Let $(X, \mathcal A, \mu)$ be a measurable space.
$u \in \mathcal M_{\bar{\mathbb R}}^+ (\mathcal A)$ where $\mathcal M_{\bar{\mathbb R}}^+ (\mathcal A)$ denotes the families of positive extended real-valued measureable functions $c:(X,\mathcal A)\rightarrow(\bar{\mathbb R} , \mathscr B(\bar{\mathbb R})$, where $\mathscr B$ denotes Borel. (Note that $\bar{\mathbb R}$ is the closure of $\mathbb R$, i.e. $[-\infty, \infty]$).
Also let $(u_n)_{n=1}^\infty$ be a series of functions in $\mathcal M_{\bar{\mathbb R}}^+ (\mathcal A)$, under the following conditions:
$$ u_n \rightarrow u ~\text{for}~ n\rightarrow \infty, ~~~~~ \lim_{n\rightarrow \infty} \int_X u_n\, d\mu = c, ~~ c\in\mathbb R $$
How can I from the above determine that $\lim_{n\rightarrow \infty} u_n$ is finite?
My thinking
We know that $u_n$ is measurable, and that the limit of the integral is finite. I feel that I can somehow use this to show that $\lim_{n\rightarrow \infty} u_n$ is finite, but I am not sure how to go about it.
I don't understand the results in this SE post, but it might be useful?
It is likely meant that $u_n\to u$ pointwise $\mu$-almost-everywhere. Here is a suggested approach using Fatou’s lemma and Tchebychev’s inequality. All my integrals will be with respect to the measure $\mu$, so $\displaystyle\int(\dots)$ means $\displaystyle\int(\dots)\, d\mu$.
By Fatou's lemma, write $$\int u = \int \lim u_n = \int \liminf u_n \le \liminf \int u_n = \lim \int u_n = c.$$ If $N\ge 1$, by Tchebychev's inequality, $\displaystyle\mu\big( \{u > N\} \big) \le \frac{1}{N}\int u \le \frac{c}{N}$. Notice that $\{ u = \infty \} = \bigcap_{N=1}^\infty \{u > N\}$.
Do you see how to take it from here?