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Suppose I have a sequence of continuous functions converging pointwise almost everywhere to a continuous function. Is it true that we have pointwise everywhere convergence?

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I changed the question. Presumably it is some sort of diagonal argument? – dcs24 Jan 22 '13 at 14:53
No. Take $f_n(x)=x^n$ on $[0,1]$. This sequence converges pointwise almost everywhere to the zero function. But, $f_n(1)$ does not converge to $0$. – David Mitra Jan 22 '13 at 15:02
I was thinking about this example, you are right, I was hoping that knowing the limit function is everywhere continuous may prevent this. – dcs24 Jan 22 '13 at 15:05

Consider the class $[f]$ of all functions which are ae equal to $f$. You can show that each such class contains at most 1 continuous function (if $f \neq g$ at some point and both are continuous, you can find a neighborhood of that point where they're unequal). So yes, you can modify the functions on a null set, but you're no longer talking about a sequence of continuous functions. I'm not sure why you'd want to do this though. Did you have an application in mind?

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Yeah, he changed the question after I commented the same above. – Thomas Andrews Jan 22 '13 at 14:56
I do have an application in mind, my functions are convergent in $L^1$ (and members of $W^{1,1}(0,1)$, so continuous) giving me a.e. pointwise in a subsequence, I need to talk about convergence at specific points, so ideally I need everywhere ptw convergence, otherwise my construction will get messy. – dcs24 Jan 22 '13 at 14:58

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