MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'll use the version of this question I posted on Stakexchange to replace the former version. To simplify the situaton, we assume the measure space we are dealing with right now is a closed interval $I$ of the real line.

When we are using the Egoroff's theorem, little attention has been paid to the exceptional set, i.e. the set $E\subset I$ on which $(f_n)$ fails to converge uniformly. The only thing we know about $E$ is that $m(E)$ can be arbitrary small. Taking the topology on the real line into account, we can also assume that $E$ is closed.

Recall that in the proof of this theorem, we constructed the exceptional set $E_{\varepsilon}$ with respect to a fixed $\varepsilon>0$ in the following way: \begin{equation}E_{\varepsilon}=\bigcup_{k=1}^{\infty}\bigcup_{i=n(k,\varepsilon)}^{\infty}\{x\in I:|f_{i}(x)-f(x)|\geq 1/k\},\end{equation} where $n(k,\varepsilon)$ is chosen so that $m(\bigcup_{i=n(k,\varepsilon)}^{\infty}\{x\in I:|f_{i}(x)-f(x)|\geq 1/k\})<\varepsilon/2^k$.

Now assume $(f_n)$ is a sequence of smooth functions, and $f\in L^1(I)$ is the limit function. Since $f$ is only determined up to a null set, the set $E_{\varepsilon}$ can only be determined up to a null set. Thus it is very natural to require $f$ to be the "best choice" to make $E_{\varepsilon}$ the as small as possible.

A satisfatory and notable case is that the family $\{E_{\varepsilon}\}$ be a sequence of nested closed intervals, thus we have $\bigcap_{\varepsilon}E_{\varepsilon}=\{x_{0}\}$, where $x_{0}\in I$. Or more generally, we can ask in which case the set $\bigcap_{\varepsilon}E_{\varepsilon}$ is a union of isolated points?

However, this is not true in general case, GTM 2 (p.38) contains an example in which case the set $I-E$ is nowhere dense, of course this coincides with the well-known result that every subset of the line can be represented as a union of a null set and a set of first category.

share|cite|improve this question
Your definition of $E_{n,k}$ doesn't mention $n$. I suppose you mean $E_{n,k} = \bigcup_{i=n}^\infty \{x \in E: |f_i(x) - f(x)| \ge 1/k \}$ – Robert Israel May 15 '11 at 18:13
Thank you. It's a mistake, I'll correct it. – Acky May 15 '11 at 18:22
@Acky: I don't know whether this question is research level or not, but if closed you may want it to be posted at – S.C. May 15 '11 at 18:33
Also $n(k)$ must depend on $\epsilon$ as well as $k$, so maybe it should be $n(k,\epsilon)$. – Robert Israel May 15 '11 at 20:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.