Application of Ulam Lemma
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I can’t understand how to derive the following conclusion from
Lemma (Ulam)
Let $(X, tau)$ be a Polish space and $mu$ a positive finite Borel measure on $X$. Then for every $epsilon >0 $ there exists a compact set $K =K(epsilon) subset X$ s.t. $ mu(X setminus K) < epsilon$
Then I have the following remark:
Let $(X,d)$ be a complete separable metric space and let $mu$ be a probability on $X$. Let $A subset X$ be a Borel set s.t. $mu( X setminus A) = 0 $.
Then $A$ is $sigma$ compact.
For me $sigma$ compact means being the union of (at most) countable compact sets.
How can it be true? Even if the statement is not precise and it means
there exists a family ${ K_n }_{n in mathbb{N}} $ of compact set each one contained in $A$ s.t.
$$ mu ( A setminus bigcup_n K_n) =0 $$
how can I prove it?
I know I can take (thanks to Ulam lemma) a sequence of compact sets $C_n$ s.t. $mu(X setminus C_n ) = mu( A setminus C_n) <1/n$ and take $$K_n := bigcup_{k=1}^n C_k$$
Then the $K_n$ are compact sets and
$$ mu ( A setminus bigcup_n K_n) =0 $$
But they are not contained in $A$. Maybe I can take $Q_n = C_n cap A$ but now they are not compact any more. Unless $A$ is closed (and it is in the case $A= text{supp} mu$).
Am I wrong?
measure-theory compactness
asked Dec 5 at 18:14
Bremen000
319110
|
show 11 more comments
up vote
1
down vote
favorite
I can’t understand how to derive the following conclusion from
Lemma (Ulam)
Let $(X, tau)$ be a Polish space and $mu$ a positive finite Borel measure on $X$. Then for every $epsilon >0 $ there exists a compact set $K =K(epsilon) subset X$ s.t. $ mu(X setminus K) < epsilon$
Then I have the following remark:
Let $(X,d)$ be a complete separable metric space and let $mu$ be a probability on $X$. Let $A subset X$ be a Borel set s.t. $mu( X setminus A) = 0 $.
Then $A$ is $sigma$ compact.
For me $sigma$ compact means being the union of (at most) countable compact sets.
How can it be true? Even if the statement is not precise and it means
there exists a family ${ K_n }_{n in mathbb{N}} $ of compact set each one contained in $A$ s.t.
$$ mu ( A setminus bigcup_n K_n) =0 $$
how can I prove it?
I know I can take (thanks to Ulam lemma) a sequence of compact sets $C_n$ s.t. $mu(X setminus C_n ) = mu( A setminus C_n) <1/n$ and take $$K_n := bigcup_{k=1}^n C_k$$
Then the $K_n$ are compact sets and
$$ mu ( A setminus bigcup_n K_n) =0 $$
But they are not contained in $A$. Maybe I can take $Q_n = C_n cap A$ but now they are not compact any more. Unless $A$ is closed (and it is in the case $A= text{supp} mu$).
Am I wrong?
measure-theory compactness
asked Dec 5 at 18:14
Bremen000
319110
Oh you mean a separable metric space.
– Charlie Frohman
Dec 5 at 18:25
I can’t understand your comment.
– Bremen000
Dec 5 at 18:52
Is $A$ measurable? If $A$ is Borel, then this follows from the inner regularity of Borel measures.
– d.k.o.
Dec 5 at 21:18
Yes $A$ is measurable. Can you expand your comment?
– Bremen000
Dec 5 at 21:23
Ok, I understand what you said. My measure is only a sigma additive function from the borel sets of $X$ info $(0,+infty)$. I do not have any regularity property.
– Bremen000
Dec 5 at 21:27
|
show 11 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I can’t understand how to derive the following conclusion from
Lemma (Ulam)
Let $(X, tau)$ be a Polish space and $mu$ a positive finite Borel measure on $X$. Then for every $epsilon >0 $ there exists a compact set $K =K(epsilon) subset X$ s.t. $ mu(X setminus K) < epsilon$
Then I have the following remark:
Let $(X,d)$ be a complete separable metric space and let $mu$ be a probability on $X$. Let $A subset X$ be a Borel set s.t. $mu( X setminus A) = 0 $.
Then $A$ is $sigma$ compact.
For me $sigma$ compact means being the union of (at most) countable compact sets.
How can it be true? Even if the statement is not precise and it means
there exists a family ${ K_n }_{n in mathbb{N}} $ of compact set each one contained in $A$ s.t.
$$ mu ( A setminus bigcup_n K_n) =0 $$
how can I prove it?
I know I can take (thanks to Ulam lemma) a sequence of compact sets $C_n$ s.t. $mu(X setminus C_n ) = mu( A setminus C_n) <1/n$ and take $$K_n := bigcup_{k=1}^n C_k$$
Then the $K_n$ are compact sets and
$$ mu ( A setminus bigcup_n K_n) =0 $$
But they are not contained in $A$. Maybe I can take $Q_n = C_n cap A$ but now they are not compact any more. Unless $A$ is closed (and it is in the case $A= text{supp} mu$).
