Prove that $\lim_{n \rightarrow \infty} \int_0^\infty f_n(x) dx = \int_0^\infty f(x) dx$.
We can find for sufficiently large $n$
$$\begin{align}\left|\int_0^\infty f_n - \int_0^\infty f\right| &\leqslant \left|\int_t^T f_n - \int_t^T f\right| \\&+ \left|\int_0^t f_n\right| + \left|\int_0^t f\right| \\&+ \left|\int_T^\infty f_n\right| + \left|\int_T^\infty f\right| \leqslant \epsilon\end{align}$$
The first term on the RHS tends to $0$ as $n \to \infty$ since $f_n$ converges uniformly to $f$ on $[t,T]$. Hence, for any $\epsilon > 0$ and fixed $t,T$ we have for sufficiently large $n$,
$$\left|\int_t^T f_n - \int_t^T f \right| < \epsilon/5$$
Each of the remaining terms on the RHS can be made smaller than $\epsilon/5$ by choosing a sufficiently small $t$ and sufficiently large $T$ (independent of $n$) using, for example, the estimate
$$\left|\int_T^\infty f_n\right| \leqslant \int_T^\infty |f_n| \leqslant \int_T^\infty |g| \leqslant \epsilon/5,$$
where the last inequality follows from the convergence of $\int_0^\infty g$.