Mixing
Is the domain of convergence of a sequence of measurable functions measurable? (general targets)
Let $(X,Sigma)$ be a measurable space, and let $Y$ be a topological space.
Let $f_n:X to Y$ be a measurable sequence, when we take on $Y$ the Borel sigma algebra. In other words the $f_n$ are measurable as mappings between the measurable spaces
$$ (X,Sigma) to (Y,B(Y)).$$
Is $E= { xin X : f_n(x) text{ is convergent in } Y }$ measurable?
This is well-known (and easy) in the case of real-valued functions. Indeed,
$$ E=bigcap_{kinmathbb{N}}bigcup_{ninmathbb{N}}bigcap_{r,s >n}left{ xin X:left|f_{r}left(xright)-f_{s}left(xright)right|<frac{1}{k}right}. $$
I think this proof could be adapted to the case where $Y$ is a complete metric space; just replace $$ left|f_{r}left(xright)-f_{s}left(xright)right|$$ with
$$ dleft(f_{r}(x),f_{s}(x)right) .$$
Does this hold for non-complete metric spaces as targets? What about when $Y$ is not a metric space?
measure-theory measurable-functions
edited Nov 21 '18 at 14:58
A. Pongrácz
5,5151827
asked Nov 21 '18 at 14:48
Asaf Shachar
5,1393941
add a comment |
Let $(X,Sigma)$ be a measurable space, and let $Y$ be a topological space.
Let $f_n:X to Y$ be a measurable sequence, when we take on $Y$ the Borel sigma algebra. In other words the $f_n$ are measurable as mappings between the measurable spaces
$$ (X,Sigma) to (Y,B(Y)).$$
Is $E= { xin X : f_n(x) text{ is convergent in } Y }$ measurable?
This is well-known (and easy) in the case of real-valued functions. Indeed,
$$ E=bigcap_{kinmathbb{N}}bigcup_{ninmathbb{N}}bigcap_{r,s >n}left{ xin X:left|f_{r}left(xright)-f_{s}left(xright)right|<frac{1}{k}right}. $$
I think this proof could be adapted to the case where $Y$ is a complete metric space; just replace $$ left|f_{r}left(xright)-f_{s}left(xright)right|$$ with
$$ dleft(f_{r}(x),f_{s}(x)right) .$$
Does this hold for non-complete metric spaces as targets? What about when $Y$ is not a metric space?
measure-theory measurable-functions
edited Nov 21 '18 at 14:58
A. Pongrácz
5,5151827
asked Nov 21 '18 at 14:48
Asaf Shachar
5,1393941
add a comment |
Let $(X,Sigma)$ be a measurable space, and let $Y$ be a topological space.
Let $f_n:X to Y$ be a measurable sequence, when we take on $Y$ the Borel sigma algebra. In other words the $f_n$ are measurable as mappings between the measurable spaces
$$ (X,Sigma) to (Y,B(Y)).$$
Is $E= { xin X : f_n(x) text{ is convergent in } Y }$ measurable?
This is well-known (and easy) in the case of real-valued functions. Indeed,
$$ E=bigcap_{kinmathbb{N}}bigcup_{ninmathbb{N}}bigcap_{r,s >n}left{ xin X:left|f_{r}left(xright)-f_{s}left(xright)right|<frac{1}{k}right}. $$
I think this proof could be adapted to the case where $Y$ is a complete metric space; just replace $$ left|f_{r}left(xright)-f_{s}left(xright)right|$$ with
$$ dleft(f_{r}(x),f_{s}(x)right) .$$
Does this hold for non-complete metric spaces as targets? What about when $Y$ is not a metric space?
measure-theory measurable-functions
edited Nov 21 '18 at 14:58
A. Pongrácz
5,5151827
asked Nov 21 '18 at 14:48
Asaf Shachar
5,1393941
Let $(X,Sigma)$ be a measurable space, and let $Y$ be a topological space.
Let $f_n:X to Y$ be a measurable sequence, when we take on $Y$ the Borel sigma algebra. In other words the $f_n$ are measurable as mappings between the measurable spaces
$$ (X,Sigma) to (Y,B(Y)).$$
Is $E= { xin X : f_n(x) text{ is convergent in } Y }$ measurable?
This is well-known (and easy) in the case of real-valued functions. Indeed,
$$ E=bigcap_{kinmathbb{N}}bigcup_{ninmathbb{N}}bigcap_{r,s >n}left{ xin X:left|f_{r}left(xright)-f_{s}left(xright)right|<frac{1}{k}right}. $$
I think this proof could be adapted to the case where $Y$ is a complete metric space; just replace $$ left|f_{r}left(xright)-f_{s}left(xright)right|$$ with
$$ dleft(f_{r}(x),f_{s}(x)right) .$$
Does this hold for non-complete metric spaces as targets? What about when $Y$ is not a metric space?
