Is a not empty subset of a unmeasurable set a unmeasurable set?












0












$begingroup$


We have a measurable space $(Omega, mathbb{A})$. I want to prove the following statement about random vectors



$$X=(X_1, ..., X_k) hbox{ is a random vector iff } X_i hbox{ is a random variable } forall i = 1,...,k $$



For this, I want to use only the following definition: $X:Omega to mathbb{R}^k$ is a random vector (random variable if $k = 1$) if $[X subset B] in mathbb{A}$, for all $B = B(q,r)= {y in mathbb{R}^k : d_{max}(q,y) < r}$, whith $(q,r) in mathbb{Q}^k times mathbb{Q}$. In other words, I want to use the fact that $mathbb{R}^k$ and its topology have a countable base. In addition, we know that:



$$B(q,r) = Pi_{i = 1}^k (q_i - r, q_i + r)$$
and



$$[X in B(q,r)] = [q_1 - r < X_1 < q_1 + r, ..., q_k - r < X_k < q_k + r] = bigcap_{i = 1}^k [q_i - r < X_i < q_i + r]$$



With this in hand, I would like to prove the most difficult implication: $(implies)$. For this, take a randoms $x in mathbb{Q}^k$ and $rin mathbb{Q}$ and suppose that $[q_i - r < X_i < q_i + r] notin mathbb{A}$ for some $i$. Since that:



$$[X in B(q,r)] subset [q_i - r < X_i < q_i + r]$$



I want to prove the following fact: if $B notin mathbb{A}$ and $A subset B$, $A neq emptyset$, then $A notin mathbb{A}$. With this, I can get some contradiction. Is it true?










share|cite|improve this question









$endgroup$












  • $begingroup$
    Let $A$ be a singleton and $Bbb A$ the Borel $sigma$-algebra
    $endgroup$
    – Alessandro Codenotti
    Jan 13 at 8:30
















0












$begingroup$


We have a measurable space $(Omega, mathbb{A})$. I want to prove the following statement about random vectors



$$X=(X_1, ..., X_k) hbox{ is a random vector iff } X_i hbox{ is a random variable } forall i = 1,...,k $$



For this, I want to use only the following definition: $X:Omega to mathbb{R}^k$ is a random vector (random variable if $k = 1$) if $[X subset B] in mathbb{A}$, for all $B = B(q,r)= {y in mathbb{R}^k : d_{max}(q,y) < r}$, whith $(q,r) in mathbb{Q}^k times mathbb{Q}$. In other words, I want to use the fact that $mathbb{R}^k$ and its topology have a countable base. In addition, we know that:



$$B(q,r) = Pi_{i = 1}^k (q_i - r, q_i + r)$$
and



$$[X in B(q,r)] = [q_1 - r < X_1 < q_1 + r, ..., q_k - r < X_k < q_k + r] = bigcap_{i = 1}^k [q_i - r < X_i < q_i + r]$$



With this in hand, I would like to prove the most difficult implication: $(implies)$. For this, take a randoms $x in mathbb{Q}^k$ and $rin mathbb{Q}$ and suppose that $[q_i - r < X_i < q_i + r] notin mathbb{A}$ for some $i$. Since that:



$$[X in B(q,r)] subset [q_i - r < X_i < q_i + r]$$



I want to prove the following fact: if $B notin mathbb{A}$ and $A subset B$, $A neq emptyset$, then $A notin mathbb{A}$. With this, I can get some contradiction. Is it true?










share|cite|improve this question









$endgroup$












  • $begingroup$
    Let $A$ be a singleton and $Bbb A$ the Borel $sigma$-algebra
    $endgroup$
    – Alessandro Codenotti
    Jan 13 at 8:30














0












0








0





$begingroup$


We have a measurable space $(Omega, mathbb{A})$. I want to prove the following statement about random vectors



$$X=(X_1, ..., X_k) hbox{ is a random vector iff } X_i hbox{ is a random variable } forall i = 1,...,k $$



For this, I want to use only the following definition: $X:Omega to mathbb{R}^k$ is a random vector (random variable if $k = 1$) if $[X subset B] in mathbb{A}$, for all $B = B(q,r)= {y in mathbb{R}^k : d_{max}(q,y) < r}$, whith $(q,r) in mathbb{Q}^k times mathbb{Q}$. In other words, I want to use the fact that $mathbb{R}^k$ and its topology have a countable base. In addition, we know that:



$$B(q,r) = Pi_{i = 1}^k (q_i - r, q_i + r)$$
and



$$[X in B(q,r)] = [q_1 - r < X_1 < q_1 + r, ..., q_k - r < X_k < q_k + r] = bigcap_{i = 1}^k [q_i - r < X_i < q_i + r]$$



