Measure for countable non-disjoint union (but arbitrary intersections are Null-Sets)
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I have a measure space $(X, xi, mu)$ and $(A_n)_{n in mathbb{N}} in xi^mathbb{N}$ with $mu(A_l ,cap,A_k) = 0$ for $l neq k$. I do want to show:
$$mu(bigcup_{n in mathbb{N}} A_n) = sum_{n in mathbb{N}}mu(A_n)$$
but seem to get stuck. I start by defining $B_n := A_n setminus bigcup_{m < n}A_m$ and get:
$$mu(bigcup_{n in mathbb{N}} A_n) = mu(dot{bigcup},B_n) = sum_{n in mathbb{N}}mu(B_n) = sum_{n in mathbb{N}}mu(A_n setminus bigcup_{m < n}A_m)$$
but fail to include the extra information $mu(A_l ,cap,A_k) = 0$ for $l neq k$ at this point. How do I proceed from here?
measure-theory
asked Jan 7 at 17:52
user7802048user7802048
382211
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add a comment |
$begingroup$
I have a measure space $(X, xi, mu)$ and $(A_n)_{n in mathbb{N}} in xi^mathbb{N}$ with $mu(A_l ,cap,A_k) = 0$ for $l neq k$. I do want to show:
$$mu(bigcup_{n in mathbb{N}} A_n) = sum_{n in mathbb{N}}mu(A_n)$$
but seem to get stuck. I start by defining $B_n := A_n setminus bigcup_{m < n}A_m$ and get:
$$mu(bigcup_{n in mathbb{N}} A_n) = mu(dot{bigcup},B_n) = sum_{n in mathbb{N}}mu(B_n) = sum_{n in mathbb{N}}mu(A_n setminus bigcup_{m < n}A_m)$$
but fail to include the extra information $mu(A_l ,cap,A_k) = 0$ for $l neq k$ at this point. How do I proceed from here?
measure-theory
asked Jan 7 at 17:52
user7802048user7802048
382211
$endgroup$
$begingroup$
Hint: What is the difference between measures of $A_n$ and $A_nsetminusbigcup_{m<n}A_m$?
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– Wojowu
Jan 7 at 17:53
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@Wojowu $A_n = (A_n setminus cup_{m<n},A_m) ,dotcup cup_{m<n} ,(A_n cap A_m)$. This gets us $mu(A_n setminus cup_{m<n},A_m) = mu(A_n) + mu(cup_{m<n} ,(A_n cap A_m)) = mu(A_n)$?
$endgroup$
– user7802048
Jan 7 at 17:58
add a comment |
$begingroup$
I have a measure space $(X, xi, mu)$ and $(A_n)_{n in mathbb{N}} in xi^mathbb{N}$ with $mu(A_l ,cap,A_k) = 0$ for $l neq k$. I do want to show:
$$mu(bigcup_{n in mathbb{N}} A_n) = sum_{n in mathbb{N}}mu(A_n)$$
but seem to get stuck. I start by defining $B_n := A_n setminus bigcup_{m < n}A_m$ and get:
$$mu(bigcup_{n in mathbb{N}} A_n) = mu(dot{bigcup},B_n) = sum_{n in mathbb{N}}mu(B_n) = sum_{n in mathbb{N}}mu(A_n setminus bigcup_{m < n}A_m)$$
but fail to include the extra information $mu(A_l ,cap,A_k) = 0$ for $l neq k$ at this point. How do I proceed from here?
