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Let $(X, \mathcal{B}, \mu)$ be a finite measure space and suppose $\{A_n\} \subseteq \mathcal{B}$ s.t. $\sum_{n=1}^\infty \mu(A_n) < \infty$. Furthermore let $\underset{n \rightarrow \infty}{\text{limsup}}$ $A_n = S$. I aim to show that $\mu(S) = 0$. Is the following proof (in particular, step 2) valid?

  1. First consider that $S = \bigcap_{n=1}^\infty \bigcup_{m=n}^\infty A_n$ is measruable since it is a countable intersection of a countable union of the measurable sets $A_n$.

  2. As an immediate consequence of the definition of $S$, we have that if $s \in S$, then $s \in A_n$ for an infinite number of the $A_n$. Moreover, $S \subseteq A_n$ for an infinite number of the $A_n$.

  3. If for sake of contradiction we had that $0 < \mu(S) = r \in \mathbb{R}^+$, then $\sum_{n=1}^\infty \mu(A_n) \ge \sum_{k=1}^\infty \mu(S) = \sum_{k=1}^\infty r = \infty$, a contradiction since $\sum_{n=1}^\infty \mu(A_n) < \infty$ by hypothesis.

Thus $\mu(S) = 0$ as desired.

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Define $B_n:=\bigcup_{m\geqslant n}A_m$. This forms a decreasing sequence to $S$. We have $\mu(B_n)\leqslant\sum_{m=n}^{+\infty}\mu(A_m)$, and the remainder of a convergent series converges to $0$.

We have $B\subset B_n$ for all $n$, so $\mu(B)\leqslant \sum_{m=n}^{+\infty}\mu(A_m)$ and we conclude.

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The finiteness of the measure $\mu$ is quite unnecessary. – Did Dec 19 '12 at 16:14
All the sets have finite measure here, by hypothesis. Note that no limit of sets is necessary in the last step, since the inclusion $S\subseteq B_n$, valid for every $n$, suffices to conclude. – Did Dec 19 '12 at 16:28

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