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?

  • 2
    $\begingroup$ It looks good, but you should stipulate that the $E_i$ are pairwise disjoint otherwise the last step is not necessarily true. $\endgroup$ – copper.hat May 20 at 0:43
  • 1
    $\begingroup$ 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. $\endgroup$ – 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

$$\int f - \epsilon < \int \phi = \int_E \phi \leq \int_E f$$

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