measure-theory
asked Jan 7 at 17:52
user7802048user7802048
382211
$endgroup$
I have a measure space $(X, xi, mu)$ and $(A_n)_{n in mathbb{N}} in xi^mathbb{N}$ with $mu(A_l ,cap,A_k) = 0$ for $l neq k$. I do want to show:
$$mu(bigcup_{n in mathbb{N}} A_n) = sum_{n in mathbb{N}}mu(A_n)$$
but seem to get stuck. I start by defining $B_n := A_n setminus bigcup_{m < n}A_m$ and get:
$$mu(bigcup_{n in mathbb{N}} A_n) = mu(dot{bigcup},B_n) = sum_{n in mathbb{N}}mu(B_n) = sum_{n in mathbb{N}}mu(A_n setminus bigcup_{m < n}A_m)$$
but fail to include the extra information $mu(A_l ,cap,A_k) = 0$ for $l neq k$ at this point. How do I proceed from here?
measure-theory
measure-theory
asked Jan 7 at 17:52
user7802048user7802048
382211
asked Jan 7 at 17:52
user7802048user7802048
382211
asked Jan 7 at 17:52
user7802048user7802048
382211
asked Jan 7 at 17:52
user7802048user7802048
382211
asked Jan 7 at 17:52
user7802048user7802048
382211
382211
$begingroup$
Hint: What is the difference between measures of $A_n$ and $A_nsetminusbigcup_{m<n}A_m$?
$endgroup$
– Wojowu
Jan 7 at 17:53
$begingroup$
@Wojowu $A_n = (A_n setminus cup_{m<n},A_m) ,dotcup cup_{m<n} ,(A_n cap A_m)$. This gets us $mu(A_n setminus cup_{m<n},A_m) = mu(A_n) + mu(cup_{m<n} ,(A_n cap A_m)) = mu(A_n)$?
$endgroup$
– user7802048
Jan 7 at 17:58
add a comment |
$begingroup$
Hint: What is the difference between measures of $A_n$ and $A_nsetminusbigcup_{m<n}A_m$?
$endgroup$
– Wojowu
Jan 7 at 17:53
$begingroup$
@Wojowu $A_n = (A_n setminus cup_{m<n},A_m) ,dotcup cup_{m<n} ,(A_n cap A_m)$. This gets us $mu(A_n setminus cup_{m<n},A_m) = mu(A_n) + mu(cup_{m<n} ,(A_n cap A_m)) = mu(A_n)$?
$endgroup$
– user7802048
Jan 7 at 17:58
$begingroup$
Hint: What is the difference between measures of $A_n$ and $A_nsetminusbigcup_{m<n}A_m$?
$endgroup$
– Wojowu
Jan 7 at 17:53
$begingroup$
Hint: What is the difference between measures of $A_n$ and $A_nsetminusbigcup_{m<n}A_m$?
$endgroup$
– Wojowu
Jan 7 at 17:53
$begingroup$
@Wojowu $A_n = (A_n setminus cup_{m<n},A_m) ,dotcup cup_{m<n} ,(A_n cap A_m)$. This gets us $mu(A_n setminus cup_{m<n},A_m) = mu(A_n) + mu(cup_{m<n} ,(A_n cap A_m)) = mu(A_n)$?
$endgroup$
– user7802048
Jan 7 at 17:58
$begingroup$
@Wojowu $A_n = (A_n setminus cup_{m<n},A_m) ,dotcup cup_{m<n} ,(A_n cap A_m)$. This gets us $mu(A_n setminus cup_{m<n},A_m) = mu(A_n) + mu(cup_{m<n} ,(A_n cap A_m)) = mu(A_n)$?
$endgroup$
– user7802048
Jan 7 at 17:58
add a comment |
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$begingroup$
Hint: What is the difference between measures of $A_n$ and $A_nsetminusbigcup_{m<n}A_m$?
$endgroup$
– Wojowu
Jan 7 at 17:53
$begingroup$
@Wojowu $A_n = (A_n setminus cup_{m<n},A_m) ,dotcup cup_{m<n} ,(A_n cap A_m)$. This gets us $mu(A_n setminus cup_{m<n},A_m) = mu(A_n) + mu(cup_{m<n} ,(A_n cap A_m)) = mu(A_n)$?
$endgroup$
– user7802048
Jan 7 at 17:58