Mixing
Is the lebesgue integral of a measurable function continuous?
$begingroup$
I was wondering if the lebesgue integral of a measurable function at least continuous? What kind of regularity on the integrand do we need for it to be absolutely continuous so that we can say its differential almost everywhere?
reference-request continuity lebesgue-integral absolute-continuity
asked Jan 19 at 17:32
user3503589user3503589
1,2841821
$endgroup$
add a comment |
$begingroup$
I was wondering if the lebesgue integral of a measurable function at least continuous? What kind of regularity on the integrand do we need for it to be absolutely continuous so that we can say its differential almost everywhere?
reference-request continuity lebesgue-integral absolute-continuity
asked Jan 19 at 17:32
user3503589user3503589
1,2841821
$endgroup$
add a comment |
$begingroup$
I was wondering if the lebesgue integral of a measurable function at least continuous? What kind of regularity on the integrand do we need for it to be absolutely continuous so that we can say its differential almost everywhere?
reference-request continuity lebesgue-integral absolute-continuity
asked Jan 19 at 17:32
user3503589user3503589
1,2841821
$endgroup$
I was wondering if the lebesgue integral of a measurable function at least continuous? What kind of regularity on the integrand do we need for it to be absolutely continuous so that we can say its differential almost everywhere?
reference-request continuity lebesgue-integral absolute-continuity
reference-request continuity lebesgue-integral absolute-continuity
asked Jan 19 at 17:32
user3503589user3503589
1,2841821
asked Jan 19 at 17:32
user3503589user3503589
1,2841821
asked Jan 19 at 17:32
user3503589user3503589
1,2841821
asked Jan 19 at 17:32
user3503589user3503589
1,2841821
asked Jan 19 at 17:32
user3503589user3503589
1,2841821
1,2841821
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The integral of an integrable function is a single number, so asking whether it's continuous makes little sense. Presumably you're asking about the continuity of $$F(x)=int_{-infty}^x f(t),dt$$or maybe $int_0^x$.
If $f$ is integrable then $F$ is absolutely continuous; this is one of the main theorems in this context. If $f$ is just locally integrable then it follows that $F(x)=int_0^x f$ defines a locally absolutely continuous function.
answered Jan 19 at 18:18
David C. UllrichDavid C. Ullrich
61k43994
$endgroup$
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%2f3079599%2fis-the-lebesgue-integral-of-a-measurable-function-continuous%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The integral of an integrable function is a single number, so asking whether it's continuous makes little sense. Presumably you're asking about the continuity of $$F(x)=int_{-infty}^x f(t),dt$$or maybe $int_0^x$.
If $f$ is integrable then $F$ is absolutely continuous; this is one of the main theorems in this context. If $f$ is just locally integrable then it follows that $F(x)=int_0^x f$ defines a locally absolutely continuous function.
answered Jan 19 at 18:18
David C. UllrichDavid C. Ullrich
61k43994
$endgroup$
add a comment |
$begingroup$
The integral of an integrable function is a single number, so asking whether it's continuous makes little sense. Presumably you're asking about the continuity of $$F(x)=int_{-infty}^x f(t),dt$$or maybe $int_0^x$.
If $f$ is integrable then $F$ is absolutely continuous; this is one of the main theorems in this context. If $f$ is just locally integrable then it follows that $F(x)=int_0^x f$ defines a locally absolutely continuous function.
answered Jan 19 at 18:18
David C. UllrichDavid C. Ullrich
61k43994
$endgroup$
add a comment |
$begingroup$
The integral of an integrable function is a single number, so asking whether it's continuous makes little sense. Presumably you're asking about the continuity of $$F(x)=int_{-infty}^x f(t),dt$$or maybe $int_0^x$.
If $f$ is integrable then $F$ is absolutely continuous; this is one of the main theorems in this context. If $f$ is just locally integrable then it follows that $F(x)=int_0^x f$ defines a locally absolutely continuous function.
answered Jan 19 at 18:18
David C. UllrichDavid C. Ullrich
61k43994
$endgroup$
The integral of an integrable function is a single number, so asking whether it's continuous makes little sense. Presumably you're asking about the continuity of $$F(x)=int_{-infty}^x f(t),dt$$or maybe $int_0^x$.
If $f$ is integrable then $F$ is absolutely continuous; this is one of the main theorems in this context. If $f$ is just locally integrable then it follows that $F(x)=int_0^x f$ defines a locally absolutely continuous function.
answered Jan 19 at 18:18
David C. UllrichDavid C. Ullrich
61k43994
answered Jan 19 at 18:18
David C. UllrichDavid C. Ullrich
61k43994
answered Jan 19 at 18:18
David C. UllrichDavid C. Ullrich
61k43994
answered Jan 19 at 18:18
David C. UllrichDavid C. Ullrich
61k43994
61k43994
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%2f3079599%2fis-the-lebesgue-integral-of-a-measurable-function-continuous%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
