Let $(T, \mathcal{A}, \mu)$ be an arbitrary measure space, and $E$ a convex subset of a Hausdorff topological vector space. Any function from $T\times E$ into $( - \infty, + \infty]$ is be called integrand. An integrand $g : T \times E \to ( -\infty , +\infty]$ is said to be respectively convex, affine, (sequentially) lower semicontinuous, (sequentially) continuous, or (sequentially) inf-compact if for every $t \in T$ the function $g(t, .)$ on $E$ has the corresponding property (recall that $g(t, .)$ is (sequentially) inf-compact on $E$ if for every $r \in \mathbb{R}$ the set of all $x \in E$ such that $g(t, x) \leq r$, is (sequentially) compact). Thus, these adjectives refer only to the behavior of the integrand in its second variable.
Let $h: T\times E \to [0, + \infty]$ be given nonnegative convex sequentially inf-compact integrand and let $\mathcal{F}$ be a set of affine sequentially continuous integrands $a: T \times E\to \mathbb{R}$. A function $f: T \to E$ is defined to be $\mathcal{F}$-scalarly measurable if $t \to a(t, f(t))$ is measurable on $T$ for every $a \in \mathcal{F}$.
We recall the definition of outer integration: the outer integral over $(T, \mathcal{A}, \mu)$ of a (possibly nonmeasurable) function $\psi: T\to (-\infty, +\infty]$ is defined by: $$ \int_{E}^{*}{\psi (t) d\mu(t)}:=\inf\{\int_{E}{\phi (t) d\mu(t)}~|~\phi:E\to \mathbb{R}\text{ integrable, },\phi\leq \psi\} $$ Problem:
Let $\{f_n\}$ be a sequence of $\mathcal{F}$-scalarly measurable functions $f_n : T\to E$ such that : $$ \sup_{n}{\int_{E}^{*}{h\big(t,f_n(t)\big)d\mu(t)}}<+\infty $$ I did not understand why :
It follows directly from the definition of outer integration that for every $n\in\mathbb{N}$ there exists an integrable function from $T$ into $\mathbb{R}$, for convenience to be denoted - perhaps somewhat oddly - as $t \to a_0(t, f_n( t))$, such that : $$ \int_{E}^{*}{h(t,f_n(t)d\mu(t)}=\int_{E}{a_0\big(t,f_n(t)\big)d\mu(t)} \qquad and \qquad a_0\big(t,f_n(t)\big)\geq h\big(t,f_n(t)\big)\geq 0\qquad \text{on } E $$