Problem 27: Show that the conclusion of Egoroff’s Theorem can fail if we drop the assumption that the domain has finite measure.

Solution: Consider sequence $f_n(x) = χ_{[n,\infty)}(x)$. Clearly $f_n \to 0$ pointwise on $\mathbb{R}$. Suppose that there existed a set $F$ such that $m(\mathbb{R} \setminus F) < \epsilon$ and $f_n \to 0$ uniformly on $F$. Since $f_n$ is an indicator function, this means we can find an $N$ such that $f_N = 0$ on $F$. This implies $F \subset (-\infty, N)$ and so $[N, \infty) \subset \mathbb{R} \setminus F$, giving $m([N, \infty)) \le m(\mathbb{R} \setminus F) < \epsilon$, which is a contradiction.

I don't understand the last line. What is being contradicted? Does $f$ not converge uniformly? Why?


It would have been clearer to write

$$\infty = m([N,\infty)) \leq m(\mathbb{R} \setminus F)<\epsilon.$$

The end result of this little calculation is that if $F$ is any set on which $f_n \to 0$ uniformly then $m(\mathbb{R} \setminus F)=\infty$.

It also would have been slightly better use of logic to fix some $\epsilon$ (since here $\epsilon$ is actually a totally unbound variable), but that's kind of nitpicking.


Lebesgue measure of $[N, +\infty)$ is infinite, so it cannot be less than $\epsilon$.

  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$
    – Guy Fsone
    Oct 19 '17 at 16:48
  • $\begingroup$ @GuyFsone I disagree, this does answer the question but I find it a bit too terse to be clear, which is why I posted a similar but somewhat longer answer. $\endgroup$
    – Ian
    Oct 19 '17 at 17:19

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