With this in hand, I would like to prove the most difficult implication: $(implies)$. For this, take a randoms $x in mathbb{Q}^k$ and $rin mathbb{Q}$ and suppose that $[q_i - r < X_i < q_i + r] notin mathbb{A}$ for some $i$. Since that:



$$[X in B(q,r)] subset [q_i - r < X_i < q_i + r]$$



I want to prove the following fact: if $B notin mathbb{A}$ and $A subset B$, $A neq emptyset$, then $A notin mathbb{A}$. With this, I can get some contradiction. Is it true?










share|cite|improve this question









$endgroup$




We have a measurable space $(Omega, mathbb{A})$. I want to prove the following statement about random vectors



$$X=(X_1, ..., X_k) hbox{ is a random vector iff } X_i hbox{ is a random variable } forall i = 1,...,k $$



For this, I want to use only the following definition: $X:Omega to mathbb{R}^k$ is a random vector (random variable if $k = 1$) if $[X subset B] in mathbb{A}$, for all $B = B(q,r)= {y in mathbb{R}^k : d_{max}(q,y) < r}$, whith $(q,r) in mathbb{Q}^k times mathbb{Q}$. In other words, I want to use the fact that $mathbb{R}^k$ and its topology have a countable base. In addition, we know that:



$$B(q,r) = Pi_{i = 1}^k (q_i - r, q_i + r)$$
and



$$[X in B(q,r)] = [q_1 - r < X_1 < q_1 + r, ..., q_k - r < X_k < q_k + r] = bigcap_{i = 1}^k [q_i - r < X_i < q_i + r]$$



With this in hand, I would like to prove the most difficult implication: $(implies)$. For this, take a randoms $x in mathbb{Q}^k$ and $rin mathbb{Q}$ and suppose that $[q_i - r < X_i < q_i + r] notin mathbb{A}$ for some $i$. Since that:



$$[X in B(q,r)] subset [q_i - r < X_i < q_i + r]$$



I want to prove the following fact: if $B notin mathbb{A}$ and $A subset B$, $A neq emptyset$, then $A notin mathbb{A}$. With this, I can get some contradiction. Is it true?







probability random-variables






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 13 at 7:41









FamFam

295




295












  • $begingroup$
    Let $A$ be a singleton and $Bbb A$ the Borel $sigma$-algebra
    $endgroup$
    – Alessandro Codenotti
    Jan 13 at 8:30


















  • $begingroup$
    Let $A$ be a singleton and $Bbb A$ the Borel $sigma$-algebra
    $endgroup$
    – Alessandro Codenotti
    Jan 13 at 8:30
















$begingroup$
Let $A$ be a singleton and $Bbb A$ the Borel $sigma$-algebra
$endgroup$
– Alessandro Codenotti
Jan 13 at 8:30




$begingroup$
Let $A$ be a singleton and $Bbb A$ the Borel $sigma$-algebra
$endgroup$
– Alessandro Codenotti
Jan 13 at 8:30










1 Answer
1






active

oldest

votes


















1












$begingroup$

Certainly your claim is false, as pointed out by Alessandro Codenotti. Use the fact that $B(q,r)$ can be written as countable union of sets of the form $[a_1,b_1] times [a_2,b_2] times cdots times [a_k,b_k]$ to prove that $X$ is measurable if each $X_i$ is.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    In fact, I want to prove that if $X$ is measurable, then $X_i$ is measurable for all $i = 1,...,k$.
    $endgroup$
    – Fam
    Jan 13 at 16:41










  • $begingroup$
    If we apply the projection continium function, I think that we can conclude!
    $endgroup$
    – Fam
    Jan 13 at 19:36










  • $begingroup$
    @Fam Yes, that is also a valid proof.
    $endgroup$
    – Kavi Rama Murthy
    Jan 13 at 23:29











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071777%2fis-a-not-empty-subset-of-a-unmeasurable-set-a-unmeasurable-set%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









