# Can we approximate an $L^1$function pointwise almost everywhere by a continous function?

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.

• @Ian I want so say that I have $h$ cts s.t $\mid h-g \mid < \epsilon$ a.e given any $\epsilon$ – Vlad Aug 26 '16 at 17:14
• 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
• 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 $\epsilon < 1/4$. – Bungo Aug 26 '16 at 18:16