Show there is a positive complete measure $m$ on a $σ$-algebra $\mathfrak{M}$ in $\mathbb{R}^k$ such that $m(W)={\rm vol}(W)$ for every $k$-cell $W$.

Question

My Background: Walter Rudin's Principles of Mathematical Analysis Chapter 1-7 & 11.

Source: The problem arises from Walter Rudin's Real & Complex Analysis:¹⁾

2.20 Theorem There exists a positive complete measure $m$ defined on a $\sigma$-algebra $\mathfrak{M}$, with the following properties:

(a) $m(W) = \operatorname{vol}(W)$ for every $k$-cell $W$.

Rudin began the proof by defining a positive linear functional

$$ \Lambda \, : \, C_c(\mathbb{R}^k) \to \mathbb{R} \, : \, \Lambda f = \lim_{n\to\infty} \Lambda_n f, $$

where $\Lambda_n f = 2^{-nk} \sum_{x\in P_n} f(x)$. ($f$ here is real, and $P_n$ is the set of all $x\in\mathbb{R}^k$ whose coordinates are integral multiples of $2^{-n}$.) Then he goes:²⁾

To prove (a), let $W$ be the open cell 2.19(4), let $E_r$ be the union of those boxes belonging to $\Omega_r$ whose closures lie in $W$, choose $f_r$ so that $\overline{E}_r \prec f_r \prec W $, and put $g_r = \max\{f_1, \ldots, f_r\}$. Our construction of $\Lambda$ shows that

$$ \operatorname{vol}(E_r) \leq \Lambda f_r \leq \Lambda g_r \leq \operatorname{vol}(W). \tag{4} $$

As $r \to \infty$, $\operatorname{vol}(E_r) \to \operatorname{vol}(W)$, and

$$ \bbox[background:#FFE600;padding:5px;]{\Lambda g_r = \int g_r \, dm} \to m(W) \tag{5} $$

by the monotone convergence theorem, since $g_r(x) \to \chi_{W}(x)$ for all $x \in \mathbb{R}^k$. Thus $m(W) = \operatorname{vol}(W)$ for every open cell $W$, and since every $k$-cell is the intersection of a decreasing seqeunce of open $k$-cells, we obtain (a).

My Question: How could $\Lambda g_r = \int g_r \, dm$ be justified without $m(E) = \operatorname{vol}(E) = \prod_{i=1}^k (\beta_i-\alpha_i)$ for every $k$-cell $E$, which is precisely what we are trying to show? By definition, $\int g_r \, dm=\sup\{\sum_{i=1}^nc_im(E_i)\}$, where $c_i, E_i$ come from a measurable function $s: \mathbb{R}^k\to\mathbb{R}$ such that $0\leq s\leq g_r$ and $s(x)=c_i$ whenever $x\in E_i$; but isn't the value of $m(E_i)$ for each $E_i\in\mathfrak{M}$ yet to be determined?

Any help would be greatly appreciated.

Answer 1

The fact that $\Lambda h = \int h \, dm$ for all $h \in C_c(X)$ is a consequence of the definition of $m$.

To see this you need to refer to Rudin's Theorem 2.14, which says that for every positive linear functional $\phi$ on $C_c(X)$ (where $X$ is locally compact and Hausdorff), there is a measure $\mu$ that represents $\phi$ in the sense that $\mu$ is defined on a sigma algebra containing the Borel sets of $X$ and $$ \phi(f) = \int_X f \, d\mu, \qquad f \in C_c(X), $$ and with various other properties.

In the argument you are focusing on, the equality $\Lambda g_r = \int g_r \, dm$ is just an application of this theorem (where $\Lambda$, a previously defined positive linear functional on $C_c(X)$, is playing the role of a $\phi$ that has been fed into Theorem 2.14, and $m$ is playing the role of the $\mu$ that is produced by that result). The inequalities relating $\Lambda g_r$ and $\operatorname{vol}$ are deduced via the order-theoretic relations between functions (encoded in the relation that Rudin denotes by $\prec$).

For what it's worth, this very functional analytic way of constructing the Lebesgue measure is somewhat peculiar to Rudin (and other analysts of his vintage). Putting the linear functional on the function space (the "integral," in other words, but wearing a mask) ahead of the underlying measure feels to some people like it is arriving at the concepts of measure and integration in the "wrong" order. If you are among such people, you might prefer another treatment. As one example, Bartle's "Elements of Integration and Lebesgue Measure" puts the measure before integration.