Egorov's theorem on intervals
up vote
2
down vote
favorite
Frm Folland's Real Analysis exercise 2.43 part (b):
Suppose that $(X, mathcal M, mu)$ is a measure space, with $mu(X) < infty$. Let $f: X × [0,1] to mathbb C$ is a function such that $f(cdot, y)$ is measurable for each $y in [0,1]$ and $f(x, cdot)$ is continuous for each $x in X$. Prove that: for any $epsilon > 0$ there is $E subset X$ such that $mu(E) < epsilon$ and $f(cdot, y) to f(cdot, 0)$ uniformly on $E^c$ as $y to 0$.
Thus is actually part (b) of the question; I managed to prove part (a) which is if $0 < epsilon, delta < 1$ then $E_{epsilon, delta} = {x : |f(x, y) - f(x, 0)| leq epsilon text{ for all } y < delta }$ is measurable. Again, I already proved this; no need for you to prove this.
I am not sure how to prove the above (part b) question. It seems like I habe to use Egoroff's theorem somehow, but we have to do something more to it, since we ae supposed to show uniform continuity not on sequences but in interval (this is very loosely stated, but I think you know what I mean. If not, never mind).
real-analysis measure-theory uniform-convergence
asked Nov 24 at 6:06
Cute Brownie
990416
add a comment |
up vote
2
down vote
favorite
Frm Folland's Real Analysis exercise 2.43 part (b):
Suppose that $(X, mathcal M, mu)$ is a measure space, with $mu(X) < infty$. Let $f: X × [0,1] to mathbb C$ is a function such that $f(cdot, y)$ is measurable for each $y in [0,1]$ and $f(x, cdot)$ is continuous for each $x in X$. Prove that: for any $epsilon > 0$ there is $E subset X$ such that $mu(E) < epsilon$ and $f(cdot, y) to f(cdot, 0)$ uniformly on $E^c$ as $y to 0$.
Thus is actually part (b) of the question; I managed to prove part (a) which is if $0 < epsilon, delta < 1$ then $E_{epsilon, delta} = {x : |f(x, y) - f(x, 0)| leq epsilon text{ for all } y < delta }$ is measurable. Again, I already proved this; no need for you to prove this.
I am not sure how to prove the above (part b) question. It seems like I habe to use Egoroff's theorem somehow, but we have to do something more to it, since we ae supposed to show uniform continuity not on sequences but in interval (this is very loosely stated, but I think you know what I mean. If not, never mind).
real-analysis measure-theory uniform-convergence
asked Nov 24 at 6:06
Cute Brownie
990416
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Frm Folland's Real Analysis exercise 2.43 part (b):
Suppose that $(X, mathcal M, mu)$ is a measure space, with $mu(X) < infty$. Let $f: X × [0,1] to mathbb C$ is a function such that $f(cdot, y)$ is measurable for each $y in [0,1]$ and $f(x, cdot)$ is continuous for each $x in X$. Prove that: for any $epsilon > 0$ there is $E subset X$ such that $mu(E) < epsilon$ and $f(cdot, y) to f(cdot, 0)$ uniformly on $E^c$ as $y to 0$.
Thus is actually part (b) of the question; I managed to prove part (a) which is if $0 < epsilon, delta < 1$ then $E_{epsilon, delta} = {x : |f(x, y) - f(x, 0)| leq epsilon text{ for all } y < delta }$ is measurable. Again, I already proved this; no need for you to prove this.
I am not sure how to prove the above (part b) question. It seems like I habe to use Egoroff's theorem somehow, but we have to do something more to it, since we ae supposed to show uniform continuity not on sequences but in interval (this is very loosely stated, but I think you know what I mean. If not, never mind).
real-analysis measure-theory uniform-convergence
asked Nov 24 at 6:06
Cute Brownie
990416
Frm Folland's Real Analysis exercise 2.43 part (b):
Suppose that $(X, mathcal M, mu)$ is a measure space, with $mu(X) < infty$. Let $f: X × [0,1] to mathbb C$ is a function such that $f(cdot, y)$ is measurable for each $y in [0,1]$ and $f(x, cdot)$ is continuous for each $x in X$. Prove that: for any $epsilon > 0$ there is $E subset X$ such that $mu(E) < epsilon$ and $f(cdot, y) to f(cdot, 0)$ uniformly on $E^c$ as $y to 0$.
