I'm reading a book on measure theory and during a section about integration the author states the following: "Note that a nonnegative measurable function is always integrable". I tried proving it, but with no success. I know it suffices to show that upper integral $\leq$ lower integral (see below). Any help would be appreciated.
Background:
i) Any measurable function $f:X\to [0,\infty]$ can be decomposed as
$$
f=\sum_{n=1}^{\infty} \frac{1}{n}\chi_{A_n} \, ,
$$
where the sets $A_n\subseteq X$ are measurable.
ii) A function whose image is a countable subset of $[-\infty,\infty]$ is called a simple function. The integral of a nonnegative simple measurable function $g$ is defined as
$$
\int g \, d\mu = \sum_{0\leq y\leq\infty} y\mu(g^{-1}(\{y\}))
$$
iii) Given a function $f:X\to[-\infty,\infty]$, the upper and lower integrals are defined as
$$
\int^{\ast} f \, d\mu := \inf\left\{ \int g \, d\mu \, ; \, g \, \text{simple}, \text{integrable}, g\geq f \, \mu\text{-a.e.} \right\}
$$
$$
\int_{\ast} f \, d\mu := \sup\left\{ \int g \, d\mu \, ; \, g \, \text{simple}, \text{integrable}, g\leq f \, \mu\text{-a.e.} \right\}
$$
In case that $f$ is a measurable function, if its upper and lower integrals are equal, we say that $f$ is integrable and write
$$
\int f \, d\mu := \int_{\ast} f \, d\mu = \int^{\ast} f \, d\mu \, .
$$
Best Answer
I guess you are reading Measure Theory and Fine Properties of Functions - Evans, and your question is in page25.
Given the "i)" -- "Decomposition of nonnegative measurable functions" in page19 you listed above, it suffices to prove the question. The basic idea of the proof is to use the decomposition to prove the inequality: $$\int_{*} fd\mu \geq \int^{*} fd\mu$$ but it's a little tricky, here is the proof:
1)$\mu(f=+\infty)>0\ $or$\ \exists n\in N_{+}\ s.t.\ \mu(A_n)=+\infty$
If $\mu(f=+\infty)>0$, any simple function $g$ s.t. $g\geq f, \mu-a.e.$ must have $\mu(g=+\infty) > 0$, and it implies: $$\int^{*}fd\mu \geq \int_{*}fd\mu \geq (+\infty) * \mu(g=+\infty) = +\infty$$ $$\Rightarrow \int^{*}fd\mu = \int_{*}fd\mu = +\infty$$ therefore f is integrable.
If $\exists n\in N_{+}\ s.t.\ \mu(A_n)=+\infty$, simple function $\frac{1}{n}\chi_{A_n} \leq f$, and therefore: $$\int^{*}fd\mu \geq \int_{*} fd\mu \geq \int \frac{1}{n}\chi_{A_n} d\mu = \frac{1}{n} * \mu(A_n) = +\infty$$ $$\Rightarrow \int^{*}fd\mu = \int_{*} fd\mu = +\infty$$ which implies $f$ is integrable.
2)$\mu(f=+\infty)=0\ $and$\ \forall n\in N_{+}\ \mu(A_n)<+\infty$
Now we can dismiss the part that $f=+\infty$ and only consider the bounded part of $f$, since the simple funcitons required in the lower and upper integral only require a.e. inequality.
Given the decomposition: $$f=\sum_{n=1}^{+\infty} \frac{1}{n} \chi_{A_n}$$ we have: $$\sum_{n=1}^k \frac{1}{n} \chi_{A_n} \leq f \quad \forall k \in N_{+}$$ and $\sum_{n=1}^k \frac{1}{n} \chi_{A_n}$ is simple function by definition, so we have: $$\begin{align} \int_{*}fd\mu &\geq \int \sum_{n=1}^k \frac{1}{n} \chi_{A_n} d\mu \\\\ & = \sum_{n=1}^{g(k)} y_n\mu(f^{-1}(B_n)) \quad \forall k \in N_{+} \end{align}$$ where $B_n = \{x|f(x)=y_n, y_n\in Image(\sum_{n=1}^k \frac{1}{n} \chi_{A_n})\}, g(k)\subseteq N_{+}, g(k)\geq k\ $ ($\{A_n\}$ are not disjoint), and $\mu(B_n)<+\infty$, since $B_n \subseteq \cup_{n=1}^k A_n$ and $\forall n\leq k,\ \mu(A_n)<+\infty$
let $k \rightarrow +\infty$, we have: $$\int_{*}fd\mu \geq \sum_{0\leq y_n\leq +\infty} y_n\mu(f^{-1}(B_n))$$ where $B_n = \{x|f(x)=y_n, y_n\in Image(\sum_{n=1}^{+\infty} \frac{1}{n} \chi_{A_n})\}$
Since $f < +\infty \ a.e.$ and $f=\sum_{n=1}^{+\infty} \frac{1}{n} \chi_{A_n}$, we have: $$\forall \epsilon > 0\ \exists N\in N_{+}\ s.t.\ \forall k>N\quad f\leq \epsilon + \sum_{n=1}^k \frac{1}{n}\chi_{A_n}\quad a.e.\ \forall k \in N_{+}$$ therefore: $$\begin{align} \int^{*}fd\mu &\leq \int \epsilon + \sum_{n=1}^k \frac{1}{n}\chi_{A_n}d\mu \\\\ &= \sum_{n=1}^{g(k)} y_n\mu(f^{-1}(B_n)) + \sum_{n=1}^{g(k)} \epsilon \mu(f^{-1}(B_n)) \quad \forall k \in N_{+} \end{align}$$
let $\epsilon \rightarrow 0^{+}$, we have: $$\int^{*}fd\mu \leq \sum_{n=1}^{g(k)} y_n\mu(f^{-1}(B_n))\quad \forall k \in N_{+}$$ then let $k\rightarrow +\infty$, we have: $$\int^{*}fd\mu \leq \sum_{0\leq y_n\leq +\infty} y_n\mu(f^{-1}(B_n))$$
Combine two inequalities above, we have: $$\int_{*} fd\mu \geq \sum_{0\leq y_n\leq +\infty} y_n\mu(f^{-1}(B_n))\geq \int^{*}fd\mu$$ which implies $f$ is integrable.
I hope you find this proof helpful.