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Show $\int_X \limsup\limits_{n\to\infty} \mathbb{1}_{A_n} d\mu=0$ for $\sum_{n\in\mathbb N} \mu(A_n)<\infty, (X,\mathcal A, \mu)$ Measure Space, $(A_n)_n$ measurable

My first attempt was to rewrite it with $\liminf\limits_{n\to\infty}$ and use Fatou's lemma until I noticed that it doesn't work for negative functions.

Any help is appreciated

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This is a purely set-theoretical result: note that $A=\limsup A_n$ is $A=\bigcap\limits_nB_n$ with $B_n=\bigcup\limits_{k\geqslant n}A_k$, in particular, for every $n$, $A\subseteq B_n$, hence $\mu(A)\leqslant\mu(B_n)$.

Now, $\mu(B_n)\leqslant\sum\limits_{k=n}^\infty\mu(A_k)$ hence $\mu(A)\leqslant\sum\limits_{k=n}^\infty\mu(A_k)$ for every $n$. As the rest of a converging series, $\sum\limits_{k=n}^\infty\mu(A_k)\to0$ when $n\to\infty$, hence $\mu(A)=0$.

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This is the 1st Borel-Cantelli's lemma. Let $N = \sum_{n\in\Bbb{N}} \mathbf{1}_{A_n}$. Then it is easy to check that

$$ \limsup_{n\to\infty} \mathbf{1}_{A_n}(x) = \mathbf{1}_{\{N(x) = \infty\}}. $$

Now the assumption tells that the set $\{N(x) = \infty\}$ has measure zero, for otherwise

$$ \sum_{n\in\Bbb{N}} \mu(A_n) = \int_X N \, d\mu$$

cannot be finite. Therefore the conclusion follows.

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