1












$begingroup$

Certainly your claim is false, as pointed out by Alessandro Codenotti. Use the fact that $B(q,r)$ can be written as countable union of sets of the form $[a_1,b_1] times [a_2,b_2] times cdots times [a_k,b_k]$ to prove that $X$ is measurable if each $X_i$ is.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    In fact, I want to prove that if $X$ is measurable, then $X_i$ is measurable for all $i = 1,...,k$.
    $endgroup$
    – Fam
    Jan 13 at 16:41










  • $begingroup$
    If we apply the projection continium function, I think that we can conclude!
    $endgroup$
    – Fam
    Jan 13 at 19:36










  • $begingroup$
    @Fam Yes, that is also a valid proof.
    $endgroup$
    – Kavi Rama Murthy
    Jan 13 at 23:29
















1












$begingroup$

Certainly your claim is false, as pointed out by Alessandro Codenotti. Use the fact that $B(q,r)$ can be written as countable union of sets of the form $[a_1,b_1] times [a_2,b_2] times cdots times [a_k,b_k]$ to prove that $X$ is measurable if each $X_i$ is.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    In fact, I want to prove that if $X$ is measurable, then $X_i$ is measurable for all $i = 1,...,k$.
    $endgroup$
    – Fam
    Jan 13 at 16:41










  • $begingroup$
    If we apply the projection continium function, I think that we can conclude!
    $endgroup$
    – Fam
    Jan 13 at 19:36










  • $begingroup$
    @Fam Yes, that is also a valid proof.
    $endgroup$
    – Kavi Rama Murthy
    Jan 13 at 23:29














1












1








1





$begingroup$

Certainly your claim is false, as pointed out by Alessandro Codenotti. Use the fact that $B(q,r)$ can be written as countable union of sets of the form $[a_1,b_1] times [a_2,b_2] times cdots times [a_k,b_k]$ to prove that $X$ is measurable if each $X_i$ is.






share|cite|improve this answer











$endgroup$



Certainly your claim is false, as pointed out by Alessandro Codenotti. Use the fact that $B(q,r)$ can be written as countable union of sets of the form $[a_1,b_1] times [a_2,b_2] times cdots times [a_k,b_k]$ to prove that $X$ is measurable if each $X_i$ is.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 13 at 23:29

























answered Jan 13 at 12:42









Kavi Rama MurthyKavi Rama Murthy

67.8k53067




67.8k53067












  • $begingroup$
    In fact, I want to prove that if $X$ is measurable, then $X_i$ is measurable for all $i = 1,...,k$.
    $endgroup$
    – Fam
    Jan 13 at 16:41










  • $begingroup$
    If we apply the projection continium function, I think that we can conclude!
    $endgroup$
    – Fam
    Jan 13 at 19:36










  • $begingroup$
    @Fam Yes, that is also a valid proof.
    $endgroup$
    – Kavi Rama Murthy
    Jan 13 at 23:29


















  • $begingroup$
    In fact, I want to prove that if $X$ is measurable, then $X_i$ is measurable for all $i = 1,...,k$.
    $endgroup$
    – Fam
    Jan 13 at 16:41










  • $begingroup$
    If we apply the projection continium function, I think that we can conclude!
    $endgroup$
    – Fam
    Jan 13 at 19:36










  • $begingroup$
    @Fam Yes, that is also a valid proof.
    $endgroup$
    – Kavi Rama Murthy
    Jan 13 at 23:29
















$begingroup$
In fact, I want to prove that if $X$ is measurable, then $X_i$ is measurable for all $i = 1,...,k$.
$endgroup$
– Fam
Jan 13 at 16:41




$begingroup$
In fact, I want to prove that if $X$ is measurable, then $X_i$ is measurable for all $i = 1,...,k$.
$endgroup$
– Fam
Jan 13 at 16:41












$begingroup$
If we apply the projection continium function, I think that we can conclude!
$endgroup$
– Fam
Jan 13 at 19:36




$begingroup$
If we apply the projection continium function, I think that we can conclude!
$endgroup$
– Fam
Jan 13 at 19:36












$begingroup$
@Fam Yes, that is also a valid proof.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 23:29




$begingroup$
@Fam Yes, that is also a valid proof.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 23:29


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071777%2fis-a-not-empty-subset-of-a-unmeasurable-set-a-unmeasurable-set%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

Human spaceflight

Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

張江高科駅