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?






probability random-variables







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$endgroup$












  • $begingroup$
    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?






probability random-variables







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?






probability random-variables







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




probability random-variables






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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











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1 Answer
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1 Answer
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active

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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


















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