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Using this result Under what condition can converge in $L^1$ imply converge a.e.?. Can one infer that we can approximate an $L^1$ a.e by a continuous function on a finite measure space?

I.e find $h$ cts s.t $\mid h-g \mid < \epsilon $ a.e

Letting $=f_{n}=g-c_{n}$ where $c_{n}$ is a sequence of continuous approximating to $g$.

Other suggestions which are simplier if this is true would be appreciated!

I might have an alternative solution which is less messy; Since the measure is finite convergece in $L^1$ implies convergnce in measure which implies there is a subseq convering a.e we can pick/find some cts function from this subsequence that are arbitrarly close a.e to out $L^1$ function.

Finite measure was added after Ian's comments!

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user123124
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    Look up Lusin's theorem. (It's not quite what you want, but it's the thing that is true that is most similar to what you want.) – Ian Aug 26 '16 at 17:10
  • @Ian I want so say that I have $h$ cts s.t $\mid h-g \mid < \epsilon$ a.e given any $\epsilon$ – user123124 Aug 26 '16 at 17:14
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    Well...look up Lusin's theorem to see what you can actually get. It is not quite as nice as you might want but it is still pretty nice. You won't do any better than it without additional assumptions. – Ian Aug 26 '16 at 17:16
  • @Ian Yea Lucin is not gonna make it for me I think, its a nice theorem tho. Whats your take on my own approch? I dont think I have any additional hypoteses. – user123124 Aug 26 '16 at 17:21
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    If you can't deduce the "pointwise approximation of an $L^1$ function by continuous functions" that you want to do from Lusin's theorem, it's probably not true. – Ian Aug 26 '16 at 17:28
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    Consider $$g(x) = \begin{cases} 1 & \text{ if }x \geq 0 \ 0 & \text{ if }x < 0 \ \end{cases}$$ By the intermediate value theorem, any continuous approximation $h$ must assume the value $1/2$ somewhere, and therefore its values must be in the interval $(1/4, 3/4)$ on a set of positive measure. So your $|h-g| < \epsilon$ a.e. is impossible if $\epsilon < 1/4$. –  Aug 26 '16 at 18:16
  • @Bungo are you sure that we cant have some Cantor function like behaviuor? If you are right it implies a flaw in my second "proof", can you find the flaw? – user123124 Aug 26 '16 at 18:26
  • Your argument is fine to show that there is a subsequence which converges a.e. to $g$. However, convergence will not in general be uniform, so there need not be an $f_n$ which satisfies $|f_n - g| < \epsilon$ a.e. However, by Egorov's theorem, convergence almost everywhere (on a set of finite measure) implies almost uniform convergence, meaning that for any $\epsilon > 0$, there is a subset $E$ such that convergence is uniform on $E$, and the measure of $E^c$ is less than $\epsilon$. –  Aug 26 '16 at 18:30
  • @Bungo right, which is basicly Lusin's theorem I guess. – user123124 Aug 26 '16 at 18:31
  • @Inepsilonwetrust Yes, the two theorems are closely related. –  Aug 26 '16 at 18:32

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No, Lusins theorem seams to be as good as it gets.

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