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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.

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    $\begingroup$ Looks nice to me. $\endgroup$ – Vim Jun 25 '16 at 1:05
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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 $$\int f - \epsilon < \int \phi =\int_{E}\phi \leq \int_{E}f$$

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