---- A [[sequence]] of functions that [[pointwise converge|converges pointwise]] need not [[uniform convergence|converge uniformly]]. The following result specifies conditions under which such *almost* happens. > [!theorem] Theorem. ([[Egorov's Theorem]]) > > - Let $(X, \Sigma, \mu)$ be a [[measure|measure space]] and $M$ a [[separable space|separable]] ($\iff$[[second-countable space|second-countable]]) [[metric space]]. > - Let $(f_{n})$ be a [[sequence]] of [[measurable function|measurable functions]] $X \to M$. > - Suppose there is a [[σ-algebra|measurable subset]] $A \in \Sigma$ with $\mu(A)<\infty$, such that $(f_{n})$ [[converges almost-everywhere|converges a.e.]] on $A$ to a function $f:X \to M$. Then: > > For all $\varepsilon>0$, there exists a [[σ-algebra|measurable subset]] $B \subset A$ such that $\mu(A-B)<\varepsilon$ and $(f_{n})$ [[uniform convergence|converges uniformly]] to $f$ on $B$. ^4a988a > [!proof]- Proof. ([[Egorov's Theorem]]) > First note that, since $M$ is [[separable space|separable]] ($=$[[second-countable space|second-countable]] as it is [[metrizable]]), $\mathcal{B}(M \times M)=\mathcal{B}(M) \otimes \mathcal{B}(M)$ and so the function $x \mapsto \big( f(x), g(x) \big) \mapsto d\big( f(x), g(x) \big)$ is [[measurable function|Borel measurable]]. > > > > > Fix $\varepsilon>0$. We'll form $B$ from those points $x \in A$ for which '$\big(f_{n}(x)\big)$ converges too slowly'. or $n,k \in \mathbb{N}$ define $E_{n,k}:=\bigcup_{m \geq n}\left\{ x \in X: d\big( f_{n}(x), f(x) \big) \geq \frac{1}{k} \right\}.$ > Note that, for a fixed $k$, $E_{1,k} \supset E_{2,k} \supset \cdots$ We have assumed $\mu(A)<\infty$, so [[measure|continuity from above]] applies: $\mu(\bigcap_{n=1}^{\infty}E_{n,k})= \lim_{n \to \infty}\mu(E_{n,k}).$The points that survive in the filtration $E_{1,k} \supset E_{2,k} \supset \cdots$ for all $n$ are those points $x$ for which $f_{n}(x) \not \to f(x)$. By the assumption of a.e. pointwise convergence, then: $\mu(\bigcap_{n=1}^{\infty}E_{n,k})=0.$ > Putting these together, we have that, for each $k$, there exists $n_{k}$ such that $\mu(E_{n_{k},k})< \frac{\varepsilon}{2^{k}}$. Define $B:=\bigcup_{k=1}^{\infty}E_{n_{k},k },$ > observing that $(f_{n}) \to f$ [[uniform convergence|uniformly]] on $A-B$.[^1] Then using [[measure|countable subadditivity]] and [[geometric series]], we have $\mu(B) \leq \sum_{k=1}^{\infty} \mu(E_{n_{k}, k}) < \sum_{k=1}^{\infty} \frac{\varepsilon}{2^{k}}=\varepsilon.$ > [^1]: Explicitly, for $\epsilon>0$, let $\frac{1}{k}<\epsilon$, then for any $n>n_{k}$ we have $d(f_{n}(x), f(x))<\epsilon$ for all $x \in A-B$. ---- #### ----- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```