Lebesgue Measure of a k-cell

Question

Working through Rudin's RCA construction (Theorem 2.20, p. 53) of the Lebesgue measure using the Riesz Representation Theorem. Rudin constructs a linear functional $\Lambda$ on $\operatorname{C}_c(\mathbb{R}^k)$ such that

$$\Lambda f := \lim\limits_{n \to \infty} 2^{-nk} \sum\limits_{x \in P_n} f(x)$$ where $P_n$ is the set of all vectors of the form $x = (a_1/2^n,...,a_k/2^n)$ for $a_1,...,a_k \in \mathbb{Z}$.

Now let $W$ be an open $k$-cell. Rudin considers the set $S_r = \{ Q \in \Omega_r: \overline{Q} \subset W\}$, where $\Omega_r$ is the set of all boxes of the form $Q = \{ x: a_i \leq x_i < a_i + 2^{-r}, a_i \in P_r, 1 \leq i \leq k\}$. He then defines $$E_r = \bigcup\limits_{Q \in S_r} Q$$ and applies Urysohn's Lemma to obtain a function $0 \leq f_r \leq 1$ such that $f[\overline{E}_r] = 1$, $\operatorname{supp}(f)\subseteq W$, and $\overline{E}_r \subseteq \operatorname{supp}(f)$. Note that $\overline{E}_r $ is compact.

He then asserts without proof that

$$\operatorname{vol}(E_r) \leq \Lambda f_r \leq \Lambda g_r \leq \operatorname{vol}(W)$$

where $g_r := \max\{f_i: 1 \leq i \leq r\}$. How do we establish this inequality? It would seem a priori that it would arise out of demonstrating $\operatorname{vol}(E_r) \leq \Lambda_n f$ for all $n$, but I have so far been unsuccessful in this matter.

Using Urysohn's Lemma we obtained that $\chi_{\overline{E}_r} \leq f \leq \chi_W$, which may yield another route.

As a secondary question, are there any other good sources for a construction of the Lebesgue measure on $\mathbb{R}^k$ using the Riesz Representation Theorem? All of the other standard texts, e.g. Royden, on the subject construct an outer Lebesgue measure and extend using the results of Caratheodory.

Answer 1

$E_r$ is a union of $P_r$-boxes, and since $W$ is a $k$-cell, so is $E_r$.

For $s \geqslant r$, consider

$$\Lambda_s(f) = 2^{-sk}\sum_{x \in P_s} f(x) \geqslant 2^{-sk} \sum_{\substack{x \in P_s\\x+[0,\,2^{-s})^k \subset E_r}} f(x) = 2^{-sk}\sum_{\substack{x \in P_s\\x+[0,\,2^{-s})^k \subset E_r}} 1 = \operatorname{vol}(E_r),$$

since the $P_s$-boxes contained in $E_r$ are disjoint, their union is $E_r$, and each has volume $2^{-sk}$.

Thus $\Lambda_s(f) \geqslant \operatorname{vol}(E_r)$ for all $s \geqslant r$, hence the same holds for $\Lambda(f)$.

I can't answer your secondary question, Rudin is the only source I know that takes this approach.

Answer 2

To prove $\Lambda g_r \leq \text{vol}(W)$; Since $f_i \prec W$ we have $f_i\leq \chi_W$ for $i=1,...,r$, therefore since $g_r=\max(f_1,...,f_r)$ it follows $g_r\leq \chi_W$ and $\Lambda g_r\leq \Lambda \chi_W$. Since $\Lambda$ is just the Riemann integral it follows that $\Lambda \chi_W$ is the volume of $W$, therefore $\Lambda g_r\leq \text{vol}(W)$