Real Analysis, Folland Problem 2.2.14 Integration of Nonnegative functions 
Problem 2.2.14 - If $f\in L^{+}$, let $\lambda(E) = \int_{E}f d\mu$ for $E\in M$. Then $\lambda$ is a measure on $M$, and for any $g\in L^{+}$, $\int g d\lambda = \int f g d\mu$.(First suppose that $g$ is simple)

Attempted proof - 
Observe that $\lambda(\emptyset) = \int_{\emptyset}f d\mu = \int 1_{\emptyset} f d\mu = \int 0 f d\mu = 0$. Let $\{E_n\}_{n\in\mathbb{N}}\subset M$ and let $F = \bigcup_{n=1}^{\infty}E_n$. Then $$\lambda(F) = \int_{F}f d\mu = \int 1_{F}f d\mu = \int \left(\sum_{n=1}^{\infty}1_{E_n}f\right)d\mu = \sum_{n=1}^{\infty}\int 1_{E_n}f d\mu \ \ \text{by proposition 2.15}\\ = \sum_{n=1}^{\infty}\int_{E_n}f d\mu = \sum_{n=1}^{\infty}\lambda(E_n)$$ Therefore $\lambda$ is a measure. Now, let $g\in L^{+}$, where $g$ is a simple with standard representation $g = \sum_{n=1}^{N}a_n 1_{E_n}$, then $$\int g d\lambda = \sum_{n=1}^{N}a_n\lambda(E_n) = \sum_{n=1}^{N}a_n\int_{E_n}f d\mu = \sum_{n=1}^{N}a_n\int f 1_{E_n}d\mu$$ $$=\int \sum_{n=1}^{N}a_n f 1_{E_n}d\mu = \int f g d\mu$$
Otherwise, there exists an increasing sequence $\{g_n \}_{n\in\mathbb{N}}\in L^{+}$ that converges to $g$, so that $\{fg_n\}_{n\in\mathbb{N}}$ converges to $fg$ and hence $$\int g d\lambda = \lim_{n\rightarrow \infty}\int g_n d\lambda = \lim_{n\rightarrow \infty}\int f g_n d\mu = \int f g d\mu$$
I am pretty sure this is correct, I just don't understand how $$\int f \chi_{F}d\mu = \int (\sum_{1}^{\infty}\chi_{E_n}f)d\mu$$
Also I do not understand the last part starting with otherwise that I found online.
 A: Congratulations! Your proof is correct. I have just added some details to make it clearer.

Problem 3.2.14 - If $f\in L^{+}$, let $\lambda(E) = \int_{E}f d\mu$ for $E\in M$. Then $\lambda$ is a measure on $M$, and for any $g\in L^{+}$, $\int g d\lambda = \int f g d\mu$.(First suppose that $g$ is simple)

Proof - 
Observe that $\lambda(\emptyset) = \int_{\emptyset}f d\mu = \int 1_{\emptyset} f d\mu = \int 0 f d\mu = 0$. Let $\{E_n\}_{n\in\mathbb{N}}\subset M$ be a disjoint family of sets and let $F = \bigcup_{n=1}^{\infty}E_n$. Then 
\begin{align*}
\lambda(F) &= \int_{F}f d\mu = \int 1_{F}f d\mu = \int \left(\sum_{n=1}^{\infty}1_{E_n}\right)fd\mu = \\ &= \int \left(\sum_{n=1}^{\infty}1_{E_n}f\right)d\mu= \sum_{n=1}^{\infty}\int 1_{E_n}f d\mu \ \ \text{by proposition 2.15}\\ &= \sum_{n=1}^{\infty}\int_{E_n}f d\mu = \sum_{n=1}^{\infty}\lambda(E_n)
\end{align*}
Therefore $\lambda$ is a measure. 
Now, let $g\in L^{+}$, where $g$ is a simple function with standard representation $g = \sum_{n=1}^{N}a_n 1_{E_n}$, then $$\int g d\lambda = \sum_{n=1}^{N}a_n\lambda(E_n) = \sum_{n=1}^{N}a_n\int_{E_n}f d\mu = \sum_{n=1}^{N}a_n\int f 1_{E_n}d\mu=\\=\int \sum_{n=1}^{N}a_n f 1_{E_n}d\mu = \int f\sum_{n=1}^{N}a_n  1_{E_n}d\mu =\int f g d\mu \tag{1}$$
Now, suppose $g \in L^+$, then  there exists an increasing sequence $\{g_n \}_{n\in\mathbb{N}}$ of simple functions that converges monotonically to $g$.  So we have 
$$\int g d\lambda = \lim_{n\rightarrow \infty}\int g_n d\lambda \tag{2}$$
Since $f \in  L^+$, we have that $\{fg_n\}_{n\in\mathbb{N}}\subset L^+$ and $\{fg_n\}_{n\in\mathbb{N}}$ converges monotonically to $fg$ , so by the Monotone Convergence Theorem, we have 
$$\lim_{n\rightarrow \infty}\int f g_n d\mu = \int f g d\mu \tag{3}$$
From $(1)$, we know that, for all $n$,
$$\int g_n d\lambda = \int f g_n d\mu$$
So we have 
$$\lim_{n\rightarrow \infty}\int g_n d\lambda = \lim_{n\rightarrow \infty}\int f g_n d\mu \tag{4} $$
From $(2)$, $(3)$ and $(4)$, we have 
$$\int g d\lambda = \lim_{n\rightarrow \infty}\int g_n d\lambda = \lim_{n\rightarrow \infty}\int f g_n d\mu = \int f g d\mu$$
A: If the $E_n$ are not disjoint, then the claim is not true. To see why it is true, though, if they are disjoint (which is the case): 
$$\chi_{\cup_{n=1}^{\infty} E_n} (x) = \begin{cases} 1, \text{ $ x \in E_{n_0}$ for some $n_0 \in \Bbb N$} \\ 0, \text{ if for all $n$, $x \notin E_n$} \end{cases}$$
If $x \in E_{n_0}$ for some $n_0$, then $x$ is not in any of the other $E_n$'s, and we get:
$$\sum_{n=1}^{\infty} \chi_{E_n}(x) = \chi_{E_{n_0}} (x) = 1 = \chi_{\cup_{n=1}^{\infty} E_n} (x)$$
Otherwise, we have $\chi_{E_n} = 0$ for all $n$, so:
$$\sum_{n=1}^{\infty} \chi_{E_n}(x) = 0 = \chi_{\cup_{n=1}^{\infty} E_n} (x)$$
So we have the equality:
$$\sum_{n=1}^{\infty} \chi_{E_n} = \chi_{\cup_{n=1}^{\infty} E_n} $$
In the other part, I think what is meant is that there exists an increasing sequence of simple, measurable nonnegative functions $(g_n)$ such that $g_n \to g$. This is because we want to use the case where $g$ is a simple nonnegative measurable function. Obviously then, $(fg_n)$ is an increasing sequence of simple, measurable nonnegative functions and it converges to $fg$. The rest follows by applying the previous case and using Lebesgue's monotone convergence theorem (or Beppo-Levi if you prefer)
