Suppose $\{f_n(x)\}$ is a sequence of extended real-valued functions. I say that this sequence converges uniformly to $f(x)$ provided for every $\epsilon > 0$ there exists $N$ such that if $n > N$ then (1) if $f(x)$ is finite, $|f_n(x) - f(x)| < \epsilon$; (2) if $f(x) = \infty$ then $f_n(x) > 1 / \epsilon$; (3) if $f(x) = -\infty$ then $f_n(x) < -1 / \epsilon$.

Under this definition I believe that Egoroff's Theorem remains true if the sequence of functions and their pointwise a.e. limit have their ranges in the extended real numbers. (The domains are assumed to be a finite measure space, of course.) Is this correct?


1 Answer 1


Yes, this is true. Egoroff's Theorem can be generalized to functions taking values in a separable metric space. You can equip $\overline{\mathbb{R}}$ with a suitable metric so that it becomes homeomorphic to $[-1,1]$, for instance you could take $d(x,y) = 2|\arctan(x) - \arctan(y)|/\pi$ with $\arctan(\pm \infty) := \pm \pi/2$. Then convergence with respect to that metric agrees with your notion.

  • $\begingroup$ Thanks. That's pretty much what I was thinking. Just wanted to make sure I wasn't deluding myself. $\endgroup$ Apr 12, 2013 at 17:40

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