# On a finite measure space, $f_n \to f$ uniform and $f$ integrable implies $f_n$ integrable and $\int_{\Omega}f_nd\mu\to\int_{\Omega}fd\mu$

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 $$\int_{\Omega}f_nd\mu\to\int_{\Omega}fd\mu.$$

So we know that $\mu(\Omega)<\infty$, $\int_{\Omega}fd\mu<\infty$ and that $$\forall \epsilon>0 \quad \exists N>0 \quad \text{s.t.} \quad n\geq N \quad \implies \quad |f_n(\omega)-f(\omega)|<\epsilon\quad\forall\omega\in\Omega$$ I think these together imply that $\int_{\Omega}|f_n-f|d\mu<\infty$ and since $$\int_{\Omega}|f_n-f|d\mu\leq\int_{\Omega}|f_n|d\mu-\int_{\Omega}|f|d\mu<\infty,$$ $\int_{\Omega}|f_n|d\mu<\infty$, 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.

If $f_n \to f$ pointwise, then the result is not true. Consider the sequence of functions $f_n \colon [0,1]\to\mathbb{R}$: $$f_n(x) := \begin{cases} n^2x, & \text{if}\ x\in [0, 1/n],\\ 2n - n^2 x, & \text{if}\ x \in (1/n, 2/n),\\ 0, & \text{if}\ x\in [2/n, 1]. \end{cases}$$ This sequence converges pointwise to $f\equiv 0$, but $\int_0^1 f_n = 1$ for every $n$.
Assume now that $f_n \to f$ uniformly on $\Omega$. Then $$\left|\int_\Omega f_n \, d\mu - \int_\Omega f\, d\mu\right| \leq \int_\Omega |f_n - f|\, d\mu \leq \mu(\Omega) \, \|f_n - f\|_\infty \to 0.$$