Thus is actually part (b) of the question; I managed to prove part (a) which is if $0 < epsilon, delta < 1$ then $E_{epsilon, delta} = {x : |f(x, y) - f(x, 0)| leq epsilon text{ for all } y < delta }$ is measurable. Again, I already proved this; no need for you to prove this.
I am not sure how to prove the above (part b) question. It seems like I habe to use Egoroff's theorem somehow, but we have to do something more to it, since we ae supposed to show uniform continuity not on sequences but in interval (this is very loosely stated, but I think you know what I mean. If not, never mind).
real-analysis measure-theory uniform-convergence
real-analysis measure-theory uniform-convergence
asked Nov 24 at 6:06
Cute Brownie
990416
asked Nov 24 at 6:06
Cute Brownie
990416
edited Nov 24 at 6:32
asked Nov 24 at 6:06
Cute Brownie
990416
asked Nov 24 at 6:06
Cute Brownie
990416
asked Nov 24 at 6:06
Cute Brownie
990416
990416
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Since Egoroff's theorem doesn't quite fit but seems related, try taking a step back to Egoroff's theorem's proof. Notice that
begin{align*}
bigcap_{nin mathbb{N}} (E_{1/m,1/ n})^c = emptyset quad forall min mathbb{N}.
end{align*}
Let $A_{m,k}=bigcap_{nleq k} (E_{1/m,1/ n})^c$. For each $min mathbb{N}$, by upper continuity, there is some $kin mathbb{N} $ for which
begin{align*}
mu(A_{m,k})<frac{epsilon}{2^m}.
end{align*}
Thus,
begin{align}
E:=bigcup_{min mathbb{N}} A_{m,k}
end{align}
has measure less than $epsilon.$ By writing out the definition of $E^c$, we should see that the convergence is uniform on $E^c$.
answered Nov 24 at 6:28
confused_wallet
473311
Yea, this solution is correct. +1
– mathworker21
Nov 24 at 6:50
Thanks, but why is $bigcap (E_{1/m, 1/n})^c = emptyset$?
– Cute Brownie
Nov 24 at 7:03
Since $f(x,cdot)$ is continuous, then $f(x,y)rightarrow f(x,0)$ as $yrightarrow 0$.
– confused_wallet
Nov 24 at 7:05
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%2f3011247%2fegorovs-theorem-on-intervals%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
up vote
2
down vote
accepted
Since Egoroff's theorem doesn't quite fit but seems related, try taking a step back to Egoroff's theorem's proof. Notice that
begin{align*}
bigcap_{nin mathbb{N}} (E_{1/m,1/ n})^c = emptyset quad forall min mathbb{N}.
end{align*}
Let $A_{m,k}=bigcap_{nleq k} (E_{1/m,1/ n})^c$. For each $min mathbb{N}$, by upper continuity, there is some $kin mathbb{N} $ for which
begin{align*}
mu(A_{m,k})<frac{epsilon}{2^m}.
end{align*}
Thus,
begin{align}
E:=bigcup_{min mathbb{N}} A_{m,k}
end{align}
has measure less than $epsilon.$ By writing out the definition of $E^c$, we should see that the convergence is uniform on $E^c$.
answered Nov 24 at 6:28
confused_wallet
473311
Yea, this solution is correct. +1
– mathworker21
Nov 24 at 6:50
Thanks, but why is $bigcap (E_{1/m, 1/n})^c = emptyset$?
– Cute Brownie
Nov 24 at 7:03
Since $f(x,cdot)$ is continuous, then $f(x,y)rightarrow f(x,0)$ as $yrightarrow 0$.
– confused_wallet
Nov 24 at 7:05
add a comment |
up vote
2
down vote
accepted
Since Egoroff's theorem doesn't quite fit but seems related, try taking a step back to Egoroff's theorem's proof. Notice that
begin{align*}
bigcap_{nin mathbb{N}} (E_{1/m,1/ n})^c = emptyset quad forall min mathbb{N}.
end{align*}
Let $A_{m,k}=bigcap_{nleq k} (E_{1/m,1/ n})^c$. For each $min mathbb{N}$, by upper continuity, there is some $kin mathbb{N} $ for which
begin{align*}
mu(A_{m,k})<frac{epsilon}{2^m}.
end{align*}
Thus,
begin{align}
E:=bigcup_{min mathbb{N}} A_{m,k}
end{align}
has measure less than $epsilon.$ By writing out the definition of $E^c$, we should see that the convergence is uniform on $E^c$.
answered Nov 24 at 6:28
confused_wallet
473311
Yea, this solution is correct. +1
– mathworker21
Nov 24 at 6:50
Thanks, but why is $bigcap (E_{1/m, 1/n})^c = emptyset$?
