Riesz-Markov Theorem Proof exercise from Stein and Shakarchi

Question

Drawn from Real Analysis: Measure Theory, Integration, and Hilbert Spaces by Elias M. Stein & Rami Shakarchi page 314 Exercise 9.

The exercise states the following,

Let $C([a,b])$ denote the vector space of continuous functions on the closed and bounded interval $[a,b]$. Suppose we are given a Borel measure $\mu$ on this interval, with $\mu([a,b])<\infty$. Then \begin{align*} f\longmapsto\ell(f)=\int_{a}^{b}f(x)\,d\mu(x) \end{align*} is a linear functional on $C([a,b])$, with $\ell$ positive in the sense that $\ell(f)\geq 0 $ if $f\geq 0$.

Prove that, conversely, for any linear functional on $C([a,b])$ that is positive in the above sense, there is a unique finite Borel measure $\mu$ so that \begin{align*} \ell(f)=\int_{a}^{b}f(x)\,d\mu(x) \end{align*} for $f\in C([a,b])$. Otherwise known as The Riesz-Markov Theorem.

[Hint: Suppose $a = 0$ and $u\geq0$. Define $F(u)$ by $F(u) = \lim_{\epsilon\rightarrow 0}\ell(f_\epsilon)$ where \begin{align*} f_\epsilon(x)=\begin{cases} 1,&\text{ if } x\in[0,u]\\ \text{linear},&\text{ if }t\in[u,u+\epsilon]\\ 0,&\text{ if }t\in[u+\epsilon,\infty) \end{cases} \end{align*} Then $F$ is increasing and right-continuous, and $\ell(f)$ can be written $\ell(f)=\int_{a}^{b}f(x)\,dF(x)$ via Theorem $3.5$.]

Theorem 3.5 Let $F$ be an increasing function on $\mathbb{R}$ that is right-continuous. Then there is a unique measure $\mu$ (also denoted by $\,dF$) on the Borel sets $\mathcal{B}$ on $\mathbb{R}$ such that $\mu((a,b]) = F(b)-F(a)$ if $a<b$.

Question: I am not sure how to use the hint to get from the fact that $\mu((a,b]) = F(b)-F(a)$ to the integral $\int_{a}^{b}f(x)\,dF(x)$. Moreover, I very much suspect that the dominated convergence theorem will play a role in proving that $\ell(f)=\int_{a}^{b}f(x)\,dF(x)$ however to me, it not clear how. Could someone provide some insight as to the purpose of the $f_\epsilon$ function? Or an outline as to prove it with Theorem $3.5$?

Remark: I am aware there are other proofs to this theorem on stack however my question is geared towards proving it in the direction of this hint.

Answer 1

I assume that you know how to prove $F$ is increasing and right-continuous.

The challenging part is to prove $\ell(f)=\int_a^b f dF$.

Below is a sketch of a proof.

• Let us first define $M_1 = \ell(\chi_[a,b])$ and $M_2 = \int_a^b dF$. (We don't know $M_1 = M_2$ yet.)
• First approximate $f$, which is continuous, by a step function $g = \sum_{k=1}^N c_k \chi_{(a_k, a_{k+1}]}$ uniformly such that $0 < g(x) - f(x) < \epsilon$. Note that both inequalities are strict. The number of points of discontinuity of $g$ is by definition finite.
• $g$ can then be approximated by a step-ish continuous function. We modify it at each point of discontinuity from the right side. For example, if $a_k$ is a point of discontinuity, we make $g$ linear between $a_k$ and $a_k+\delta < a_{k+1}$. Because the inequalities that we required for $g$ were strict, we can choose $\delta$ small enough such that the new function, let's call it $g_1$, still satisfies the condition $0 < g_1(x) - f(x) < \epsilon$ whether we are dealing with a "step-up" or a "step-down".
• Now we let $\delta$ tend to 0 and get a series of continuous functions $g_1, g_2, \dots$.
• We have the following.
  • $0 \le \ell(g_i) - \ell(f) \le \epsilon M_1 $
  • $\lim_{i\rightarrow\infty}\ell(g_i) = \sum_{k=1}^N c_k \left(F(a_{k+1}) - F(a_k)\right) = \sum_{k=1}^N \int_{a_k}^{a_{k+1}} c_k dF$
  • $0 \le \sum_{k=1}^N \int_{a_k}^{a_{k+1}} c_k dF - \int_a^b f dF \le \epsilon M_2 $
• Combining these relationships, we have $ \big| \ell(f) - \int_a^b f dF \big| \le \epsilon \big|M_1-M_2\big| $.
• Since $\epsilon$ is arbitrary, $\ell(f) = \int_a^b f dF$ for any $f$.