measure-theory measurable-functions
measure-theory measurable-functions
edited Nov 21 '18 at 14:58
A. Pongrácz
5,5151827
asked Nov 21 '18 at 14:48
Asaf Shachar
5,1393941
edited Nov 21 '18 at 14:58
A. Pongrácz
5,5151827
asked Nov 21 '18 at 14:48
Asaf Shachar
5,1393941
edited Nov 21 '18 at 14:58
A. Pongrácz
5,5151827
edited Nov 21 '18 at 14:58
A. Pongrácz
5,5151827
edited Nov 21 '18 at 14:58
A. Pongrácz
5,5151827
5,5151827
asked Nov 21 '18 at 14:48
Asaf Shachar
5,1393941
asked Nov 21 '18 at 14:48
Asaf Shachar
5,1393941
asked Nov 21 '18 at 14:48
Asaf Shachar
5,1393941
5,1393941
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
This is a really instructive question: apparently, completeness is the key!
First, let $X=Y=[0,1]$ with the usual topology. We define an example for a sequence of functions $f_n: Xrightarrow Y$, which converges everywhere point-wise. Then we ruin $Y$.
Let $f_1$ be constant $0$ on $[0,1/2[$ and constant $1/2$ on $[1/2,1]$.
Let $f_2$ be constant $0$ on $[0,1/4[$, constant $1/4$ on $[1/4,1/2[$, constant $1/2$ on $[1/2,3/4[$, and constant $3/4$ on $[3/4,1]$.
In general, divide $[0,1]$ into $2^n$ intervals of equal length (closed from the left and open from the right, except for the last one which is closed), and let $f_n$ be the step function whose value on each interval is the left endpoint.
It is clear that this sequence of functions converges point-wise to the identity function on $[0,1]$.
We keep these functions, but modify $Y$ now. Pick any non-measurable set $Ssubseteq [0,1]$, and let $D$ consist of those rational numbers in $[0,1]$ whose denominator is a power of $2$.
Let $Y$ be the topological subspace of $[0,1]$ induced by $Scup D$. This is of course a metric space (as it is a subspace of a metric space), but it is no longer complete!
Define the same sequence of functions as above; they are still measurable (and also functions $Xrightarrow Y$, because we were careful enough to put all elements of $D$ in $Y$).
However, the domain of convergence is $Y= Scup D$: a sequence in $Y$ is convergent if and only if it is convergent as a sequence in $[0,1]$ with limit in $Y$.
As $S$ is not measurable and $D$ is countable, we have that $Y$ is not measurable (as a subset of $X$).
answered Nov 21 '18 at 20:51
A. Pongrácz
5,5151827
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%2f3007811%2fis-the-domain-of-convergence-of-a-sequence-of-measurable-functions-measurable%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
This is a really instructive question: apparently, completeness is the key!
First, let $X=Y=[0,1]$ with the usual topology. We define an example for a sequence of functions $f_n: Xrightarrow Y$, which converges everywhere point-wise. Then we ruin $Y$.
Let $f_1$ be constant $0$ on $[0,1/2[$ and constant $1/2$ on $[1/2,1]$.
Let $f_2$ be constant $0$ on $[0,1/4[$, constant $1/4$ on $[1/4,1/2[$, constant $1/2$ on $[1/2,3/4[$, and constant $3/4$ on $[3/4,1]$.
In general, divide $[0,1]$ into $2^n$ intervals of equal length (closed from the left and open from the right, except for the last one which is closed), and let $f_n$ be the step function whose value on each interval is the left endpoint.
It is clear that this sequence of functions converges point-wise to the identity function on $[0,1]$.
We keep these functions, but modify $Y$ now. Pick any non-measurable set $Ssubseteq [0,1]$, and let $D$ consist of those rational numbers in $[0,1]$ whose denominator is a power of $2$.
Let $Y$ be the topological subspace of $[0,1]$ induced by $Scup D$. This is of course a metric space (as it is a subspace of a metric space), but it is no longer complete!
Define the same sequence of functions as above; they are still measurable (and also functions $Xrightarrow Y$, because we were careful enough to put all elements of $D$ in $Y$).
However, the domain of convergence is $Y= Scup D$: a sequence in $Y$ is convergent if and only if it is convergent as a sequence in $[0,1]$ with limit in $Y$.
As $S$ is not measurable and $D$ is countable, we have that $Y$ is not measurable (as a subset of $X$).
answered Nov 21 '18 at 20:51
A. Pongrácz
5,5151827
add a comment |
This is a really instructive question: apparently, completeness is the key!
First, let $X=Y=[0,1]$ with the usual topology. We define an example for a sequence of functions $f_n: Xrightarrow Y$, which converges everywhere point-wise. Then we ruin $Y$.
Let $f_1$ be constant $0$ on $[0,1/2[$ and constant $1/2$ on $[1/2,1]$.
Let $f_2$ be constant $0$ on $[0,1/4[$, constant $1/4$ on $[1/4,1/2[$, constant $1/2$ on $[1/2,3/4[$, and constant $3/4$ on $[3/4,1]$.
