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Here is a statement of Egoroff's Theorem:

Egoroff's Theorem. Suppose that $\mu(X) < \infty$, and $f, f_1, f_2, f_3, \ldots$ are all measurable complex-valued functions on $X$ such that $f_n \to f$ almost everywhere. Then for every $\varepsilon > 0$, there exists $E \subseteq X$ such that $\mu(E) < \varepsilon$ and $f_n \to f$ uniformly on $E^c$ (the complement of $E$ in $X$).

Is it possible to replace the "$\mu(E) < \varepsilon$" condition by "$\mu(E) = 0$" in the statement and still have the theorem be true? I suspect that the answer is no, but I am having trouble thinking of an appropriate counterexample.

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  • $\begingroup$ Patrick Da Silva has already given an example, but if you want a more extended discussion about Egoroff's Theorem, see the essay I posted in this thread: math.stackexchange.com/questions/15088 (moments later) Oops, this essay is mostly on Luzin's Theorem. Nonetheless, it should still be fairly relevant. $\endgroup$ – Dave L. Renfro Feb 16 '12 at 15:00
  • $\begingroup$ Many times the condition $\mu(E)\lt \epsilon$ is all that we need. See here for a example. $\endgroup$ – leo Feb 16 '12 at 15:06
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Let $X = [0, \pi/2)$ and $f_n(x) = \frac 1n \tan(x)$. Clearly $f_n \to 0 = f(x)$. Suppose $\mu(E) = 0$ with $E \subseteq X$. Then since for every $\delta > 0$, $\mu( (\pi/2 - \delta, \pi/2) ) = \delta > 0$, then $\mu( (\pi/2 - \delta, \pi/2) \cap E^c ) = \delta > 0$ (assuming $\delta < \pi/2$ obviously) and thus you have a sequence of points $x_m$ such that $x_m \in E^c$ and $x_m \to \pi/2$. But $f_n(x_m) \to \infty$ as $m \to \infty$, so that $\sup_{X \cap E^c} |f_n(x) - f(x)|$ can never even be finite, so you can't possibly hope it goes to zero as $n \to \infty$. Therefore convergence is not uniform.

Hope that helps,

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  • $\begingroup$ Actually I didn't need $\tan$ : I just needed a function that went to infinity at some point and then multiply it by $1/n$ and take a neighborhood close enough to it. I could've taken $g(x) = 1/x$, $X = (0,1)$ and $f_n(x) = (1/n) g(x)$, a similar argument would've worked too. $\endgroup$ – Patrick Da Silva Feb 16 '12 at 4:38
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Here is a simpler example $f_{n}:(0,1)\rightarrow \mathbb{R}$ such that $f_{n}(x)=x^n$, $f_{n}(x)\rightarrow 0$ but $f_{n}$ cannot converges uniformly outside any set of measure zero.

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  • $\begingroup$ You replace my issue at infinity by an issue of discontinuity... it's a matter of choice whether it is simpler to have an infinity-kind of discontinuity or a gap. But still you gave the other point of view, so I'm happy to +1 :) (A bit late though, but I admit I didn't read it back then I guess) $\endgroup$ – Patrick Da Silva Jun 10 '13 at 3:08

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