Showing that supremum function is integrable Let $g_1(\omega),g_2(\omega),...$ be integrable functions defined on $\Omega$ with $g_n\rightarrow g$ and $g$ is integrable and also $\lim \int g_n=\int g$ . Define $h(\omega)= \sup_n g_n(\omega)$. How do we show that $h$ is integrable?
I was trying to use fatou's lemma, but I'm stuck.
 A: In general, the claim does not hold true as the following counterexample shows:
Consider $((0,1],\mathcal{B}(0,1])$ endowed with the Lebesgue measure. For $a_n := \frac{1}{n^2}$ we define
$$g_n(\omega) := \begin{cases} \frac{1}{n}, & \omega \notin [a_n,a_{n-1}), \\ \frac{1}{\omega}, & \omega \in [a_n,a_{n-1}), \end{cases} \qquad \omega \in (0,1].$$
Obviously, $g_n \to 0$ almost surely and $$\begin{align*} \int g_n(\omega) \, d\omega &\leq \frac{1}{n} + \int_{\frac{1}{n^2}}^{\frac{1}{(n-1)^2}} \underbrace{\frac{1}{\omega}}_{\leq n^2} \, d\omega \\ &\leq \frac{1}{n} + n^2 \left( \frac{1}{(n-1)^2} - \frac{1}{n^2} \right) \\ &\stackrel{n \to \infty}{\to} 0 \end{align*}$$
On the other hand,
$$\sup_{n \in \mathbb{N}} g_n(\omega) = \frac{1}{\omega} \notin L^1.$$
A: Counterexample: Take $\Omega=[0,\infty)$ and consider the sequence of characteristic functions: $1_I, 1_{I/2+1}, 1_{I/2+1+1/2}, 1_{I/3+2}, 1_{I/3+2+1/3}, 1_{I/3+2+2/3}...$
Where $1_A$ is the characteristic function of the set $A$ and I is the unit interval. 
