I came up with a proof of Egorov's Theorem which I can't find in my books or on the net, which makes me think it's wrong, which means I have a misunderstanding somewhere which needs to be hammered out.

Egorov's Theorem: Let $(X,M,\mu)$ be a finite measure space and $f_n$ a sequence of measurable functions on $X$ that converges pointwise a.e. on $X$ to a function $f$ that is finite a.e. on $X$. Then for each $\epsilon > 0$, there is a measurable subset $X_{\epsilon}$ of $X$ for which

$f_n\rightarrow f$ uniformly on $X_{\epsilon}$ and $\mu(X\backslash X_{\epsilon})<\epsilon$


Let $\epsilon > 0$.

So first I let $E_0$ be the set of points at which $f_n$ does not converge to $f$, union the set of points at which $f$ is infinite. Then $\mu(E_0)=0$. Now define

$$A_N = \{x\in X\backslash E_0 : |f_N(x)-f(x)|<\epsilon\},$$

and set $E_k = \bigcap_{N=k}^{\infty}A_N$. Then since the $A_N$ are measurable and $E_k$ is the countable intersection of measurable sets, then the $E_k$ are measurable. Furthermore the $E_k$ are ascending, since as $k$ grows one is intersecting fewer and fewer of the $A_N$.

Thus since $\mu(E_0)=0$ (to justify the first equality) and by the continuity of measure (to justify the third), we have:

$$\mu(X)=\mu(X\backslash E_0)=\mu(\bigcup_{k=1}^{\infty}E_k)=\lim_{k\rightarrow\infty}\mu(E_k).$$

Where the second equality holds since for every $x\in X- E_0$ there exists some $K$ such that for all $n > K$, $|f_n(x)-f(x)| < \epsilon$ which is what it means for $x\in E_K$; the reverse inclusion being trivial.

Therefore since $\lim_{k\rightarrow\infty}\mu(E_k)=\mu(X)$, we can find a $K$ such that $\mu(X\backslash E_K)=\mu(X)-\mu(E_K)<\epsilon$, with $f_n\rightarrow f$ uniformly on $E_K$.

Is this correct?

  • $\begingroup$ Why is $f_n\to f$ uniformly on $E_K$? $\endgroup$ – Norbert May 25 '13 at 18:51
  • $\begingroup$ you're right Norbert it isn't. $\endgroup$ – Thoth May 25 '13 at 19:01

I don't think this is correct, no. I don't see why $f_n\to f$ uniformly on $E_K$. Certainly it's true that on $E_K$, for $n\ge K$ and any $x$, $|f_n(x) - f(x)| < \varepsilon$, but $\varepsilon$ was fixed at the beginning; this says nothing about convergence. Because what happens if you make epsilon smaller? If this just raised $K$, then you'd have uniform converge; but the problem is that changing $\varepsilon$ also changes all your sets and so changes the set $E_K$ -- you're not talking about convergence on the same set anymore!

So no, this doesn't appear to be right.

(A note: A warning sign that made me suspect it was probably wrong was when you used $\varepsilon$ to define how close $f_n$ and $f$ had to be -- $\varepsilon$ is a measure, so taking it as a distance is unlikely to be meaningful. Certainly clever arguments that involve doing things like that sometimes work, but it's definitely a warning sign.)

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  • $\begingroup$ Ah yes I see what you mean. Yah I think you're right. So the lesson learned is that I don't want the size of my set to depend on epsilon. $\endgroup$ – Thoth May 25 '13 at 18:57

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