Given $\{f_k\}$, a sequence of bounded, complex-valued measurable functions, prove $\lim_{k \to \infty} \int_{\Omega} f_k dx = \int_{\Omega} f dx$

Question

So I am given the following to show:

Suppose $\{f_k\}$ is a sequence of bounded, complex-valued measurable functions defined on $\Omega$, with $\Omega$ having finite measure. Assume $f_k \to f$ uniformly. Prove $\lim_{k \to \infty} \int_{\Omega} f_k dx = \int_{\Omega} f dx$

I'm a bit lost at where to start. Our professor only covered integrals with real-valued, non-negative functions, and the textbook we are following only has a paragraph covering Lebesgue integrals with complex-valued functions defined specifically on $\mathbb{R^d}$. Any hints or advice on how to work with complex-valued sequences of functions would be greatly appreciated!

Answer 1

Hint: If $\varepsilon>0$ and if $N\in\mathbb N$ is such that$$(\forall n\in\mathbb{N})(\forall x\in\Omega):n\geqslant N\implies\bigl\lvert f(x)-f_n(x)\bigr\rvert<\varepsilon,$$then$$n\geqslant N\implies\left\lvert\int_\Omega f(x)\,\mathrm dx-\int_\Omega f(x)\,\mathrm dx\right\rvert<\varepsilon\times m(\Omega),$$where $m$ stands for the Lebesgue measure.

Answer 2

Given $\epsilon >0$ there exists $n$ such that $|f_k(x)-f(x)| <\epsilon$ for all $x \in \Omega$ for all $ k \geq n$. Now $|\int_{\Omega} f_k(x)\, dx -\int_{\Omega} f(x)\, dx| \leq \epsilon M$ for $k \geq n$ where $M$ is the measure of $\Omega$.