Am I wrong?
measure-theory compactness
asked Dec 5 at 18:14
Bremen000
319110
I can’t understand how to derive the following conclusion from
Lemma (Ulam)
Let $(X, tau)$ be a Polish space and $mu$ a positive finite Borel measure on $X$. Then for every $epsilon >0 $ there exists a compact set $K =K(epsilon) subset X$ s.t. $ mu(X setminus K) < epsilon$
Then I have the following remark:
Let $(X,d)$ be a complete separable metric space and let $mu$ be a probability on $X$. Let $A subset X$ be a Borel set s.t. $mu( X setminus A) = 0 $.
Then $A$ is $sigma$ compact.
For me $sigma$ compact means being the union of (at most) countable compact sets.
How can it be true? Even if the statement is not precise and it means
there exists a family ${ K_n }_{n in mathbb{N}} $ of compact set each one contained in $A$ s.t.
$$ mu ( A setminus bigcup_n K_n) =0 $$
how can I prove it?
I know I can take (thanks to Ulam lemma) a sequence of compact sets $C_n$ s.t. $mu(X setminus C_n ) = mu( A setminus C_n) <1/n$ and take $$K_n := bigcup_{k=1}^n C_k$$
Then the $K_n$ are compact sets and
$$ mu ( A setminus bigcup_n K_n) =0 $$
But they are not contained in $A$. Maybe I can take $Q_n = C_n cap A$ but now they are not compact any more. Unless $A$ is closed (and it is in the case $A= text{supp} mu$).
Am I wrong?
measure-theory compactness
measure-theory compactness
asked Dec 5 at 18:14
Bremen000
319110
asked Dec 5 at 18:14
Bremen000
319110
edited Dec 5 at 21:33
asked Dec 5 at 18:14
Bremen000
319110
asked Dec 5 at 18:14
Bremen000
319110
asked Dec 5 at 18:14
Bremen000
319110
319110
Oh you mean a separable metric space.
– Charlie Frohman
Dec 5 at 18:25
I can’t understand your comment.
– Bremen000
Dec 5 at 18:52
Is $A$ measurable? If $A$ is Borel, then this follows from the inner regularity of Borel measures.
– d.k.o.
Dec 5 at 21:18
Yes $A$ is measurable. Can you expand your comment?
– Bremen000
Dec 5 at 21:23
Ok, I understand what you said. My measure is only a sigma additive function from the borel sets of $X$ info $(0,+infty)$. I do not have any regularity property.
– Bremen000
Dec 5 at 21:27
|
show 11 more comments
Oh you mean a separable metric space.
– Charlie Frohman
Dec 5 at 18:25
I can’t understand your comment.
– Bremen000
Dec 5 at 18:52
Is $A$ measurable? If $A$ is Borel, then this follows from the inner regularity of Borel measures.
– d.k.o.
Dec 5 at 21:18
Yes $A$ is measurable. Can you expand your comment?
– Bremen000
Dec 5 at 21:23
Ok, I understand what you said. My measure is only a sigma additive function from the borel sets of $X$ info $(0,+infty)$. I do not have any regularity property.
– Bremen000
Dec 5 at 21:27
Oh you mean a separable metric space.
– Charlie Frohman
Dec 5 at 18:25
Oh you mean a separable metric space.
– Charlie Frohman
Dec 5 at 18:25
I can’t understand your comment.
– Bremen000
Dec 5 at 18:52
I can’t understand your comment.
– Bremen000
Dec 5 at 18:52
Is $A$ measurable? If $A$ is Borel, then this follows from the inner regularity of Borel measures.
– d.k.o.
Dec 5 at 21:18
Is $A$ measurable? If $A$ is Borel, then this follows from the inner regularity of Borel measures.
– d.k.o.
Dec 5 at 21:18
Yes $A$ is measurable. Can you expand your comment?
– Bremen000
Dec 5 at 21:23
Yes $A$ is measurable. Can you expand your comment?
– Bremen000
Dec 5 at 21:23
Ok, I understand what you said. My measure is only a sigma additive function from the borel sets of $X$ info $(0,+infty)$. I do not have any regularity property.
– Bremen000
Dec 5 at 21:27
Ok, I understand what you said. My measure is only a sigma additive function from the borel sets of $X$ info $(0,+infty)$. I do not have any regularity property.
– Bremen000
Dec 5 at 21:27
|
show 11 more comments
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Oh you mean a separable metric space.
– Charlie Frohman
Dec 5 at 18:25
I can’t understand your comment.
– Bremen000
Dec 5 at 18:52
Is $A$ measurable? If $A$ is Borel, then this follows from the inner regularity of Borel measures.
– d.k.o.
Dec 5 at 21:18
Yes $A$ is measurable. Can you expand your comment?
– Bremen000
Dec 5 at 21:23
Ok, I understand what you said. My measure is only a sigma additive function from the borel sets of $X$ info $(0,+infty)$. I do not have any regularity property.
– Bremen000
Dec 5 at 21:27