14:09
@MartinSleziak Thanks for the links. I think I came up with a proof based upon them, but mine appears slightly different, so I was wondering if you'd take a quick look at them...Just give me a few minutes to type it up....
First recall Lusin's theorem: If $f : E \to \Bbb{R}$ is measurable, then for every $\epsilon > 0$, there is a continuous function $g : \Bbb{R} \to \Bbb{R}$ and a closed set $F \subseteq E$ for which $f=g$ on $F$ and $m(E-F) < \epsilon$.
Recall the following theorem about integrability: If $f : E \to [-\infty, \infty]$ is measurable and integrable, then for each $\epsilon >0$, there is a $\delta > 0$ for which if $A \subseteq E$ is measurable and $m(A) < \delta$, then $\int_A |f| < \epsilon$.
