Are Cat. II sets measurable (for a given outer measure)
$begingroup$
Given a complete metric space $X$. Let $mu^*(E)=0$ if $E$ is of Cat. I, $mu^*(E)=1$ if $E$ is of Cat. II.
I have proved that $mu^*$ is an outer measure.
What are the $mu^*$-measurable sets?
I've proved that all Cat. I sets are measurable. But I found no way to prove/disprove that the Cat. II sets are measurable.
Proof of All Cat. I sets are measurable:
Suppose $A$ is Cat. I, then the Carathéodory condition becomes $mu^*(S)gemu^*(Scap A)+mu^*(Scap A^C)=0+mu^*(Scap A^C)$ which is always true (subsets of Cat I are Cat I sets), thus Cat. I sets are measurable.
general-topology outer-measure
asked Dec 11 '18 at 9:56
PeptideChainPeptideChain
464311
$endgroup$
add a comment |
$begingroup$
Given a complete metric space $X$. Let $mu^*(E)=0$ if $E$ is of Cat. I, $mu^*(E)=1$ if $E$ is of Cat. II.
I have proved that $mu^*$ is an outer measure.
What are the $mu^*$-measurable sets?
I've proved that all Cat. I sets are measurable. But I found no way to prove/disprove that the Cat. II sets are measurable.
Proof of All Cat. I sets are measurable:
Suppose $A$ is Cat. I, then the Carathéodory condition becomes $mu^*(S)gemu^*(Scap A)+mu^*(Scap A^C)=0+mu^*(Scap A^C)$ which is always true (subsets of Cat I are Cat I sets), thus Cat. I sets are measurable.
general-topology outer-measure
asked Dec 11 '18 at 9:56
PeptideChainPeptideChain
464311
$endgroup$
$begingroup$
Are open and dense subsets measurable? Maybe also post your proof that first category subsets are, too.
$endgroup$
– Henno Brandsma
Dec 11 '18 at 22:17
add a comment |
$begingroup$
Given a complete metric space $X$. Let $mu^*(E)=0$ if $E$ is of Cat. I, $mu^*(E)=1$ if $E$ is of Cat. II.
I have proved that $mu^*$ is an outer measure.
What are the $mu^*$-measurable sets?
I've proved that all Cat. I sets are measurable. But I found no way to prove/disprove that the Cat. II sets are measurable.
Proof of All Cat. I sets are measurable:
Suppose $A$ is Cat. I, then the Carathéodory condition becomes $mu^*(S)gemu^*(Scap A)+mu^*(Scap A^C)=0+mu^*(Scap A^C)$ which is always true (subsets of Cat I are Cat I sets), thus Cat. I sets are measurable.
general-topology outer-measure
asked Dec 11 '18 at 9:56
PeptideChainPeptideChain
464311
$endgroup$
Given a complete metric space $X$. Let $mu^*(E)=0$ if $E$ is of Cat. I, $mu^*(E)=1$ if $E$ is of Cat. II.
I have proved that $mu^*$ is an outer measure.
What are the $mu^*$-measurable sets?
I've proved that all Cat. I sets are measurable. But I found no way to prove/disprove that the Cat. II sets are measurable.
Proof of All Cat. I sets are measurable:
Suppose $A$ is Cat. I, then the Carathéodory condition becomes $mu^*(S)gemu^*(Scap A)+mu^*(Scap A^C)=0+mu^*(Scap A^C)$ which is always true (subsets of Cat I are Cat I sets), thus Cat. I sets are measurable.
general-topology outer-measure
general-topology outer-measure
asked Dec 11 '18 at 9:56
PeptideChainPeptideChain
464311
asked Dec 11 '18 at 9:56
PeptideChainPeptideChain
464311
edited Dec 12 '18 at 7:11
PeptideChain
asked Dec 11 '18 at 9:56
PeptideChainPeptideChain
464311
asked Dec 11 '18 at 9:56
PeptideChainPeptideChain
464311
asked Dec 11 '18 at 9:56
PeptideChainPeptideChain
464311
464311
$begingroup$
Are open and dense subsets measurable? Maybe also post your proof that first category subsets are, too.
$endgroup$
– Henno Brandsma
Dec 11 '18 at 22:17
add a comment |
$begingroup$
Are open and dense subsets measurable? Maybe also post your proof that first category subsets are, too.
$endgroup$
– Henno Brandsma
Dec 11 '18 at 22:17
$begingroup$
Are open and dense subsets measurable? Maybe also post your proof that first category subsets are, too.
$endgroup$
– Henno Brandsma
Dec 11 '18 at 22:17
$begingroup$
Are open and dense subsets measurable? Maybe also post your proof that first category subsets are, too.
$endgroup$
– Henno Brandsma
Dec 11 '18 at 22:17
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There are 4 types of sets in a general space $X$ (a set is first category ($C^I$) or if not it's second category ($C^{II}$) :
$A$ $C^I$ and $A^complement$ $C^I$.
$A$ $C^I$ and $A^complement$ $C^{II}$.
$A$ $C^{II}$ and $A^complement$ $C^I$.
$A$ $C^{II}$ and $A^complement$ $C^{II}$.
