Show $f$ is in $L^1$ (d$\mu$) space and $_X\int f$ d$\mu=\lim_{n\to \infty}\int_X f_n d\mu$

Suppose $f$ is in $L^1$($\mu$). Prove that for each $\epsilon > 0$ , there exists a $\delta > 0$ so that the $\int |f|\mathrm d\mu$ < $\epsilon$ over the set $E$ whenever $\mu(E) < \delta$. This has been proved. Now, Suppose we have a sequence of non-negative integrable functions $\{f_n\}$ for which the above property is satisfied uniformly for $\epsilon$ and $\delta$, and that the measure space is finite. Suppose $\{f_n\}$ converges to $f$ a.e.

Show $f$ is in $L^1$(d$\mu$) space and $\int_X f\mathrm d\mu= \lim_{n\to \infty}\int_X f_n\mathrm d\mu$.

• Do you know Egoroff's theorem? – Davide Giraudo May 10 '13 at 15:39
• yes, but not sure how that relates to showing that f is in the $L^1$(d$\mu$) space – Real Anal May 10 '13 at 15:48
• That's what I don't understand: I only modified formatting, and in your first sentence, you write Suppose $f$ is in $L^1$. In your last sentence, Show $f$ is in $L^1(\mu)$. – Davide Giraudo May 10 '13 at 15:56
• I just edited the question to show it as it was presented to me. – Real Anal May 10 '13 at 16:06
• I'm guessing that $L^1$(d$\mu$) is a subset of $L^1$($\mu$) and maybe i could use Egorov – Real Anal May 10 '13 at 16:12