The following is a problem posed in Stein and Stakarchi's Real Analysis text :
Suppose $f \geq 0$, and $f$ is integrable. If $\alpha > 0$ and $E_{\alpha} = \{x : f(x) > \alpha\}$, prove that $m(E_{\alpha}) \leq \frac{1}{\alpha}\int f$.
A proof of this has already been provided in Proving Tchebychev's Inequality, but I'll restate the argument here :
If $x \in E_{\alpha}$, then $f(x)\chi_{E_{\alpha}}(x) > \alpha\chi_{E_{\alpha}}(x)$. If $x \notin E_{\alpha}$, then $f(x)\chi_{E_{\alpha}}(x) = \alpha\chi_{E_{\alpha}}(x) = 0$. Together, we have that $f(x)\chi_{E_{\alpha}}(x) \geq \alpha\chi_{E_{\alpha}}(x) \geq 0$ for all $x \in \mathbb{R}$. Thus, by monotonicity of integration, we have $m(E_{\alpha}) = \int \chi_{E_{\alpha}} \leq \frac{1}{\alpha} \int f\chi_{E_{\alpha}} \leq \frac{1}{\alpha} \int f$, where the last inequality follows from the fact that $\chi_{E_{\alpha}} \leq 1$ for all $x \in \mathbb{R}$.
My textbook states the monotonicity property of integration as follows : If $0 \leq f \leq g$ for two non-negative measurable functions $f$ and $g$, then $\int f \leq \int g$.
In the above mentioned solution, it's clear that $\alpha\chi_{E_{\alpha}}(x)$ is a measurable function : Show a set E is measurable iff its characteristic function is measurable. gives that $\chi_{E_{\alpha}}$ is measurable, and a real scalar multiple of a measurable function is measurable. But, my question is, how does the author know that $f(x)\chi_{E_{\alpha}}(x)$ is a measurable function ?
To attempt to answer this question, elsewhere in my text, I found that the product of two measurable functions is measurable if the two functions involved are both finite-valued. $\chi_{E_{\alpha}}$ is certainly finite-valued, but can we assume here that $f$ is finite-valued ? Are integrable functions assumed to be finite-valued in general ?
Thanks!
Every measurable function (in particular, $f$) is a pointwise limit of a sequence $(s_n)$ of measurable functions which are finite-valued (this is certainly a theorem in Stein and Stakarchi's text, though I don't know where). Now $(s_n\chi_{E_\alpha})$ converges pointwise to $f\chi_{E_\alpha}$, and any pointwise limit of a sequence of measurable functions is again measurable.