I'm self-learning Real Analysis using Real Analysis of Folland, and I got stuck on this problem.
Let $(X, \mathcal{M}, \mu)$ be a measure space with $\mu(X) < \infty$, and let $(X, \overline{\mathcal{M}}, \overline{\mu})$ be its completion. Suppose $f: X \rightarrow \mathbb{R}$ is bounded. Prove that $f$ is $\overline{\mathcal{M}}$-measurable iff there exist sequences $\{\phi_n\}$ and $\{\omega_n\}$ of $\mathcal{M}$-measurable simple functions such that $\phi_n\ \le f \le \omega_n$ and $\int(\omega_n - \phi_n)d \mu < n^{-1}$. In this case, $\lim{\int \phi_n d\mu} = \lim{\int \omega_n d\mu} = \int{fd\overline{\mu}}$
I thought about those functions which uses in Rienmann integral, but this problem only relates to Lebesgue integral, so I don't think they can help. Can anyone help me or give me some clue so I can solve this problem? Thanks so much your help. I really appreciate.
Here's a proof of the converse without assuming that $\phi_n, \psi_n$ are simple.
Let $\phi_n, \psi_n$ be $\mathcal{M}$-measurable functions such that $\phi_n \leq f \leq \psi_n$ and $0 \leq \int \psi_n - \phi_n \,d\mu < n^{-1}$.
Then $$\lim \int \psi_n - \phi_n \,d\mu = 0$$ By Fatou's lemma, $$0 \leq \int \liminf \, (\psi_n - \phi_n) \,d\mu \leq \lim \int \psi_n - \phi_n \,d\mu = 0 $$ which implies that $\liminf \, (\psi_n - \phi_n) = 0$ $\mu$-a.e.
For $\mu$-a.e., $$0 = \liminf \, (\psi_n - \phi_n) \geq \liminf \, \psi_n - \limsup \phi_n$$ From $\phi_n \leq f \leq \psi_n$, we get $$\limsup \, \phi_n \leq f \leq \liminf \, \psi_n \leq \limsup \, \phi_n$$ and so $f = \limsup \, \phi_n$ $\mu$-a.e.
Since $\limsup \, \phi_n$ is $\overline{\mathcal{M}}$-measurable and $f = \limsup \, \phi_n$ for $\overline{\mu}$-a.e., we conclude that $f$ is also $\overline{\mathcal{M}}$-measurable.