $int_E|f| < +infty$.
In this case there is no problem with $x mapsto 2xsinfrac{1}{x^2}$ as this can be continuously extended at $0$ and so is both Riemann and Lebesgue integrable on $[0,1]$ and $[-1,0]$. Recall that a bounded function that is Riemann integrable on a bounded interval is always Lebesgue integrable.
On the other hand, $x mapsto frac{2}{x} cos frac{1}{x^2}$ is not Lebesgue integrable on $[0,1]$. If it were we could apply the change-of-variables $x to frac{1}{sqrt{t}}$ and have
$$int_1^infty frac{|cos t|}{t} , dt = int_0^1 frac{2}{x} left|cos frac{1}{x^2}right| , dx < +infty$$
However,
$$int_1^infty frac{|cos t|}{t} , dt > sum_{k=1}^inftyint_{kpi}^{kpi + pi} frac{|cos t|}{t} , dt > sum_{k=1}^infty frac{1}{kpi +pi}int_{kpi}^{kpi + pi} |cos t| , dt = frac{2}{pi}sum_{k=1}^{infty} frac{1}{k+1} \ = +infty $$
since the harmonic series on the RHS is divergent.
Note that the improper (Riemann) integral, without the absolute value present,
$$int_0^1 frac{2}{x} cos frac{1}{x^2} , dx $$
is convergent, but this does not enforce Lebesgue integrability.
answered Nov 22 '18 at 1:40
RRL
49.3k42573
The precise meaning that a function $f$ is Lebesgue integrable on some measurable set $E$ is that both positive and negative parts, $f^+(x) = max(f(x),0)$ and $f^-(x) = max(-f(x),0)$, must have finite-valued integrals
$$int_E f^+ < +infty, ,, int_E f^- < +infty$$
Since $f^+$ and $f^-$ are nonnegative and $|f| = f^+ +f^-$, Lebesgue integrability requires that $int_E|f| < +infty$.
In this case there is no problem with $x mapsto 2xsinfrac{1}{x^2}$ as this can be continuously extended at $0$ and so is both Riemann and Lebesgue integrable on $[0,1]$ and $[-1,0]$. Recall that a bounded function that is Riemann integrable on a bounded interval is always Lebesgue integrable.
On the other hand, $x mapsto frac{2}{x} cos frac{1}{x^2}$ is not Lebesgue integrable on $[0,1]$. If it were we could apply the change-of-variables $x to frac{1}{sqrt{t}}$ and have
$$int_1^infty frac{|cos t|}{t} , dt = int_0^1 frac{2}{x} left|cos frac{1}{x^2}right| , dx < +infty$$
However,
$$int_1^infty frac{|cos t|}{t} , dt > sum_{k=1}^inftyint_{kpi}^{kpi + pi} frac{|cos t|}{t} , dt > sum_{k=1}^infty frac{1}{kpi +pi}int_{kpi}^{kpi + pi} |cos t| , dt = frac{2}{pi}sum_{k=1}^{infty} frac{1}{k+1} \ = +infty $$
since the harmonic series on the RHS is divergent.
Note that the improper (Riemann) integral, without the absolute value present,
$$int_0^1 frac{2}{x} cos frac{1}{x^2} , dx $$
is convergent, but this does not enforce Lebesgue integrability.
answered Nov 22 '18 at 1:40
RRL
49.3k42573
edited Nov 22 '18 at 1:45
answered Nov 22 '18 at 1:40
RRL
49.3k42573
answered Nov 22 '18 at 1:40
RRL
49.3k42573
answered Nov 22 '18 at 1:40
RRL
49.3k42573
49.3k42573
add a comment |
add a comment |
The derivative of $F(x)= x^2 sin (1/x^2)$ if $f$, except at $x=0$, where the former is zero and the latter is not defined.
Now if $f$ were integrable, then $F$ would be absolutely continuous.
We show that this is not the case.
For $j=1,2,dots$ let $a_j= frac{1}{sqrt{pi j}}$ and $b_j = frac{1}{sqrt{pi( j + 1/2)}}$.
Then $F(a_j)=0$ and $F(b_j)=pi (j+1/2)$.
To conclude, for any $j_1$:
$sum_{j>j_1}| a_j - b_j| < a_{j_1}to 0$ as $j_1toinfty$, while
$sum_{j>j_1} |F(a_j)-F(b_j)| = sum_{j>j_1} pi (j+1/2)= infty$.
answered Nov 22 '18 at 2:11
Fnacool
4,976511
add a comment |
The derivative of $F(x)= x^2 sin (1/x^2)$ if $f$, except at $x=0$, where the former is zero and the latter is not defined.
Now if $f$ were integrable, then $F$ would be absolutely continuous.
We show that this is not the case.
For $j=1,2,dots$ let $a_j= frac{1}{sqrt{pi j}}$ and $b_j = frac{1}{sqrt{pi( j + 1/2)}}$.
Then $F(a_j)=0$ and $F(b_j)=pi (j+1/2)$.
To conclude, for any $j_1$:
$sum_{j>j_1}| a_j - b_j| < a_{j_1}to 0$ as $j_1toinfty$, while
$sum_{j>j_1} |F(a_j)-F(b_j)| = sum_{j>j_1} pi (j+1/2)= infty$.
answered Nov 22 '18 at 2:11
Fnacool
4,976511
add a comment |
The derivative of $F(x)= x^2 sin (1/x^2)$ if $f$, except at $x=0$, where the former is zero and the latter is not defined.
Now if $f$ were integrable, then $F$ would be absolutely continuous.
We show that this is not the case.
For $j=1,2,dots$ let $a_j= frac{1}{sqrt{pi j}}$ and $b_j = frac{1}{sqrt{pi( j + 1/2)}}$.
Then $F(a_j)=0$ and $F(b_j)=pi (j+1/2)$.
To conclude, for any $j_1$:
$sum_{j>j_1}| a_j - b_j| < a_{j_1}to 0$ as $j_1toinfty$, while
$sum_{j>j_1} |F(a_j)-F(b_j)| = sum_{j>j_1} pi (j+1/2)= infty$.
answered Nov 22 '18 at 2:11
Fnacool
4,976511
The derivative of $F(x)= x^2 sin (1/x^2)$ if $f$, except at $x=0$, where the former is zero and the latter is not defined.
Now if $f$ were integrable, then $F$ would be absolutely continuous.
We show that this is not the case.
For $j=1,2,dots$ let $a_j= frac{1}{sqrt{pi j}}$ and $b_j = frac{1}{sqrt{pi( j + 1/2)}}$.
Then $F(a_j)=0$ and $F(b_j)=pi (j+1/2)$.
To conclude, for any $j_1$:
$sum_{j>j_1}| a_j - b_j| < a_{j_1}to 0$ as $j_1toinfty$, while
$sum_{j>j_1} |F(a_j)-F(b_j)| = sum_{j>j_1} pi (j+1/2)= infty$.
answered Nov 22 '18 at 2:11
Fnacool
4,976511
answered Nov 22 '18 at 2:11
Fnacool
4,976511
answered Nov 22 '18 at 2:11
Fnacool
4,976511
answered Nov 22 '18 at 2:11
Fnacool
4,976511
4,976511
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3007873%2ffunction-is-not-lebesgue-integrable%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
It's worth commenting that this thing integrates to $x^2 sin x^{-2}$, and is Riemann integrable ...
– Richard Martin
Nov 21 '18 at 15:43
The problem with the Lebesgue integral is the wild oscillation of $f'$ near zero.
– Richard Martin
Nov 21 '18 at 15:45