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?










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$endgroup$












  • $begingroup$
    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


















0












$begingroup$


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?










share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















0












0








0





$begingroup$


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?










share|cite|improve this question









$endgroup$




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






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share|cite|improve this question











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share|cite|improve this question










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




















  • $begingroup$
    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












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