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...

  • $\begingroup$ 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

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).

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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