Real Analysis, Folland Proposition 2.7 If $\{f_j\}$ is a sequence of $\overline{\mathbb{R}}$-valued measurable functions on $(X,M)$, then the functions 
$$\begin{aligned}
g_1(x) = \sup_{j}f_j(x), \ \ \ \  g_3(x) = \lim_{j\rightarrow \infty}\sup f_j(x)
\end{aligned}$$
$$\begin{aligned}
 g_2(x) = \inf_{j}f_j(x), \ \ \ \ g_4(x) = \lim_{j\rightarrow \infty}\inf f_j(x)
\end{aligned}$$
are all measurable functions. If $f(x) = \lim_{j\rightarrow \infty}f_j(x)$ exists for every $x\in X$, then $f$ is measurable.
proof: $$\begin{aligned}
g_1^{-1}((a,\infty)) = \bigcup_{1}^{\infty}f_j^{-1}((a,\infty)), \ \ \ \ \ g_2^{-1}((-\infty,a)) = \bigcup_{1}^{\infty}f_j^{-1}((-\infty,a))
\end{aligned}$$
so $g_1$ and $g_2$ are measurable by proposition 2.3. Now we can define $$\begin{aligned}
g_3(x) = \lim_{j\rightarrow \infty}\sup f_j(x) = \inf_{k\geq 1}\left( \sup_{j \geq k} f_j(x)\right) \ \ \ \ \ g_4(x) = \lim_{j\rightarrow \infty}\inf f_j(x) = \sup_{k\geq 1}\left(\inf_{j \geq k} f_j(x)\right)
\end{aligned}$$
so, $g_3$ and $g_4$ are measurable.
I am not sure if this is right, any suggestions is greatly appreciated.
 A: One has
\begin{align*}
\limsup f_j(x) \geq a
&\iff \inf \sup_{j \geq k} f_j(x) \geq a\\
&\iff \forall k: \quad \sup_{j \geq k} f_j(x) \geq a \\
&\iff \forall k, \forall \epsilon > 0, \exists j \geq k: \quad f_j(x) \geq a - \epsilon\\
&\iff \forall k, \forall n > 0, \exists j :  \quad f_j(x) \geq a - \frac{1}{n}\\
&\iff x \in \bigcap_{k=1}^{\infty} \bigcap_{n = 1}^{\infty} \bigcup_{j=k}^{\infty} f_j^{-1}\left(\left[a - \frac{1}{n}, \infty\right)\right)
\end{align*}
This shows that
$$g_3^{-1}([a,\infty)) = \bigcap_{k=1}^{\infty} \bigcap_{n = 1}^{\infty} \bigcup_{j=k}^{\infty} f_j^{-1}\left(\left[a - \frac{1}{n}, \infty\right)\right)$$
Now by assumption, each $f_j^{-1}([a - \frac{1}{n}, \infty))$ is measurable. So are their countable intersections and unions. Reflecting every inequality or working with $-f_j$ will give you the measurability of $g_4$.
A: @Wolfy , Your proof is correct and, in fact, it is a simple and clear proof.  I copy it here, just to add some details, to make it even clearer.

If $\{f_j\}$ is a sequence of $\overline{\mathbb{R}}$-valued measurable functions on $(X,M)$, then the functions 
  $$\begin{aligned}
g_1(x) = \sup_{j}f_j(x), \ \ \ \  g_3(x) = \lim_{j\rightarrow \infty}\sup f_j(x)
\end{aligned}$$
  $$\begin{aligned}
 g_2(x) = \inf_{j}f_j(x), \ \ \ \ g_4(x) = \lim_{j\rightarrow \infty}\inf f_j(x)
\end{aligned}$$
  are all measurable functions. If $f(x) = \lim_{j\rightarrow \infty}f(x)$ exists for every $x\in X$, then $f$ is measurable.

Proof: 
Part 1: $$\begin{aligned}
g_1^{-1}((a,\infty)) = \bigcup_{1}^{\infty}f_j^{-1}((a,\infty)), \ \ \ \ \ g_2^{-1}((-\infty,a)) = \bigcup_{1}^{\infty}f_j^{-1}((-\infty,a))
\end{aligned}$$
so $g_1$ and $g_2$ are measurable by proposition 2.3. 
Part 2:
Now we can define $g_{1,k}(x)=\sup_{j \geq k} f_j(x)$ and $g_{2,k}(x)=\inf_{j \geq k} f_j(x)$. 
For all $k$, we apply Part 1 to the sequence $\{f_j\}_{j\geq k}$, and we have that,  $g_{1,k}$ and $g_{2,k}$ are measurable functions.
Now, 
$$\begin{aligned}
g_3(x) &= \lim_{j\rightarrow \infty}\sup f_j(x) = \inf_{k\geq 1}\left( \sup_{j \geq k} f_j(x)\right)= \inf_{k\geq 1}\left(g_{1,k}(x) \right)\\ 
g_4(x) &= \lim_{j\rightarrow \infty}\inf f_j(x) = \sup_{k\geq 1}\left(\inf_{j \geq k} f_j(x)\right)=\sup_{k\geq 1}\left( g_{2,k}(x) \right)
\end{aligned}$$
so, applying Part 1, to the sequences $\{g_{1,k}\}_k$ and $\{g_{2,k}\}_k$, we conclude that  $g_3$ and $g_4$ are measurable.
