Integral on a measurable set is almost the integral on the whole set

This is problem 16 Chapter 2 in Folland's Real Analysis.

"If $$f\in L^+$$ and $$\int f<\infty$$, for every $$\epsilon>0$$, we can find $$E$$ measurable s.t. $$\mu(E)<\infty$$ and $$\int_E f>\int f -\epsilon.$$"

THere is a solution online using Monotone COnvergent theorem. However, I use a solution that is more simple. So I think something must be wrong.

By definition, there exists a simple function $$0\leq \phi = \sum_1^n a_i \chi_{E_i} \leq f$$ s.t. $$\int f -\epsilon < \int \phi$$ and here $$E_i$$ are disjoint and $$a_i$$ are nonzero. Hence $$\int f -\epsilon < \int \sum_1^n a_i \chi_{E_i} = \sum \int_{ E_i}a_i \leq \sum \int_{E_i} f= \int_{\cup E_i} f$$

Then the set we need is $$\cup E_i$$. Suppose $$\mu (\cup E_i)=\infty$$. Then $$\int \phi=\infty$$ which is a contradiction.

This proof goes directly from definitions. So I think this is too good to be true. Am I wrong somewhere?

• It looks good, but you should stipulate that the $E_i$ are pairwise disjoint otherwise the last step is not necessarily true. – copper.hat May 20 at 0:43
• You should probably also mention that in your simple function decomposition, the $a_i$'s are nonzero, or handle the case where one or more of the $a_i$'s may be zero. Aside from that it looks fine to me. – Bungo May 20 at 1:33

Let $$f\in L^{+}$$ and $$\int f < \infty$$. Let $$\epsilon > 0$$, by definition of $$\int f$$, there exists a simple function $$\phi = \sum_{n}a_n \chi_{E_n}$$ such that $$0\leq \phi \leq f$$ and
$$\int f - \epsilon < \int \phi$$
We can assume wlog that for all $$n$$, $$a_n > 0$$. Note we have a finite family of disjoint measurable sets $$\{E_n\}_{n}$$. Let $$E = \bigcup E_n$$ then $$E\in M$$. Since
$$\int \phi \leq \int f < \infty$$
and for each $$n$$, $$a_n > 0$$, we have for each $$n$$, $$\mu(E_n) < \infty$$ and so $$\mu(E) < \infty$$. Since $$0 \leq \phi \leq f$$, we have that $$\int_E \phi \leq \int_E f$$ and
$$\int f - \epsilon < \int \phi = \int_E \phi \leq \int_E f$$