# Proof of Egoroff's Theorem

Let $\{f_n \}$ be a sequence of measurable functions, $f_n \to f$ $\mu$-a.e. on a measurable set $E$, $\mu(E) < \infty$. Let $\epsilon>0$ be given. Then $\forall \space n \in \mathbb{N} \space \exists A_n \subset E$ with $\mu(A_n) <\frac{\epsilon}{2^n}$ and $\exists N_n$ such that $\forall \space x \notin A_n$ and $k \ge N_n \space |f_k(x) - f(x)| < \epsilon$. That is: if we define $A = \cup_{n=1}^{\infty} A_n$ with $\mu(A) < \epsilon$ then ${f_n}$ converges uniformly on $E \setminus A$.

$\mathbf{Proof}$: (taken from Royden's Real Analysis)

Let $A = \cup_{n=1}^{\infty} A_n \Rightarrow A \subset E$ and $\mu(A) < \sum_{n=1}^{\infty} \frac{\epsilon}{2^n} = \epsilon. \mathbf{Q1}$.

choose $n_0$ such that $\frac{1}{n_0} < \epsilon$. If $x \notin A$ and $k \ge N_{n_0}$ we have $\space |f_k(x) - f(x)| < \frac{1}{n_0} < \epsilon \space$. $\square$

$\mathbf{Q1}$: First off, I don't see how $\sum_{n=1}^{\infty} \frac{\epsilon}{2^n} = \epsilon$. It's a geometric series: $\epsilon \sum_{n=1}^{\infty} \frac{1}{2^n} = \epsilon \frac{1}{1-\frac12} = 2 \epsilon$. Am I wrong?

$\mathbf{Q2}$: The idea behind Egoroff is in order to turn almost sure convergence into uniform convergence on $E$ we only need to take away a really small set, right? Interestingly, as $\epsilon \to 0$ (that is $f_n$ is getting closer to $f$), the measure of the set $A$ is getting proportionally smaller ($\mu(A) \to 0$). So are we ultimately taking away a set of zero measure?

• A1: you're summing up from $n=1$, not $n=0$, so the sum is $\varepsilon$. – Damian Sobota May 21 '13 at 19:49
• What Royden edition is it? – Gaston Burrull May 21 '13 at 20:15
• @GastónBurrull I don't know. I got the book from the internet – shimee May 22 '13 at 5:29

A2: You are correct, that for arbitrarily small $\epsilon$ there is a set $A$, such that $\mu(A)<\epsilon$, where uniform convergence fails. So the measure of $A$ can be arbitrarily small. However, also keep in mind that uniform convergence requires a finite $N$ such that given $\delta>0$, for all $k>N$, $|f_k - f|<\delta$. Imagine that as $\epsilon$ gets smaller and smaller, for a fixed $\delta$ this $N$ may get larger and larger. Then in the limit as $\epsilon\to 0$, $N \to\infty$ and uniform convergence would fail.
• That's actually quite fascinating. Is it possible to find $\sup N < \infty$ for which we can get the uniform convergence (and for any larger $N$ it fails)? My guess would be no. – shimee May 22 '13 at 5:38
A1: we begin the sum at $1$, so the result is $\varepsilon$ (even if we had got $2\varepsilon$ it wouldn't matter).
A2: the problem comes from the dependence of the set $A$ and $\varepsilon$. Things are more concrete when we think about $X=[0,1]$ and $f_n(x)=x^n$.