Lebesgue integral uniform convergence

Let $f_n, f \colon [a,b] \to \mathbb{R}.$

Show that, if $f_n \to f$ uniformly, then the Lebesgue integrals are equal, i.e. $\int f = \lim \int f_n$.

This is clearly true for continuous functions, but how do I handle the case of non-continuous functions?

Do you know how to do this argument for Riemann integrable functions? It's the same one for Lebesgue integrable functions. Hint: if $f_n \rightarrow f$ uniformly, then for all $\epsilon > 0$ and all sufficiently large $n$, $|f_n(x) - f(x)|$ is measurable and less than $\epsilon$ for all $x \in [a,b]$. What does that tell you about the Lebesgue integral $\int_{[a,b]} f_n -f$?

Note that the hypothesis that the total measure of the space be finite is essential. It is easy to construct a sequence of integrable continuous functions $f_n: [0,\infty) \rightarrow [0,\infty)$ which converges uniformly to $0$ but so that $\int_{[0,\infty)} f \rightarrow \infty$.

It follows from

$\left| \int f \, dm - \int f_n \, dm \right| \le \int \left|f - f_n\right| \, dm \le (b-a) \Vert f - f_n \Vert_\infty$