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To prove: Let $S$ be an arbitrary, non-empty set and let $\Sigma_0$ be an algebra on $S$. Let $\mu:\Sigma_0 \rightarrow [0,\infty]$ be a countably additive map. Show that for every decreasing sequence of sets $H_n \in \Sigma_0$ with $\cap_{n=1}^\infty H_n = \varnothing$ we have that $\lim_{n \rightarrow \infty}\mu(H_n) = 0$.

My proof:Let $H_n$ to be a decreasing sequence of sets in $\Sigma_0$ with $\bigcap_{n=1}^\infty H_n = \varnothing$. Define $H'_n:= H_n \setminus H_{n+1}$. These $H'_n$ can be seen as the disjunct components of our sequence. Then we have $H_n = \bigcup_{k=n}^\infty H'_k$. Therefore some manipulation gives: \begin{align*} \mu(H_n) &= \mu\left( \bigcup_{k=n}^\infty H'_k \right) \\ &= \sum_{k=n}^\infty \mu(H'_k) \\ &= \sum_{k=n}^\infty \mu(H_k \setminus H_{k+1}) \\ &= \sum_{k=n}^\infty \mu(H_k) - \sum_{k=n}^\infty \mu(H_{k+1}). \end{align*}

If we then let $n \rightarrow \infty$ we see that the right hand side goes to zero, which shows that $\lim_{n \rightarrow \infty} \mu(H_n) = 0$.

My problem is I haven't used that $\cap_{n=1}^\infty H_n = \varnothing$, which is pretty essential. What am I doing wrong?!

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    You do use the intersection being empty: $H_n = \bigcup_{k=n}^\infty H'_k$ is only true in that case; otherwise you'd miss members in the intersection. And what about infinite measure?2012-09-10

2 Answers 2

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The statement you are trying to prove is false. Let $S$ be $\mathbb{R}$, let $\Sigma_0$ be the Borel sets, let $\mu$ be the Lebesgue measure, and let $H_n$ be the interval $(n,\infty)$. There are also counterexamples with measures that send $\emptyset$ to 0 and all other sets to $\infty$. To make the result correct you have to assume that $\mu(S)$ is finite. One error in your proof is in the last line of the displayed equation; you cannot write $\infty - \infty$.

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    Ah, thanks guys! You're right that I assumed the measure was finite, something I'll need to be careful of.2012-09-10
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What reason is there to think that $\sum_{k=n}^\infty \mu(H_k) - \sum_{k=n}^\infty \mu(H_{k+1})$ tends to $0$? This sum is nothing but $\mu(H_n)$, by cancellation, so you've not really made any progress. (Possibly you're reasoning as if $\sum \mu(H_k) <\infty$, but there's no reason for this to hold.)

As others have suggested, you need an additional hypothesis. I suggest $\mu(H_1)<\infty$. Then you have

$\sum_{k\geq1} \mu(H_k^\prime) = \mu(\bigcup_{k\ge 1} H_k^\prime) = \mu(H_1)<\infty,$

so necessarily $\mu(H_n) = \sum_{k\geq n}\mu(H_k^\prime)\to 0$.

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    Yeah, that was just silly me assuming the measure was finite. I had a feeling something wasn't right, but thanks for pointing out where I went wrong.2012-09-10