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Sorry for my bad english.
Let $(E, A, \mu)$ be a measured space. Let $(A_n)_{n \geq 1}$ be a sequence of sets of $A$.
If we consider $\sum_{n \geq 1} \mu (A_n) < \infty$, we would like to show that $\mu (\lim \sup (A_n)) = 0$. I wonder how to do it ?
Someone could help me ? Thank you in advance.
Recall $\limsup A_{n}=\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_{k}$. Then $$ \mu(\limsup A_{n})\leq \mu(\bigcup_{k=m}^{\infty}A_{k})\leq \sum_{k=m}^{\infty}\mu(A_{k}) $$ for all $m$. Then, since $\sum_{k=1}^{\infty}\mu(A_{k})<\infty$, $\lim_{m\to\infty}\sum_{k=m}^{\infty}\mu(A_{k})=0$ and we deduce $$\mu(\limsup A_{n})=0$$
We have $$\sum_{n\ge k}\mu(A_n)\ge\mu\left(\bigcup_{n\ge k}A_n\right)\ge\mu\left(\limsup A_n\right)$$ Since $\sum_{n\ge k}\mu(A_n)\rightarrow0$ as $k\rightarrow\infty$, the result follows.
Here is another proof. Recall that, $$ \limsup A_n=\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_k, $$ namely it is a set with the property that for every $\omega \in\limsup A_n$, for every $n$, there exists a $k\geq n$ such that $\omega \in A_k$, that is to say, $\omega$ belongs to infinitely many of the sets $\{A_n\}_{n=1}^{\infty}$.
With this in mind, let us define a sequence of functions $\{f_n\}$ according to $$ f_n(\omega) = \sum_{k=1}^n \mathcal{X}_{A_n} $$ where $\mathcal{X}_{A_n}$ is the characteristic function of the set $A_n$. Observe that since each $A_n$ is a measurable set, $\mathcal{X}_{A_n}$ is a measurable function for every $n$, so does $f_n$. Now, by monotone convergence theorem, since $0 \leq f_n(\omega) \nearrow \sum_{k=1}^{\infty}\mathcal{X}_{A_n}(\omega)$ $$ \int\left( \sum_{k=1}^{\infty}\mathcal{X}_{A_n}\right)d\mu = \lim_{n\to\infty}\int f_n d\mu = \lim_{n\to\infty}\sum_{k=1}^n \mu(A_k) = \sum_{k=1}^{\infty}\mu(A_k)<\infty. $$ Hence, for almost every $\omega$, $\sum_{k=1}^{\infty}\mathcal{X}_{A_k}(\omega)<\infty$, that means for almost every $\omega$, $\omega$ must belong to only finitely many of the sets $A_n$'s. Hence, $$ \mu\left(\limsup A_n\right)=0 $$ as desired.
