Vitali convergence theorem

Remarks

Proof idea for 1 implies 2: uniform integrability bounds $\int \abs{f_n}$, so $f$ is integrable by Fatou's lemma. Split $\int \abs{f_n - f}$ into the part on a set where $f_n \to f$ uniformly, given by Egorov's theorem, and the part on a small set. Uniform integrability makes the second part small, uniformly in $n$. For 2 implies 1, a sequence converging in $L^1$ is uniformly integrable, because finitely many integrable functions are, and the tail is close to $f$.

Example: if $\sup_n \int \abs{f_n}^2 \, d\mu < \infty$ and $f_n \to f$ almost everywhere, then the family is uniformly integrable and $f_n \to f$ in $L^1$.

The finiteness of $\mu$ is needed in this form. On $\R$ with Lebesgue measure, $f_n = \frac{1}{n} \mathbf{1}_{[0, n]}$ satisfies the condition of uniform integrability, since $\abs{f_n} \le 1$, and tends to $0$ everywhere, but $\int \abs{f_n} = 1$ for all $n$. The theorem also holds with convergence in measure in place of convergence almost everywhere.

Going deeper

For a student

The theorem refines the dominated convergence theorem: a dominated sequence is uniformly integrable, but not conversely. It is the form of the convergence theorem used in probability, where one shows uniform integrability with a moment bound such as $\sup_n \mathbb{E}[\abs{Y_n}^p] < \infty$ for some $p > 1$. There are versions for $L^p$ and for infinite measure spaces, where a condition of tightness is added.