3
$\begingroup$

Egorov's Theorem

Let $(f_n)$ be a sequence of measurable functions converging pointwise almost everywhere to a real-valued function $f$ on a measurable set $D$ of finite measure. Then for $\epsilon>0$, there is a measurable set $E\subset D$ such that $m(E)<\epsilon$ and such that $(f_n)$ converges uniformly to $f$ on $D\setminus E$.

Proof given in Carother's Real Analysis

Suppose $f_n\to f$ everywhere on $D$. For each $n,k$, consider

$$E(n,k)=\bigcup_{m=n}^{\infty}\left\{x\in D: |f_m(x)-f(x)|\geq \frac{1}{k}\right\}$$

If $k$ is fixed and $n\to\infty$, these sets $E(n,k)$ decrease. $\color{blue}{\text{In fact, $\cap_{n=1}^{\infty}=\emptyset$, since $f_n\to f$ everywhere on $D$.}} $ Since $m(D)<\infty$, we have $m(E(n,k))\to 0$ as $n\to \infty$. $\color{blue}{\text{Then we may choose a subsequence $(n_k)$ for which $m(E(n_k,k))<\frac{\epsilon}{2^k}$.}}$ Now if we set $E=\cup_{k=1}^{\infty}E(n_k,k)$, $\color{blue}{\text{then $m(E)<\epsilon$}}$. What's more, for $x\notin E$, we have $x\notin E(n_k,k)$ for all $k$. In particular $|f_m(x)-f(x)|<\frac{1}{k}$ for all $m\geq n_k$. $\color{blue}{\text{Hence $f_n$ converges uniformly to $f$ on $D\setminus E$.}}$

The parts in blue are the parts where I have some questions about.

Questions:

  • Why does $f_n\to f$ make the infinite intersection of the $E(n,k)$ have to be empty?
  • How are we guaranteed the existence of a subsequence (does this follow from the fact that $f_n\to f$?)
  • Why is $m(E)<\epsilon$? (Is this because $E(n,k)$ is decreasing?)
  • Why is the convergence uniform? Isn't it just pointwise? The proof still hasn't shown that $\sup_{x\in E}|f_m(x)-f(x)|<\frac{1}{k}$, has it?
$\endgroup$
2
  • $\begingroup$ The second blue phrase is not a complete sentence. Something missing? $\endgroup$
    – BaronVT
    Commented Feb 27, 2015 at 17:36
  • 1
    $\begingroup$ @BaronVT: Thanks for the catch! I've fixed it. $\endgroup$ Commented Feb 27, 2015 at 17:38

2 Answers 2

1
$\begingroup$
  • Show that, for every $x$, there is a set $E(n,k)$ that $x$ is not in. In particular, for fixed $k$, let $N$ be such that $|f_n(x) - f(x)| < \dfrac1k$ if $n \geq N$.

  • For each fixed $k$, $m(E(n,k)) \to 0$. That is, for every $\epsilon > 0$, there is $n$ (depending on which $k$ we have fixed, so call it $n_k$ instead of $N$ like we did in the last paragraph) so that $m(E(n,k)) < \epsilon$ if $n \geq n_k$. Replace $\epsilon$ by $\epsilon/2^k$.

  • Since $E$ is the countable union of these sets, its measure is less than or equal to the sum of the individual measures. What is the sum of $\epsilon / 2^k$ for $k = 1,2,\dots$?

  • This follows from $x\not\in E(n_k,k)$ for all $k$ (for fixed $k$, choose any $x$ such that $x \not\in E(n_k,k)$ for all $k$, then $|f_m(x) - f(x)| < \dfrac1k$ if $m \geq n_k$). In particular, $E$ is where the convergence is not uniform.

$\endgroup$
1
$\begingroup$
  1. Why does $f_n\to f$ make the infinite intersection of the $E(n,k)$ have to be empty?

Because, for all $x$, there exist a $n_x$ such that for all $ m>n, |f_m(x)-f(x)| < \frac{1}{k}$. Then $E(n_0,k)$ doesn't contain x. There is no x that is in all the $E(n,k)$

  1. How are we guaranteed the existence of a subsequence (does this follow from the fact that $f_n\to f$?)

It follow from the fact that $m(E(n,k)) \to 0$

  1. Why is $m(E)<\epsilon$? (Is this because $E(n,k)$ is decreasing?)

It's because each E(n,k) has measure $< \frac{\epsilon}{2^k}$

$$M( \cup_k E(n_k,k) ) \leq \sum_k M( E(n_k,k) ) \leq \epsilon \sum_k \frac{1}{2^k}$$

  1. Why is the convergence uniform? Isn't it just pointwise? The proof still hasn't shown that $\sup_{x\in E}|f_m(x)-f(x)|<\frac{1}{k}$, has it?

It has. Notice that n_k doesn't depend on x, it's the same for all the x in $E^c$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .