Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer

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.

share|improve this answer
    
The finiteness of the measure $\mu$ is quite unnecessary. –  Did Dec 19 '12 at 16:14
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.