Exercise 1.3.9 (Riemann integrable functions are Lebesgue measurable)

Question

Let $f:[a,b]	o\mathbb{R}^d$ be a Riemann integrable function ($f(x) = 0$ for $x
otin[a,b])$. Show that $f$ is Lebesgue measurable.

Elu Thingol · Accepted Answer

First of all, it must be clear that any piecewise constant function is automatically simple. By Darboux criteria of the Riemann integrability we have
    $$\inf \left\{ p.c.\int \overline{f} : \begin{array}{l} \overline{f} 	ext{ is p.c.} \ \overline{f} \geq f \end{array} ight\} = \sup \left\{ p.c.\int \underline{f} : \begin{array}{l} \underline{f} 	ext{ is p.c.} \ f \geq \underline{f} \end{array} ight\}$$
    By axiom of choice we can pick a sequence $\left(\overline{f}_night)_{n\in\mathbb{N}}$ and $\left(\underline{f}_night)_{n\in\mathbb{N}}$ of piecewise constant (and therefore simple) functions such that for all $n \in \mathbb{N}$ we have
    $$p.c.\int \overline{f} - p.c.\int \underline{f} = \sum_{i \in I} \overline{c}_i m(I_i) - \sum_{j\in J} \underline{c}_j m(J_j) = \sum_{l \in L} |\overline{c}_l - \underline{c}_l| \cdot m(I_l) \leq \sum_{l \in L} |\overline{c}_l - \underline{c}_l| \cdot (b-a) \stackrel{!}{\leq} \frac{1}{n}$$
    where $(\overline{c}_l)_{l \in L}$ and $(\underline{c}_l)_{l \in L}$ are the piecewise constant representations of $\overline{f}_n$ and $\underline{f}_n$ with respect to the common partition $(I_l)_{l\in L}$ of $[a,b]$. Taking limits as $n	o\infty$ we see from $\overline{f}_n(x) \geq \underline{f}_n(x)$ that
    $$\forall x \in \mathbb{R}^d:\lim_{n	o\infty} \left( \overline{f}_n(x) - \underline{f}_n(x) ight) = 0$$
    But this implies that $\overline{f_n}$ and $\underline{f_n}$ are simple functions which pointwise converge to $f$ by taking limits of:
    $$\forall x \in \mathbb{R}^d: \overline{f}_n(x) \geq f(x) \geq \underline{f}_n(x) $$
    Thus, $f$ is Lebesgue measurable by definition.

Exercise 1.3.9 (Riemann integrable functions are Lebesgue measurable)

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Exercise 1.3.9 (Riemann integrable functions are Lebesgue measurable)

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