= Solution
The <dominated convergence theorem> says that if measurable $f_n\to f$ almost everywhere and $|f_n|\le g$ for an integrable $g$, then $f$ is integrable, $\int|f_n-f|\to0$, and consequently $\int f_n\to\int f$.
First derive <Fatou's lemma> from monotone convergence. For nonnegative $u_n$, the measurable functions $v_n=\inf_{j\ge n}u_j$ increase to $\liminf u_n$, and $\int v_n\le\inf_{j\ge n}\int u_j$. Monotone convergence therefore yields $\int\liminf u_n\le\liminf\int u_n$.
Now $f$ is measurable and $|f|\le g$ almost everywhere, so is integrable. Apply <Fatou's lemma> to $u_n=2g-|f_n-f|\ge0$, which converges almost everywhere to $2g$:
$$
\int2g\le\liminf_n\left(\int2g-\int|f_n-f|\right)
=\int2g-\limsup_n\int|f_n-f|.
$$
The last integrals are nonnegative, so they tend to zero. Finally $|\int f_n-\int f|\le\int|f_n-f|\to0$. This argument also works for complex-valued functions.
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