2 The Reisz Theorem

Destination: MeasureTheory/Integral/RieszMarkovKakutam/ComplexRMK
Principal reference: Theorem 6.19 of [Walter Rudin, Real and Complex Analysis.][Rud87].

The main statement is:

If \(X\) is a locally compact Hausdorff space, then every bounded linear functional \(\Phi \) on \(C_0(X)\) is represented by a unique regular complex Borel measure \(\mu \), in the sense that

\begin{equation*} \Phi f = \int _X f \, d\mu \end{equation*}

for every \(f \in C_0(X)\). Moreover, the norm of \(\Phi \) is the total variation of \(\mu \):

\begin{equation*} \| \Phi \| = |\mu |(X). \end{equation*}

Definition 1 Variation of a Vector-Valued Measure

✓

Let \((X, \mathcal{A})\) be a measurable space and let \(Y\) be a Banach space. For a vector-valued measure \(\mu : \mathcal{A} \to Y\), the variation of \(\mu \) is the set function \(|\mu |: \mathcal{A} \to [0, +\infty ]\) defined by

\begin{equation*} |\mu |(E) = \sup \left\{ \sum _{i=1}^n \| \mu (E_i)\| _Y : \{ E_1, E_2, \ldots , E_n\} \text{ is a finite partition of } E \text{ in } \mathcal{A} \right\} \end{equation*}

for each \(E \in \mathcal{A}\).

Equivalently, the above definition can be written as:

\begin{equation*} |\mu |(E) = \sup \left\{ \sum _{i=1}^n \| \mu (E_i)\| _Y : E_i \in \mathcal{A}, \, E_i \cap E_j = \emptyset \text{ for } i \neq j, \, \bigcup _{i=1}^n E_i \subseteq E \right\} \end{equation*}

Theorem 2 Rudin 6.12 (polar representation of a complex measure)

Let \(\mu \) be a complex measure on a \(\sigma \)-algebra \(\mathfrak {M}\) in \(X\). Then there is a measurable function \(h\) such that \(|h(x)| = 1\) for all \(x \in X\) and such that

\begin{equation} d\mu = h \, d|\mu |. \end{equation}

Proof ▶

This rather depends on how the integral with respect to a complex measure is defined. See [Rudin, Theorem 6.12] for details.

Definition 3

Let \(X\) be a locally compact Hausdorff space. Associated to every bounded linear functional \(\Phi \) on \(C_0(X)\) we define a regular complex Borel measure \(\mu \) which we call the Riesz Measure associated to \(\Phi \).

TO DO: insert details from the proof of the exact definition.

In order to prove the main result we divide the result into several smaller results.

Theorem 4 Rudin 3.14

For \(1 \leq p {\lt} \infty \), \(C_c(X)\) is dense in \(L^p(\mu )\).

Proof ▶

Define \(S\) as in Theorem 3.13. If \(s \in S\) and \(\varepsilon {\gt} 0\), there exists a \(g \in C_c(X)\) such that \(g(x) = s(x)\) except on a set of measure \({\lt} \varepsilon \), and \(|g| \leq \| s\| _\infty \) (Lusin’s theorem). Hence

\begin{equation} \| g - s\| _p \leq 2\varepsilon ^{1/p}\| s\| _\infty . \end{equation}

Since \(S\) is dense in \(L^p(\mu )\), this completes the proof.

Theorem 5 Rudin 6.13

Suppose \(\mu \) is a positive measure on \(\mathfrak {M}\), \(g \in L^1(\mu )\), and

\begin{equation} \lambda (E) = \int _E g \, d\mu \quad (E \in \mathfrak {M}). \end{equation}

Then

\begin{equation} |\lambda |(E) = \int _E |g| \, d\mu \quad (E \in \mathfrak {M}). \end{equation}

Proof ▶

By Theorem 2, there is a function \(h\), of absolute value 1, such that \(d\lambda = h \, d|\lambda |\). By hypothesis, \(d\lambda = g \, d\mu \). Hence

\[ h \, d|\lambda | = g \, d\mu . \]

This gives \(d|\lambda | = \bar{h}g \, d\mu \). (Compare with Theorem 1.29.) Since \(|\lambda | \geq 0\) and \(\mu \geq 0\), it follows that \(\bar{h}g \geq 0\) a.e. \([\mu ]\), so that \(\bar{h}g = |g|\) a.e. \([\mu ]\). □

Theorem 6 Rudin 6.16

Suppose \(1 \leq p {\lt} \infty \), \(\mu \) is a \(\sigma \)-finite positive measure on \(X\), and \(\Phi \) is a bounded linear functional on \(L^p(\mu )\). Then there is a unique \(g \in L^q(\mu )\), where \(q\) is the exponent conjugate to \(p\), such that

\begin{equation} \Phi (f) = \int _X fg \, d\mu \quad (f \in L^p(\mu )). \end{equation}

Moreover, if \(\Phi \) and \(g\) are related as in (1), we have

\begin{equation} \| \Phi \| = \| g\| _q. \end{equation}

In other words, \(L^q(\mu )\) is isometrically isomorphic to the dual space of \(L^p(\mu )\), under the stated conditions.