– Cute Brownie
Nov 24 at 7:03
Since $f(x,cdot)$ is continuous, then $f(x,y)rightarrow f(x,0)$ as $yrightarrow 0$.
– confused_wallet
Nov 24 at 7:05
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Since Egoroff's theorem doesn't quite fit but seems related, try taking a step back to Egoroff's theorem's proof. Notice that
begin{align*}
bigcap_{nin mathbb{N}} (E_{1/m,1/ n})^c = emptyset quad forall min mathbb{N}.
end{align*}
Let $A_{m,k}=bigcap_{nleq k} (E_{1/m,1/ n})^c$. For each $min mathbb{N}$, by upper continuity, there is some $kin mathbb{N} $ for which
begin{align*}
mu(A_{m,k})<frac{epsilon}{2^m}.
end{align*}
Thus,
begin{align}
E:=bigcup_{min mathbb{N}} A_{m,k}
end{align}
has measure less than $epsilon.$ By writing out the definition of $E^c$, we should see that the convergence is uniform on $E^c$.
answered Nov 24 at 6:28
confused_wallet
473311
Since Egoroff's theorem doesn't quite fit but seems related, try taking a step back to Egoroff's theorem's proof. Notice that
begin{align*}
bigcap_{nin mathbb{N}} (E_{1/m,1/ n})^c = emptyset quad forall min mathbb{N}.
end{align*}
Let $A_{m,k}=bigcap_{nleq k} (E_{1/m,1/ n})^c$. For each $min mathbb{N}$, by upper continuity, there is some $kin mathbb{N} $ for which
begin{align*}
mu(A_{m,k})<frac{epsilon}{2^m}.
end{align*}
Thus,
begin{align}
E:=bigcup_{min mathbb{N}} A_{m,k}
end{align}
has measure less than $epsilon.$ By writing out the definition of $E^c$, we should see that the convergence is uniform on $E^c$.
answered Nov 24 at 6:28
confused_wallet
473311
edited Nov 24 at 6:57
answered Nov 24 at 6:28
confused_wallet
473311
answered Nov 24 at 6:28
confused_wallet
473311
answered Nov 24 at 6:28
confused_wallet
473311
473311
Yea, this solution is correct. +1
– mathworker21
Nov 24 at 6:50
Thanks, but why is $bigcap (E_{1/m, 1/n})^c = emptyset$?
– Cute Brownie
Nov 24 at 7:03
Since $f(x,cdot)$ is continuous, then $f(x,y)rightarrow f(x,0)$ as $yrightarrow 0$.
– confused_wallet
Nov 24 at 7:05
add a comment |
Yea, this solution is correct. +1
– mathworker21
Nov 24 at 6:50
Thanks, but why is $bigcap (E_{1/m, 1/n})^c = emptyset$?
– Cute Brownie
Nov 24 at 7:03
Since $f(x,cdot)$ is continuous, then $f(x,y)rightarrow f(x,0)$ as $yrightarrow 0$.
– confused_wallet
Nov 24 at 7:05
Yea, this solution is correct. +1
– mathworker21
Nov 24 at 6:50
Yea, this solution is correct. +1
– mathworker21
Nov 24 at 6:50
Thanks, but why is $bigcap (E_{1/m, 1/n})^c = emptyset$?
– Cute Brownie
Nov 24 at 7:03
Thanks, but why is $bigcap (E_{1/m, 1/n})^c = emptyset$?
– Cute Brownie
Nov 24 at 7:03
Since $f(x,cdot)$ is continuous, then $f(x,y)rightarrow f(x,0)$ as $yrightarrow 0$.
– confused_wallet
Nov 24 at 7:05
Since $f(x,cdot)$ is continuous, then $f(x,y)rightarrow f(x,0)$ as $yrightarrow 0$.
– confused_wallet
Nov 24 at 7:05
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%2f3011247%2fegorovs-theorem-on-intervals%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