In general, divide $[0,1]$ into $2^n$ intervals of equal length (closed from the left and open from the right, except for the last one which is closed), and let $f_n$ be the step function whose value on each interval is the left endpoint.
It is clear that this sequence of functions converges point-wise to the identity function on $[0,1]$.
We keep these functions, but modify $Y$ now. Pick any non-measurable set $Ssubseteq [0,1]$, and let $D$ consist of those rational numbers in $[0,1]$ whose denominator is a power of $2$.
Let $Y$ be the topological subspace of $[0,1]$ induced by $Scup D$. This is of course a metric space (as it is a subspace of a metric space), but it is no longer complete!
Define the same sequence of functions as above; they are still measurable (and also functions $Xrightarrow Y$, because we were careful enough to put all elements of $D$ in $Y$).
However, the domain of convergence is $Y= Scup D$: a sequence in $Y$ is convergent if and only if it is convergent as a sequence in $[0,1]$ with limit in $Y$.
As $S$ is not measurable and $D$ is countable, we have that $Y$ is not measurable (as a subset of $X$).
answered Nov 21 '18 at 20:51
A. Pongrácz
5,5151827
add a comment |
This is a really instructive question: apparently, completeness is the key!
First, let $X=Y=[0,1]$ with the usual topology. We define an example for a sequence of functions $f_n: Xrightarrow Y$, which converges everywhere point-wise. Then we ruin $Y$.
Let $f_1$ be constant $0$ on $[0,1/2[$ and constant $1/2$ on $[1/2,1]$.
Let $f_2$ be constant $0$ on $[0,1/4[$, constant $1/4$ on $[1/4,1/2[$, constant $1/2$ on $[1/2,3/4[$, and constant $3/4$ on $[3/4,1]$.
In general, divide $[0,1]$ into $2^n$ intervals of equal length (closed from the left and open from the right, except for the last one which is closed), and let $f_n$ be the step function whose value on each interval is the left endpoint.
It is clear that this sequence of functions converges point-wise to the identity function on $[0,1]$.
We keep these functions, but modify $Y$ now. Pick any non-measurable set $Ssubseteq [0,1]$, and let $D$ consist of those rational numbers in $[0,1]$ whose denominator is a power of $2$.
Let $Y$ be the topological subspace of $[0,1]$ induced by $Scup D$. This is of course a metric space (as it is a subspace of a metric space), but it is no longer complete!
Define the same sequence of functions as above; they are still measurable (and also functions $Xrightarrow Y$, because we were careful enough to put all elements of $D$ in $Y$).
However, the domain of convergence is $Y= Scup D$: a sequence in $Y$ is convergent if and only if it is convergent as a sequence in $[0,1]$ with limit in $Y$.
As $S$ is not measurable and $D$ is countable, we have that $Y$ is not measurable (as a subset of $X$).
answered Nov 21 '18 at 20:51
A. Pongrácz
5,5151827
This is a really instructive question: apparently, completeness is the key!
First, let $X=Y=[0,1]$ with the usual topology. We define an example for a sequence of functions $f_n: Xrightarrow Y$, which converges everywhere point-wise. Then we ruin $Y$.
Let $f_1$ be constant $0$ on $[0,1/2[$ and constant $1/2$ on $[1/2,1]$.
Let $f_2$ be constant $0$ on $[0,1/4[$, constant $1/4$ on $[1/4,1/2[$, constant $1/2$ on $[1/2,3/4[$, and constant $3/4$ on $[3/4,1]$.
In general, divide $[0,1]$ into $2^n$ intervals of equal length (closed from the left and open from the right, except for the last one which is closed), and let $f_n$ be the step function whose value on each interval is the left endpoint.
It is clear that this sequence of functions converges point-wise to the identity function on $[0,1]$.
We keep these functions, but modify $Y$ now. Pick any non-measurable set $Ssubseteq [0,1]$, and let $D$ consist of those rational numbers in $[0,1]$ whose denominator is a power of $2$.
Let $Y$ be the topological subspace of $[0,1]$ induced by $Scup D$. This is of course a metric space (as it is a subspace of a metric space), but it is no longer complete!
Define the same sequence of functions as above; they are still measurable (and also functions $Xrightarrow Y$, because we were careful enough to put all elements of $D$ in $Y$).
However, the domain of convergence is $Y= Scup D$: a sequence in $Y$ is convergent if and only if it is convergent as a sequence in $[0,1]$ with limit in $Y$.
As $S$ is not measurable and $D$ is countable, we have that $Y$ is not measurable (as a subset of $X$).
answered Nov 21 '18 at 20:51
A. Pongrácz
5,5151827
edited Nov 21 '18 at 21:06
answered Nov 21 '18 at 20:51
A. Pongrácz
5,5151827
answered Nov 21 '18 at 20:51
A. Pongrácz
5,5151827
answered Nov 21 '18 at 20:51
A. Pongrácz
5,5151827
5,5151827
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%2f3007811%2fis-the-domain-of-convergence-of-a-sequence-of-measurable-functions-measurable%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
