This is the problem:
Let $m(X)< \infty$, let $\lambda_n > 0, n\in \mathbb N$, and suppose $\{{f_n}\}$ is a sequence of extended real-valued measurable functions defined on $X$ that satisfies $\sum_{n=1}^{\infty} m\{|f_n|>\lambda_n\} < \infty$. Prove that $\lim \sup_n|f_n|/\lambda _n \leq1 $.
I am trying to understand a solution to this problem, which starts as follows:
Suppose $\overline{\lim}\frac{|f_{n}|}{\lambda_n}\geq 1$, therefore: $ \exists A \subseteq X: \overline{\lim} \frac{|f_n|(x)}{\lambda_n} \geq 1, \forall x \in A$ such that $ m(A)>0$ (I understand this part)
Let $B_n = \{| f_n| > \lambda _n\}$ (As I understand this is the set of x's for which $|f_n(x)| > \lambda _n$ ? )
This is the part that I don't understand:
$A\subseteq \bigcup_{j=1}^\infty \bigcap_{n=j}^{\infty}B_n =:\underline\lim B_n $
$\Rightarrow m(B_n)\not \rightarrow$ $0$ (meaning $m(B_n)$ doesn't converge to 0)
$\Rightarrow \sum m(B_n) \rightarrow \infty $ and this is a contradiction because we know that $\sum_{n=1}^{\infty} m\{|f_n|>\lambda_n\} < \infty$
The main statement that bugs me is why: $A\subseteq \bigcup_{j=1}^\infty \bigcap_{n=j}^{\infty}B_n =:\underline\lim B_n $. Because that according to my knowledge, in general: $\lim\inf B_n\subseteq \lim\sup B_n$. So I don't see a reason for A to be a subset of $\underline\lim B_n$.
Thanks.
This is an application of the Borel-Cantelli lemma. Consider the event
$$E_n = \{x{}:{} |f_n(x)| > \lambda_n \} = \left\{ x{}:{} \frac{|f_n(x)|}{\lambda_n} > 1 \right\}.$$
Since $\sum_n m(E_n) < \infty$, by the Borel-Cantelli lemma,
$$ m\left(\limsup_n E_n \right) = 0, $$
that is, the probability that $E_n$ occurs infinitely often in $0$, therefore, $0 \leq \lambda_n^{-1}|f_n| \leq 1$ infinitely often, almost surely. As a result,
$$ \limsup_n \lambda_n^{-1}|f_n| \leq 1, \text{ a.s.} $$