I have the following exercise:
Let $\Omega$ be a open subset of $\mathbb R^3$ and $\left( z_n : \Omega \to \mathbb R \mid n \in \mathbb N \right)$ a sequence of measurable functions convergent almost everywhere to $z : \Omega \to \mathbb R$. Assume also $$\int_\Omega e^{z_n (x)} \mathrm d x \le 100 \quad \text{for every } n \in \mathbb N .$$ Is the function $e ^z : \Omega \to \mathbb R$ integrable on $\Omega$?
As far as I know, if $\Omega$ is a bounded interval and the functions $z_n$ are bounded, then $e^z$ would be integrable since $$\int_\Omega e^{z(x)} \mathrm d x = \lim_{n \to +\infty} \int_\Omega e^{z_n(x)} \mathrm d x$$ since $e^{z_n} \to e^z$ almost everywhere. I am using the dominated convergence theorem.
Now le me consider the thing differently. If $\left( z_n \mid n \in \mathbb N \right)$ is a monotonic sequence of integrable functions, then again $$\int_\Omega e^{z(x)} \mathrm d x = \lim_{n \to +\infty} \int_\Omega e^{z_n(x)} \mathrm d x \le 100.$$ This is possible using the Beppo-Levi Theorem.
By the way, what can be said in general? Can you provide if possible some counterexample?
$\int_{\Omega}e^{z(x)}dx\leq \lim \inf\int_{\Omega}e^{z_n(x)}dx\leq 100$ by Fatou's Lemma which shows that $e^{z(x)}$ is integrable on $\Omega$.
Ref.: https://en.wikipedia.org/wiki/Fatou%27s_lemma