If a function $f: A \rightarrow \mathbb{R}$ on a bounded set $A$ is Riemann integrable, how do you prove that $f$ is measurable?
From partitions by $n$-dimensional intervals, I can make simple functions $\phi_1 \leq \phi_2 \leq \phi_3 \leq \dots \leq f$ such that $\lim_{n \to \infty}\int_A \phi_n = \int_A f$.
It suffice to say that $\lim_{n \to \infty}\phi_n=f$ almost everywhere but I don't know how to prove.
Is there a simple proof of the measurability of $f$, not limited to above approach?
Your argument for construction of $\phi_n$'s should also give you a decreasing sequence $\{\psi_n\}$ of simple functions such that $f \leq \phi_n$ and $\int \psi_n \to \int f $. From this you get $\sup_n \phi_n \leq f \leq \inf_n \psi_n$ and $\int [\inf_n \psi_n-\sup_n \phi_n ] =0$. This implies $ \inf_n \psi_n-\sup_n \phi_n =0$ almost everywhere from which Lebesgue measurability of $f$ is immediate.