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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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    $\begingroup$ 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.) $\endgroup$
    – Ian
    Aug 26, 2016 at 17:10
  • $\begingroup$ @Ian I want so say that I have $h$ cts s.t $\mid h-g \mid < \epsilon$ a.e given any $\epsilon$ $\endgroup$
    – user123124
    Aug 26, 2016 at 17:14
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    $\begingroup$ 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. $\endgroup$
    – Ian
    Aug 26, 2016 at 17:16
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    $\begingroup$ 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. $\endgroup$
    – Ian
    Aug 26, 2016 at 17:28
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    $\begingroup$ 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 $\epsilon < 1/4$. $\endgroup$
    – user169852
    Aug 26, 2016 at 18:16

1 Answer 1

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

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