Exercise 1.4.50 (Function measurability criterion)

Question

Let $(X,\mathcal{B},\mu)$ be a finite complete measure space, and let $f:X	o\mathbb{R}$ be a bounded function. If the upper integral
    $$\overline{\int_{X} f \,\mathrm{d}{\mu}} = \inf \left\{\int_{X} g \,\mathrm{d}{\mu}: \begin{array}{ll} 	ext{$g$ simple}\ f \leq g \end{array} ight\} $$
    and lower integral
    $$\underline{\int_{X} f \,\mathrm{d}{\mu}} = \sup \left\{\int_{X} h \,\mathrm{d}{\mu}: \begin{array}{ll} 	ext{$h$ simple}\ h \leq f \end{array} ight\} $$
    agree, then $f$ is measurable.

Elu Thingol · Accepted Answer

$
ewcommand{\intx}[1]{\int_{X} #1 \mathop{}\!\mathrm{d}{\mu}}$
By definition of supremum/infimum we can find a sequence of simple functions $\left(hight)_{n\in\mathbb{N}}$ and $\left(gight)_{n\in\mathbb{N}}$ such that
$$\lim_{n	o\infty} \int_{X} h_n \,\mathrm{d}{\mu} = I(f) = \lim_{n	o\infty} \intx{g_n}$$
where $I(f)$ denotes the value of the upper/lower integral. Notice that this implies
$$\lim_{n	o\infty} \intx{g_n} - \lim_{n	o\infty} \int_{X} h_n \,\mathrm{d}{\mu} = \lim_{n	o\infty} \intx{\left(g_n - h_night)}  = 0.$$
On one hand, $g_n-h_n$ is a non-negative function, since $g_n\geq f \geq h_n$; thus,
$$\intx{\lim_{n	o\infty}(g_n-h_n)} \geq 0.$$
On the other hand, by Fatou's lemma (Corollary 1.4.46) we have
$$0 = \lim_{n	o\infty} \intx{\left(g_n - h_night)} \geq \intx{\lim_{n	o\infty}(g_n-h_n)}.$$
Thus, $\intx{\lim_{n	o\infty}(g_n-h_n)} = 0$, and by \href{./exercise_1-4-35}{Exercise 1.4.35}(viii) we must have $\lim_{n	o\infty}(g_n-h_n) = 0$ almost everywhere. Thus, we squeezed $f$ between $\left(gight)_{n\in\mathbb{N}}$ and $\left(hight)_{n\in\mathbb{N}}$, i.e., $f$ is a pointwise limit of simple functions $\left(gight)_{n\in\mathbb{N}}$ and $\left(hight)_{n\in\mathbb{N}}$. By \href{./exercise_1-4-29}{Exercise 1.4.29} (vi), $f$ must be measurable itself.

Exercise 1.4.50 (Function measurability criterion)

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Exercise 1.4.50 (Function measurability criterion)

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