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Let $\{f_{n}\}$ be a sequence of functions in $C[0,1]$ such that $f_{n}\to f$ pointwise. Then $f$ has at most finitely (or probably countably) many discontinuities. Is this statement TRUE or FALSE? Any kind of help will be appreciated.

Indrajit
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1 Answers1

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By Egorov's theorem, pointwise convergence almost everywhere implies uniform convergence on a set of measure arbitrarily close to $1$. Therefore $f$ is continuous except possibly on a set of measure $0$.

Edit: Disregard this, it's wrong. I apologize. See Aram's link below in comments, it answers the question. The link is Give an example of a sequence of continuous functions which converges on a compact set to a function that has an infinite number of discontinuities., from which we can see both that the above is wrong (by choosing C to be a nowhere dense set of positive measure, having the sequence converge to the characteristic function of C) and that the answer to your question is false, by noting that a set of positive measure is uncountable.

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Matt Samuel
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  • Thanks. But measure zero sets may not be countable, for example the Cantor set. If this happens then my statement is FALSE. Could give a counterexample then? – Indrajit Nov 01 '14 at 19:09
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    Egorov's theorem does not describe the "geometry" of the "small set" where uniform convergence does not happen, so it may be something very irregular (e.g. a dense susbset) which totally scrambles the continuity. – Milly Nov 01 '14 at 19:12
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    Wait, by the Baire-Osgood theorem the set of discontinuitys is a set of the first category, and sets of first category can have measure greater than zero.... in fact look at this http://math.stackexchange.com/questions/661221/give-an-example-of-a-sequence-of-continuous-functions-which-converges-on-a-compa – aram Nov 01 '14 at 19:13
  • You do have that the restriction of $f$ to $X \setminus N$ is continuous. But this says nothing about $f$ on $X$ as a whole. That is, if $f$ is continuous on $X$ then it is continuous on any subspace, but $f$ being continuous on a subspace of $X$ does not imply that it is continuous on the whole space. This is still true even if the subspace in question is dense, perhaps surprisingly. – Ian Nov 01 '14 at 19:17
  • @Aram Thanks for the link. It answers my question. – Indrajit Nov 01 '14 at 19:21