# Integration of Nonnegative funtion, Real Analysis Chapter 2 problem 16

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

Proof: Let $\epsilon > 0$, and $f\in L^{+}$ where $f$ is simple, we can define $$\int f d\mu = \sup\{\int \phi d\mu: 0\leq \phi \leq f, \phi \ \ \text{simple}\}$$ Set $E_n = \{x:f(x) > 1/n\}$ where $n\in\mathbb{N}$, then $$\int_{E_n}f d\mu = \int f 1_{E_n}d\mu = \sum_{1}^{N}a_j\mu(E_j)$$ and so given an $\epsilon\in (0,\infty)$ there exists $n\in\mathbb{N}$ such that $$\int_{E_n}f d\mu > \int f d\mu - \epsilon$$

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

First of all you haven't proved that $\mu(E_n) <\infty$. This is true as $f$ is positive, $$\int f d\mu > \int_{E_n} f d\mu > \int_{E_n} \frac 1n d\mu = \frac 1n \mu(E_n).$$
To show your last claim, you can use the monotone convergence theorem: let $f_n = f 1_{E_n}$. Then $\{f_n\}$ is increasing and $f_n \to f$ pointwisely. So $$\int_{E_n} f d\mu = \int f_n d\mu \to \int fd\mu.$$ Thus the last claim can be clarified by writing down the above limit definition.