Dominated convergence theorem

Remarks

Proof idea: $\abs{f} \le g$ almost everywhere, so $f$ is integrable. The functions $2g - \abs{f_n - f}$ are non-negative almost everywhere and tend to $2g$. Fatou's lemma gives $\int 2g \le \int 2g - \limsup_n \int \abs{f_n - f}$, so $\limsup_n \int \abs{f_n - f} = 0$. The second limit follows from $\abs{\int f_n - \int f} \le \int \abs{f_n - f}$.

Example: $\int_0^1 x^n \, dx \to 0$, since $x^n \to 0$ for $x \in [0, 1)$ and $\abs{x^n} \le 1$, which is integrable on $[0, 1]$.

The dominating function is needed. For $f_n = n \mathbf{1}_{(0, 1/n)}$ on $\R$, $f_n \to 0$ everywhere but $\int f_n = 1$. Here $\sup_n f_n$ behaves like $1/x$ near $0$ and is not integrable.

Going deeper

For a student

This is the most used convergence theorem of analysis. It justifies differentiating under the integral sign, exchanging sums and integrals, and passing to the limit in many approximation arguments. For finite measure spaces, a constant bound is enough, which gives the bounded convergence theorem. The Vitali convergence theorem replaces domination by the weaker uniform integrability and gives a necessary and sufficient condition.