7
$\begingroup$

Egorov's Theorem: Let $(f_n)$ be a sequence of measurable functions converging pointwise a.e. to a real-valued function f on a measurable set $D$ of finite meausre. Then, given $\epsilon>0$, there is a measurable set $E \subset D$ such that $m(E)<\epsilon$ and such that $(f_n)$ converges uniformly to $f$ on $D$\ $E$.

Can somebody give me an example of where this fails if $m(D)= \infty$?

My book goes on to say that almost uniform convergence always implies a.e. pointwise convergence, can somebody help me to understand why this is true also?

Thanks!

$\endgroup$
2
  • 1
    $\begingroup$ See the example in the question here (which btw I found by googling "counterexample egorov's theorem". You could try that first next time.) $\endgroup$ Commented Jun 19, 2018 at 22:18
  • 1
    $\begingroup$ See the wikipedia page for egorov's theorem! $\endgroup$
    – SBK
    Commented Jun 19, 2018 at 22:18

2 Answers 2

8
$\begingroup$

$I_{(n,\infty )} \to 0$ almost everywhere on $(0,\infty )$ with Lebesgue measure. If $f_n \to 0$ uniformly on $E^{c}$ then $E^{c} \subset (0,N]$ for some $N$ which makes $m(E)=\infty $.

$\endgroup$
4
$\begingroup$

Consider $(0, +\infty)$ with the Lebesgue measure $\lambda$. Define $f_n : (0, +\infty) \to \mathbb{R}$ as $f_n(x) = \sum_{k=1}^n \frac{x^k}{k!}$.

Then $(f_n)_n$ converges pointwise to the exponential function $e^x = \sum_{k=1}^\infty \frac{x^k}{k!} $ on $(0, +\infty)$.

Assume $f_n \to \exp$ uniformly on a measurable set $E \subseteq (0, +\infty)$. Then for $\varepsilon = 1$ there exists $n \in \mathbb{N}$ such that

$$\sum_{k=n+1}^\infty \frac{x^k}{k!} = \left|e^x - \sum_{k=1}^n \frac{x^k}{k!}\right| = \left|e^x - f_n(x)\right| < 1$$

for all $x \in E$. Since $\frac{x^{n+1}}{n+1}\le \sum_{k=n+1}^\infty \frac{x^k}{k!}$, we conclude that $\frac{x^{n+1}}{n+1} < 1$ for all $x \in E$ so $E \subseteq \left(0, \sqrt[n+1]{(n+1)!}\right)$.

Therefore $E^c$ always has infinite measure so we cannot pick $E$ such that $\lambda(E^c) < \varepsilon$.

$\endgroup$

You must log in to answer this question.

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