Assume $E \subset R_n$ and $f \in L(E)$. Show that for every $\epsilon > 0$ there is $\delta > 0$ such that $\int \limits _{E_0}|f| < \epsilon$ for every measurable $E_0 \subset E$ such that $|E_0| < \delta$. We are given that $|f| = \lim \limits_{k = \infty} f_k$ where each $f_k : E \rightarrow [0,\infty]$ is simple, Lebesgue measurable, and $f_k \leq f_{k+1}$.
This is confusing off the bat because the only $\epsilon - \delta$ definition that I know is: for $\epsilon > 0$ there is $\delta > 0$ such that $|f(x) - L| < \epsilon$ and $|x - x_0|< \delta$.
With that being said, my approach to this problem was to manipulate $\int \limits _{E_0}|f|$ and get some form of $a*|E_0|$, where $a$ is a constant. I know that $\int\limits _{E_0} a = a |E_0|$. The problem is I can't seem to get $\int \limits _{E_0}|f|$ to be the integral of a constant. Any ideas/suggestions?
EDIT: I think that I should start with $|\int \limits_{E_0}f(x) - L| <\epsilon$ and manipulate that first to get $\int \limits _{E_0}|f|<\epsilon$. Is this right?
EDIT (Attempted simple case soln, see user 15464 answer): For $\epsilon >0$ there is $\delta > 0$ such that $|E| < \delta$ if $\int \limits_E \phi < \epsilon$ $ \int \limits_E \phi < \epsilon \Rightarrow \int \limits_{E_i} a_i < \epsilon $ where $a_i$ is a value of $\phi$ on $E_i$ which is a disjoint subset of $E$. $ \int \limits_{E_i} a_i < \epsilon \Rightarrow a_i |E_i| < \epsilon \Rightarrow |E_i| < \frac {\epsilon}{a_i } $ then I can take the sum of all the subsets ($m \in N$) $ \sum \limits_{i=1}^{m} |E_i| = |E| < \sum \limits_{i=1}^{m} \frac {\epsilon}{a_i } = \frac {\epsilon}{\sum \limits_{i=1}^{m} a_i } = \epsilon \sum \limits_{i=1}^{m} \frac {1}{a_i } $ finally $\delta = \epsilon \sum \limits_{i=1}^{m} \frac {1}{a_i } > 0$