Let $D \subset \mathbb{R}$ be a measurable set of finite measure. Suppose that $f : D \to \mathbb{R}$ is a bounded function. Prove that $$\sup\left\{\int_D \varphi \mid \varphi \leq f \text{ and } \varphi \text{ simple}\right\} = \inf\left\{\int_D \psi \mid f \leq \psi \text{ and } \psi \text{ simple}\right\}$$ iff $f$ is measurable.
I was wondering if I could get a hint.
Suppose the inf and the sup are equal. Then for any $\epsilon > 0$ there exist simple functions $\varphi \le f \le \psi$ with the property that $\displaystyle \int_D \psi < \int_D \varphi + \epsilon$. Use this fact to construct sequences $\varphi_n$ and $\psi_n$ of simple functions with the property that $\varphi_n \le f \le \psi_n$ and $\displaystyle \int_D \psi_n - \varphi_n < 4^{-n}$. Let $m^*$ denote Lebesgue outer measure and $m$ Lebesgue measure. Then $$ m^*(\{f - \varphi_n > 2^{-n}\}) \le m^*(\{\psi_n - \varphi_n > 2^{-n}\}) = m(\{\psi_n - \varphi_n > 2^{-n}\}) $$ because $f \le \psi_n$. Chebyshev's inequality gives you $$ m(\{\psi_n - \varphi_n > 2^{-n}\}) \le 2^n \int_D \psi_n - \varphi_n < 2^{-n}. $$ Write $E_n = \{f - \varphi_n > 2^{-n}\}$. Then $m^*(E_n) < 2^{-n}$, and by subadditivity $m^*(\cup_{n \ge N} E_n) \le 2^{1-N}$. The monotonicity of $m^*$ implies $$ m^* \left( \bigcap_{N \ge 1} \bigcup_{n \ge N} E_n \right) = 0. $$
On edit: can you use this fact to show $f$ is measurable?