Suppose $\mu(X)<\infty$, $\{f_n\}$ is a sequence of bounded complex measurable functions on $X$, and $f_n \to f$ uniformly on $X$. Prove that $$\lim_{n\to \infty} \int_X f_n d\mu =\int_X f d\mu.$$
Is the condition of uniform convergence necessary? If it is, is it because $f_n$ are complex valued functions? (I haven't had complex analysis yet.)
I think if $\{f_n\}$ is bounded, then we can find a $g$ such that $|f_n|<g$. Since $g$ can be a constant function, $g\in L^1(\mu)$. Then by Lebesgue's dominated convergence theorem, we can get the result.
Let $a_n:= \sup \{|f_n(x)-f(x)|: x \in X\}$. Then $a_n \to 0$ and
$|\int_X f_n d \mu - \int_X f d \mu| \le \int_X |f_n -f| d \mu \le\int_X a_n d \mu =a_n \mu(X)$.