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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 $$\delta = \epsilon \sum \limits_{i=1}^{m} \frac {1}{a_i } > 0$$

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1 Answer 1

Notice that if $\phi$ is a simple function, then $\phi$ takes on only finitely many values, so in particular $|\phi|$ is bounded. Can you use this to prove the result in the case of a simple function?

Once you have the result for a simple function, and you want to extend it to the general $f$, choose a simple function $\phi$ with $\int |f - \phi| < \epsilon/2$. Then choose $\delta$ such that if $|E| < \delta$, then $\int_E |\phi| < \epsilon/2$. Write

$$\int_E |f| \leq \int_E |f - \phi| + \int_E |\phi|$$

and bound each term separately.

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I just added my attempted solution to the simple case. Am I on the right track? –  rioneye Oct 18 '12 at 5:19
    
You seem to be overthinking this. Let $a$ be the maximum value of $\phi$. Then if $|E| < \epsilon/a$, $\int_E \phi \leq a|E| < \epsilon$. –  user15464 Oct 18 '12 at 21:11

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