In a complete space type 1 does not exist, as this would imply $X$ is first category.
You've already shown that type 2 is measurable.
Type 3. is exactly the complement of sets of type 2, so also measurable.
(also easy to show directly, by the same argument as for 2).
Type 4. is not measurable as we can take $S=X$ and get a contradiction $1ge 2$ in the Carathéodory condition.
So the measurable sets are those such that $A$ or its complement is first category (but not both as that is impossible in this situation).
answered Dec 12 '18 at 22:49
Henno BrandsmaHenno Brandsma
110k348118
$endgroup$
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%2f3035129%2fare-cat-ii-sets-measurable-for-a-given-outer-measure%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
$begingroup$
There are 4 types of sets in a general space $X$ (a set is first category ($C^I$) or if not it's second category ($C^{II}$) :
$A$ $C^I$ and $A^complement$ $C^I$.
$A$ $C^I$ and $A^complement$ $C^{II}$.
$A$ $C^{II}$ and $A^complement$ $C^I$.
$A$ $C^{II}$ and $A^complement$ $C^{II}$.
In a complete space type 1 does not exist, as this would imply $X$ is first category.
You've already shown that type 2 is measurable.
Type 3. is exactly the complement of sets of type 2, so also measurable.
(also easy to show directly, by the same argument as for 2).
Type 4. is not measurable as we can take $S=X$ and get a contradiction $1ge 2$ in the Carathéodory condition.
So the measurable sets are those such that $A$ or its complement is first category (but not both as that is impossible in this situation).
answered Dec 12 '18 at 22:49
Henno BrandsmaHenno Brandsma
110k348118
$endgroup$
add a comment |
$begingroup$
There are 4 types of sets in a general space $X$ (a set is first category ($C^I$) or if not it's second category ($C^{II}$) :
$A$ $C^I$ and $A^complement$ $C^I$.
$A$ $C^I$ and $A^complement$ $C^{II}$.
$A$ $C^{II}$ and $A^complement$ $C^I$.
$A$ $C^{II}$ and $A^complement$ $C^{II}$.
In a complete space type 1 does not exist, as this would imply $X$ is first category.
You've already shown that type 2 is measurable.
Type 3. is exactly the complement of sets of type 2, so also measurable.
(also easy to show directly, by the same argument as for 2).
Type 4. is not measurable as we can take $S=X$ and get a contradiction $1ge 2$ in the Carathéodory condition.
So the measurable sets are those such that $A$ or its complement is first category (but not both as that is impossible in this situation).
answered Dec 12 '18 at 22:49
Henno BrandsmaHenno Brandsma
110k348118
$endgroup$
add a comment |
$begingroup$
There are 4 types of sets in a general space $X$ (a set is first category ($C^I$) or if not it's second category ($C^{II}$) :
$A$ $C^I$ and $A^complement$ $C^I$.
$A$ $C^I$ and $A^complement$ $C^{II}$.
$A$ $C^{II}$ and $A^complement$ $C^I$.
$A$ $C^{II}$ and $A^complement$ $C^{II}$.
In a complete space type 1 does not exist, as this would imply $X$ is first category.
You've already shown that type 2 is measurable.
Type 3. is exactly the complement of sets of type 2, so also measurable.
(also easy to show directly, by the same argument as for 2).
Type 4. is not measurable as we can take $S=X$ and get a contradiction $1ge 2$ in the Carathéodory condition.
So the measurable sets are those such that $A$ or its complement is first category (but not both as that is impossible in this situation).
answered Dec 12 '18 at 22:49
Henno BrandsmaHenno Brandsma
110k348118
$endgroup$
There are 4 types of sets in a general space $X$ (a set is first category ($C^I$) or if not it's second category ($C^{II}$) :
$A$ $C^I$ and $A^complement$ $C^I$.
$A$ $C^I$ and $A^complement$ $C^{II}$.
$A$ $C^{II}$ and $A^complement$ $C^I$.
$A$ $C^{II}$ and $A^complement$ $C^{II}$.
In a complete space type 1 does not exist, as this would imply $X$ is first category.
You've already shown that type 2 is measurable.
Type 3. is exactly the complement of sets of type 2, so also measurable.
(also easy to show directly, by the same argument as for 2).
Type 4. is not measurable as we can take $S=X$ and get a contradiction $1ge 2$ in the Carathéodory condition.
So the measurable sets are those such that $A$ or its complement is first category (but not both as that is impossible in this situation).
answered Dec 12 '18 at 22:49
Henno BrandsmaHenno Brandsma
110k348118
answered Dec 12 '18 at 22:49
Henno BrandsmaHenno Brandsma
110k348118
answered Dec 12 '18 at 22:49
Henno BrandsmaHenno Brandsma
110k348118
answered Dec 12 '18 at 22:49
Henno BrandsmaHenno Brandsma
110k348118
110k348118
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%2f3035129%2fare-cat-ii-sets-measurable-for-a-given-outer-measure%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
$begingroup$
Are open and dense subsets measurable? Maybe also post your proof that first category subsets are, too.
$endgroup$
– Henno Brandsma
Dec 11 '18 at 22:17