# Real Analysis, Folland problem 2.2.16 Integration of Nonnegative functions

If $f\in L^+$ and $\int f < \infty$, for every $\epsilon > 0$ there exists $E\in M$ such that $\mu(E) < \infty$ and $\int_E f > (\int f) - \epsilon$.

Attempted proof - 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 $$\int f - \epsilon < \int \phi$$ Note, we have a finite family of disjoint measurable sets $\{E_n\}_{n}$. Let $E = \bigcup_{n}E_n$ then $E\in M$ and for each $n$ $\mu(E_n) < \infty$ this $\mu(E) < \infty$. Note also we have that $\int \phi \leq \int f < \infty$ thus $$\int f - \epsilon < \int_{E}\phi \leq \int_{E}f$$

I am not sure if this is correct any suggestions is greatly appreciated.

• Looks nice to me.
– Vim
Commented Jun 25, 2016 at 1:05

Your proof is correct. It just need a small adjustment: to make explicit that the representation $\sum_{n}a_n \chi_{E_n}$ of $\phi$ being used satisfies the condition: for all $n$, $a_n>0$. I have also improved the wording in the end of the proof.

If $f\in L^+$ and $\int f < \infty$, for every $\epsilon > 0$ there exists $E\in M$ such that $\mu(E) < \infty$ and $\int_E f > (\int f) - \epsilon$.

Proof - 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 without loss of generality that, for all $n$, $a_n>0$ (just exclude any value of $n$ for which $a_n=0$).

Note, we have a finite family of disjoint measurable sets $\{E_n\}_{n}$. Let $E = \bigcup_{n}E_n$ 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 $\int_E \phi \leq \int_E f$ thus $$\int f - \epsilon < \int \phi =\int_{E}\phi \leq \int_{E}f$$

• why you exclude $a_n$ which are 0? Commented Oct 1, 2023 at 11:36
• @MeetPatel, Because if $a_n =0$, then this term does not contribute to $\phi$. The only terms whose coefficient is different to 0 contributes to $\phi$. Commented Oct 1, 2023 at 11:43
• Why we need to remove them? Commented Oct 1, 2023 at 11:44
• @MeetPatel A term like $0 \chi_{E_k}$ **does not** contribute to $\phi$, moreover $E_k$ may have infinite measure, without changing $\phi$. Adding those terms to $\phi$ would only make the wording of the proof more complicate and less elegant, because then in the definition of the family $\{E_n\}_n$, we would need to explicitly exclude the values of $n$ for which $a_n=0$, to ensure that all $E_n$ have finite measures. Commented Oct 1, 2023 at 12:03
• Since $0\cdot \infty=0$ .So despite of having $\mu (E_k)=\infty$ we can have $\int_E \phi <\infty$ . Got it Thank you Commented Oct 1, 2023 at 12:10