> [!lemma|*] Schéffe's Lemma > If $(f_{n}) \subseteq \mathcal{L}^{1}(\Omega, \mathcal{F}, \mu)^{+}$ and $f_{n} \to f ~\mathrm{a.e.}$, then $\mu(|f_{n}-f|) \to 0 \iff \mu (f_{n}) \to \mu(f).$For general functions $(f_{n}) \subseteq \mathcal{L}^{1}(\Omega, \mathcal{F},\mu)$, $\mu(|f_{n}-f|) \to 0 \iff \mu (|f_{n}|) \to \mu(|f|).$ > >[!proof]- First Part > > $[\Rightarrow]$ if trivial. > > $[\Leftarrow]$ Note that $|f_{n}-f| \le f+f_{n}$, so we may apply Fatou to $f+f_{n}-|f_{n}-f| \ge0$. This gives $\begin{align*} \mu(2f)&= \mu\left[ \liminf_{n}(f+f_{n}-|f-f_{n}|) \right] &\mathrm{a.e.}\text{ convergence}\\ &\le \liminf_{n}\Big[ \mu(f + f_{n} - |f-f_{n}|) \Big] &\text{Fatou}\\ &= \mu(f) + \lim_{ n \to \infty } \mu(f_{n}) - \limsup_{n}\mu(|f-f_{n}|)\\ &= 2\mu(f)-\limsup_{n}\mu(|f-f_{n}|) &\text{assumption (RHS)} \end{align*}$ Hence $\limsup_{n}\mu(|f-f_{n}|) \to 0$.