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I need help with this problem:

Let $\mathcal{M}$ be a sigma algebra of subsets of $X$ and the set function $\mu:\mathcal{M}\to [0,\infty)$ be finitely additive. Prove that $\mu$ is a measure if and only when $\{A_k\}_{k=1}^\infty$ is a descending sequence of sets in $\mathcal{M}$, then $$ \mu\left(\bigcap_{k=1}^\infty A_k\right) = \lim_{k\to \infty} \mu(A_k).$$

Below is my attempt:

Clearly one direction follows from continuity of measure.

Now suppose that the display equation above is true. Let $E=\cup_{k=1}^\infty E_k$, where the $E_k$ are disjoint. Set $A_k=E\setminus \cup_{k=1}^n E_k$. Then $A_k \supseteq A_{k+1}$. Also, $\cap_{k=1}^\infty A_k = \emptyset$. $\mu$ is also finitely additive, so $$ \mu(A_k)= \mu(E)- \sum_{n=1}^k \mu(E_n).$$ So $$\begin{align*} \mu(E) & = \lim_{k\to \infty} \mu(A_k)+\lim_{k\to \infty} \sum_{n=1}^k \mu(E_n)\\ & = \mu \left(\cap_{k=1}^\infty A_k\right) + \sum_{k=1}^\infty \mu(E_k)\\ & = \sum_{k=1}^\infty \mu(E_k) \end{align*} $$

Please, does this look right?

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There is typo in the line before the last, it should be $\mu(\bigcap_{k=1}^{+\infty}A_k)$. I think you have justify that $\mu(\emptyset)=0$ (even if it's obvious). Note that the fact that the measure is finite is needed to get the first direction, so it may be more detailed. –  Davide Giraudo Feb 20 '12 at 14:19
    
Thanks. I've fixed it. So If I justify that $\mu(\emptyset)=0$, it'll be fine? –  Bill Feb 20 '12 at 14:24
    
And if you give more details for the first direction, yes! –  Davide Giraudo Feb 20 '12 at 14:26
    
ok. Thanks very much. –  Bill Feb 20 '12 at 14:28
    
You're welcome. –  Davide Giraudo Feb 20 '12 at 14:28
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2 Answers 2

up vote 2 down vote accepted

First, we assume that $\mu$ is a measure. Let $\{A_k\}\subset \mathcal M$ a decreasing sequence of measurable subsets of $X$. Put $B_k:=X\setminus A_k$ for $k\geq 0$ and $C_k:=B_k-B_{k-1}$ for $k\geq 1$, $C_0:=B_0$. Then the sets $C_k$ are pairwise disjoint, $\bigcup_{k\geq 0}C_k=\bigcup_{k\geq 0}B_k=\bigcup_{k\geq 0}A_k^c=X\setminus \bigcap_{k\geq 0}A_k$ and since $\mu(B_k)$ is finite we have $\mu(C_k)=\mu(B_k)-\mu(B_{k-1})$. Hence $$\mu\left(\bigcap_{k\geq 0}A_k\right)=\mu(X)-\mu\left(X\setminus \bigcap_{k\geq 0}A_k\right)=\mu(X)-\mu\left(\bigcup_{k\geq 0}C_k\right)\\\ =\mu(X)-\sum_{k\geq 1}(\mu(B_k)-\mu(B_{k-1}))-\mu(B_0)=\mu(X)-\lim_{k\to \infty}\mu(B_k)=\lim_{k\to \infty}\mu(A_k).$$

For the converse what you did is fine,I would just add that $\mu(\emptyset)=\mu(\emptyset\cup\emptyset)=\mu(\emptyset)+\mu(\emptyset)$ and since $ \mu(\emptyset)<\infty$ we have $\mu(\emptyset)=0$.

The fact that the measure was finite was necessary here, for example if we take $\mu(A)=+\infty$ for all $A\in\mathcal M$ we won't get a measure since we need $\mu(\emptyset)=0$.

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Every measure satisfies $\mu(\emptyset)=0$. But the finiteness of measures is necessary. Lebesgue measure and the family of sets $[n.\infty)$ provide a counterexample that the condition doesn't work for infinite measures. –  Michael Greinecker Feb 20 '12 at 14:49
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The assumptions is false $A_{k}=[k,\infty)$ is decreasing however the measure of the intersection is zero when the measure of every set is $\infty$. You have to assume that the measure of one set is not $\infty$.

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It's the case since the range of $\mu$ is supposed to be contained on $[0,+\infty)$. –  Davide Giraudo Feb 20 '12 at 18:24
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