> [!theorem] > > Let $(X, \cm, \mu)$ be a finite [[Measure Space|measure space]], and $\seq{f_n}$ and $f$ be [[Borel Measurable Function|Borel measurable functions]] such that $f_n \to f$ [[Almost Everywhere|a.e.]], then for every $\varepsilon > 0$, there exists $E \subset X$ with $\mu(E) < \varepsilon$ such that $f_n \to f$ uniformly on $E^c$. > > *Proof*. Assume without loss of generality that $f_n \to f$ everywhere. For $k, n \in \nat$, let > $ > E_n(k) = \bigcup_{m \ge n}\bracs{x: \abs{f_m(x) - f(x)} \ge \frac{1}{k}} > $ > then for any fixed $k \in \nat$, $E_n(k) \downto \emptyset$, so $\mu(E_n(k)) \downto 0$ as $n \to \infty$. > > Let $\varepsilon > 0$ and $k \in \nat$, choose $n_k \in \nat$ such that $\mu(E_{n_k}(k)) < \varepsilon \cdot 2^{-k}$. Let $E = \bigcup_{k \in \nat}E_{n_k}(k)$, then $\mu(E) < \varepsilon$, and $\abs{f_n(x) - f(x)} < 1/k$ for all $n > n_k$ and $x \in E^c$. Therefore $f_n \to f$ uniformly on $E^c$.