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The question

Let $(X,\mathcal{A},\mu)$ be a measure space. Let $A_n$ be a sequence of sets in $\mathcal{A}$.

Define $A := \{ x \in X $ such that for all but finitely many $n \in \mathbb{N}$ it holds that $x ∈ A_n$ $ \}$

Show that $\lim_{n\to\infty}\inf \mu (A_n) \geq \mu(A)$

My attempt

$ \mu (A_n) = \int_X \chi_{A_n} d\mu$ and so by Fatou's lemma:
$\lim_{n\to\infty}\inf \mu (A_n) = \lim_{n\to\infty}\inf \int_X \chi_{ A_n} \geq \int_X \lim_{n\to\infty}\inf \chi_{A_n} d\mu$

Now all I need to show is that $\lim_{n\to\infty}\inf \chi_{A_n}(x) = \chi_A(x)$ a.e

Consider $x\in A$ then eventually $x\in A_n \forall n $ eventually and so $\chi_{A_n}(x) = 1 \forall n $ and so $\lim_{n\to\infty}\inf \chi_{A_n}(x) = \lim_{n\to\infty} 1 = 1= \chi_A(x)$

Now consider $x\not\in A$ then $\forall N \in \mathbb{N} \exists n >N$ such that $x \not\in A_n$ and so $\inf_{m \geq n} \chi_{A_n}(x) = 0 \forall n$ and hence $\lim_{n\to\infty}\inf \chi_{A_n}(x) = \lim_{n\to\infty} 0 = 0= \chi_A(x)$

Ando so the required follows.

However this feels awfully complicated and I was wondering if anyone has any tips for something simpler

OskarSzarowicz
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    Have you made the observation that $$A=\liminf {n \rightarrow \infty} A{n}=\bigcup_{n \geq 1} \bigcap_{j \geq n} A_{j}$$ holds? This observation makes life a lot easier. – Mark Apr 24 '21 at 10:25
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    Does this answer your question? Fatou's lemma and measurable sets – Mark Apr 24 '21 at 10:29
  • @Mark Thank you very much the union intersection representation was very helpful! I think your notation might be off unfortunately as the $A_n$ are subsets of the real line and thus lim inf's are not defined onto them (to the best of my knowledge :) ) – OskarSzarowicz Apr 24 '21 at 18:08
  • The liminf and liminf can also be defined for subsets. In my previous comment, the intersection and union is its definition. See also https://math.stackexchange.com/questions/2341089/show-that-lim-inf-a-n-subset-lim-sup-n-a-n. – Mark Apr 25 '21 at 16:04

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