I try to prove the following.
Let ($\Omega, \mathcal{A}, \mu)$ be a finite measure space and suppose $f_n:\Omega\to\mathbb{R}$, $n\in\mathbb{N}$ and $f:\Omega\to\mathbb{R}$ are measurable functions. In addition, assume $f$ is integrable. Show that, if $f_n\to f$ uniform, then, for large $n$, $f_n$ is integrable and \begin{equation} \int_{\Omega}f_nd\mu\to\int_{\Omega}fd\mu. \end{equation}
So we know that $\mu(\Omega)<\infty$, $\int_{\Omega}fd\mu<\infty$ and that \begin{equation} \forall \epsilon>0 \quad \exists N>0 \quad \text{s.t.} \quad n\geq N \quad \implies \quad |f_n(\omega)-f(\omega)|<\epsilon\quad\forall\omega\in\Omega \end{equation} I think these together imply that $\int_{\Omega}|f_n-f|d\mu<\infty$ and since \begin{equation} \int_{\Omega}|f_n-f|d\mu\leq\int_{\Omega}|f_n|d\mu-\int_{\Omega}|f|d\mu<\infty, \end{equation} $\int_{\Omega}|f_n|d\mu<\infty$, thus $f_n$ is integrable for large enough n. I hope that the second result also follows from something like this, and that this argument is correct. Any hints or tips are greatly appreciated, thank you.