$f$ is the pointwise limit of measurable functions $f_n$. Show $\lambda(f^{-1}(a,\infty]) \leq \liminf_{n\to \infty }\lambda(f_n^{-1}[a, \infty])$ Let $A \subseteq \mathbb{R}$ be measurable. Let $f_n : A \rightarrow [-\infty, \infty]$ be measurable $\forall n \in \mathbb{N}$ with $\lim_{n \rightarrow \infty} f_n(x) = f(x), \forall x \in A$.
I have shown already that this implies that f is also measurable.
I am left to show that $\lambda(f^{-1} (a, \infty]) \le \liminf_{n \rightarrow \infty} \lambda(f_n^{-1} [a, \infty])$ for any $a \in \mathbb{R}$. So far I have shown that:
\begin{align*}
\lambda(f^{-1} (a, \infty]) &= \lambda(\{x \in A \mid f(x) \in (a, \infty] \}) \\
&= \lambda(\{x \in A \mid \lim_{n \rightarrow \infty} f_n(x) \in (a, \infty] \}) \\
&= \lambda(\{x \in A \mid \liminf_{n \rightarrow \infty} f_n(x) \in (a, \infty] \}) \\
&\le \lambda(\{x \in A \mid \liminf_{n \rightarrow \infty} f_n(x) \in [a, \infty] \}) \\
&= \lambda(\liminf_{n \rightarrow \infty} f_n^{-1}[a, \infty]))
\end{align*}
From here I am stuck, I don't know how to justify taking the $\liminf$ out, and I'm not even sure that you can do so.
Any hints are appreciated! Thanks
 A: First of all note that if $x\in f^{-1}(a,\infty]$ then there must be an $m\in\mathbb{N}$ such that $x\in f_n^{-1}[a,\infty]$ for all $n\geq m$, because otherwise we will get a contradiction to the pointwise convergence. So that means $f^{-1}(a,\infty]\subseteq\cup_{m=1}^\infty\cap_{n=m}^\infty f_n^{-1}([a,\infty]):=F$. Now we can use the monotonicity property of measure, so $\lambda(f^{-1}(a,\infty])\leq \lambda(F)$. 
Now, for each $m\in\mathbb{N}$ let $F_m=\cap_{n=m}^\infty f_n^{-1}[a,\infty]$. Then obviously $F=\cup_{m=1}^\infty F_m$. Also note that $F_m$ is an increasing sequence, we have $F_m\subseteq F_{m+1}$ for all $m\in\mathbb{N}$. Hence by a basic property of measures we conclude that $\lambda(F)=\lim_{m\to\infty}\lambda(F_m)$. We will use this result a bit later.
For each $m\in\mathbb{N}$ we know that $\lambda(F_m)\leq\lambda(f_n^{-1}[a,\infty])$ for all $n\geq m$, this is by monotonicity of measure. A lower bound can't be bigger than the infimum so $\lambda(F_m)\leq \inf_{n\geq m}\lambda(f_n^{-1}[a,\infty])$. Since this holds for all $m$ we conclude that $\lim_{m\to\infty}\lambda(F_m)\leq\lim_{m\to\infty}\inf_{n\geq m}\lambda(f_n^{-1}[a,\infty])$. But note that the last expression is simply $\liminf_{n\to\infty}\lambda(f_n^{-1}[a,\infty])$ by definition. So combining that with the previous results we conclude that:
$\lambda(f^{-1}(a,\infty])\leq \lambda(F)=\lim_{m\to\infty}\lambda(F_m)\leq\liminf_{n\to\infty}\lambda(f_n^{-1}[a,\infty])$
