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: $$ \int_{T}^{*}{\psi (t) d\mu(t)}:=\inf\{\int_{T}{\phi (t) d\mu(t)}~|~\phi:T\to \mathbb{R}\text{ integrable, },\phi\leq \psi\} $$ 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: $$ \int_{T}^{*}{\psi (t) d\mu(t)}=\int_{T}{\phi (t) d\mu(t)} $$

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