$ \int_{E}^{*}{\psi (t) d\mu(t)}=\int_{E}{\phi (t) d\mu(t)} $

Question

Let $(T, \mathcal{A}, \mu)$ be an arbitrary measure space.

The outer integral over $(T, \mathcal{A}, \mu)$ of a (possibly nonmeasurable) function $\psi: T\to (-\infty, +\infty]$ is defined by: $$ \int_{T}^{*}{\psi (t) d\mu(t)}:=\inf\{\int_{T}{\phi (t) d\mu(t)}~|~\phi:T\to \mathbb{R}\text{ integrable, },\psi\leq \phi\} $$ Problem:

For a function $\psi:T\to \mathbb{R}$ such that $\int_{T}^{*}{\psi (t) d\mu(t)}<\infty$. Can we say that there exists an integrable function $\phi:T\to\mathbb{R}$ such that: $$ \int_{T}^{*}{\psi (t) d\mu(t)}=\int_{T}{\phi (t) d\mu(t)} $$

Answer 1

Yes. By definition of the inf, we can find a sequence of integrable functions $\phi_n$ such that $\psi \le \phi_n$ for every $n$, and $\int \phi_n \to \int^* \psi$. Set $\phi = \liminf \phi_n$, which is again integrable and satisfies $\psi \le \phi$. As such, $\int^* \psi \le \int \phi$. On the other hand, by Fatou's lemma, $\int \phi \le \liminf \int \phi_n = \int^* \psi.$