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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 A. We have $B_n = \bigcap_{k \geq n} A_k$ and $C_n = \bigcup_{k \geq n} A_k$. We have also $\lim \sup (A_n) = \bigcap_{n \geq 1} C_n$ and $\lim \inf (A_n) = \bigcup_{n \geq 0} B_n$.

I want to show that :

a) $\mu (\lim \inf (A_n)) \leq \lim \inf (\mu(A_n))$

b) If we suppose $\mu(\bigcup_{n \geq 1} A_n) < \infty$, $\mu (\lim \sup (A_n)) \geq \lim \sup (\mu(A_n))$.

For the question a) I know that $\mu (\lim \inf (A_n)) = \mu (\bigcup B_n)) = \sum \mu(B_n) = \sum \mu (\bigcap (A_k))$... I don't see how to prove the inequality. I'm a beginner.

Someone could help me ? Thank you in advance...

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  • $\begingroup$ The $B_n$-s are not disjoint (they are actually an increasing sequence of sets), therefore the identity $\mu\left(\bigcup_n B_n\right)=\sum_n \mu(B_n)$ never holds (unless $\mu(B_n)$ is $0$ for all $n$). $\endgroup$ – user228113 Apr 23 '17 at 16:41
  • $\begingroup$ $\mu(\cup_n B_n) \le \sum_n\mu(B_n)$. $\endgroup$ – oliverjones Apr 23 '17 at 16:48
  • $\begingroup$ For (a) use (i) $B_n\subset B_{n+1}$ for all $n$; and (ii) $B_n\subset A_k$ for all $k\ge n$ and all $n$. $\endgroup$ – John Dawkins Apr 23 '17 at 17:04
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First let's come up with the following result:

Let $B_i=\cap_{k \ge i}A_k \in \mathscr A$, and thus $B_i \uparrow B$. Then $\mu (B) = \lim_{n\rightarrow \infty}\mu(B_n)$

Proof. $Let D_1=B_1,D_2=B_2-B_1,D_3=B_3-(B_1 \cup B_2),..., D_i=B_i-(\cup_{j=1}^{i-1}B_j)$. The $D_i$ are pairwise disjoint, $D_i \subset B_i$ for each $i$, and $\cup_{i =1}^{\infty}B_i=\cup_{i =1}^{\infty}D_i$. Hence: $$\mu(B)=\mu(\cup_{i =1}^{\infty}B_i)=\mu(\cup_{i =1}^{\infty}D_i)=\sum_{i =1}^{\infty}\mu(D_i) $$ $$=\lim_{n\rightarrow \infty} \sum_{i=1}^{n}\mu(D_i)=\lim_{n \rightarrow \infty} \mu(\cup_{i=1}^n D_i)=\lim_{n\rightarrow \infty}\mu(\cup_{i =1}^{n}B_i)=\lim_{n\rightarrow \infty}\mu(B_n)$$


Now notice that $$\liminf_{n \rightarrow \infty} (A_n)=\cup_{n \ge 1}B_n=B$$ And $B_n \subset A_k, \forall k \ge n$, thus $\mu(B_n ) \le \mu( A_k), \forall k \ge n$, therefore $$\lim_{n \rightarrow \infty}\mu(B_n) \le \lim_{n \rightarrow \infty}(\inf_{k \ge n}\mu(A_k))=\liminf_{n \rightarrow \infty}\mu(A_n)$$

Thus $$\mu(\liminf_{n \rightarrow \infty} A_n)=\mu(B)=\lim_{n \rightarrow \infty}\mu(B_n) \le \liminf_{n \rightarrow \infty}\mu(A_n)$$

The other one is similar (the proof is not difficult, but it's just so much typing, esp. with MathJax, so I'll save it).

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  • $\begingroup$ @MélanieDelaCheminée you're welcome $\endgroup$ – Yujie Zha Apr 24 '17 at 11:16
  • $\begingroup$ Of course, It's done ! :-) $\endgroup$ – Mélanie De la Cheminée Apr 24 '17 at 17:40

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