Proof ▶

Rudin 6.16: Duality of \(L^1\) and \(L^∞\) (not in Mathlib https://leanprover.zulipchat.com/#narrow/channel/217875-Is-there-code-for-X.3F/topic/Lp.20duality/near/495207025)

Lemma 7

Let \(X\) be a locally compact Hausdorff space, and let \(\Phi \) be a bounded linear functional on \(C_0(X)\). Suppose that \(\mu \), \(\nu \) are regular complex Borel measure such that

\begin{equation*} \Phi f = \int _X f \, d\mu = \int _X f \, d\nu . \end{equation*}

Then \(\mu = \nu \).

Proof ▶

Suppose \(\mu \) is a regular complex Borel measure on \(X\) and \(\int f \, d\mu = 0\) for all \(f \in C_0(X)\). By Theorem 2 there is a Borel function \(h\), with \(|h| = 1\), such that \(d\mu = h \, d|\mu |\). For any sequence \(\{ f_n\} \) in \(C_0(X)\) we then have

\begin{equation} |\mu |(X) = \int _X (\bar{h} - f_n)h \, d|\mu | \leq \int _X |\bar{h} - f_n| \, d|\mu |, \tag {3} \end{equation}

and since \(C_c(X)\) is dense in \(L^1(|\mu |)\) (Theorem 4), \(\{ f_n\} \) can be so chosen that the last expression in (3) tends to 0 as \(n \to \infty \). Thus \(|\mu |(X) = 0\), and \(\mu = 0\). It is easy to see that the difference of two regular complex Borel measures on \(X\) is regular. This shows that at most one \(\mu \) corresponds to each \(\Phi \).

Lemma 8

Consider a given bounded linear functional \(\Phi \) on \(C_0(X)\). Assume \(\| \Phi \| = 1\). (Update statement to be the general case.) We shall construct a positive linear functional \(\Lambda \) on \(C_c(X)\), such that

\begin{equation} |\Phi (f)| \leq \Lambda (|f|) \leq \| f\| \quad (f \in C_c(X)), \tag {4} \end{equation}

where \(\| f\| \) denotes the supremum norm.

Proof ▶

Assume \(\| \Phi \| = 1\), without loss of generality.

So all depends on finding a positive linear functional \(\Lambda \) that satisfies (4). If \(f \in C_c^+(X)\) [the class of all nonnegative real members of \(C_c(X)\)], define

\begin{equation} \Lambda f = \sup \{ |\Phi (h)| : h \in C_c(X), |h| \leq f\} . \tag {9} \end{equation}

Then \(\Lambda f \geq 0\), \(\Lambda \) satisfies (4), \(0 \leq f_1 \leq f_2\) implies \(\Lambda f_1 \leq \Lambda f_2\), and \(\Lambda (cf) = c\Lambda f\) if \(c\) is a positive constant. We have to show that

\begin{equation} \Lambda (f + g) = \Lambda f + \Lambda g \quad (f \text{ and } g \in C_c^+(X)), \tag {10} \end{equation}

and we then have to extend \(\Lambda \) to a linear functional on \(C_c(X)\).

Fix \(f\) and \(g \in C_c^+(X)\). If \(\varepsilon {\gt} 0\), there exist \(h_1\) and \(h_2 \in C_c(X)\) such that \(|h_1| \leq f\), \(|h_2| \leq g\), and

\begin{equation} \Lambda f \leq |\Phi (h_1)| + \varepsilon , \quad \Lambda g \leq |\Phi (h_2)| + \varepsilon . \tag {11} \end{equation}

There are complex numbers \(\alpha _i\), \(|\alpha _i| = 1\), so that \(\alpha _i \Phi (h_i) = |\Phi (h_i)|\), \(i = 1, 2\). Then

\begin{align} \Lambda f + \Lambda g & \leq |\Phi (h_1)| + |\Phi (h_2)| + 2\varepsilon \\ & = \Phi (\alpha _1 h_1 + \alpha _2 h_2) + 2\varepsilon \\ & \leq \Lambda (|h_1| + |h_2|) + 2\varepsilon \\ & \leq \Lambda (f + g) + 2\varepsilon , \end{align}

so that the inequality \(\geq \) holds in (10).

Next, choose \(h \in C_c(X)\), subject only to the condition \(|h| \leq f + g\), let \(V = \{ x : f(x) + g(x) {\gt} 0\} \), and define

\begin{align} h_1(x) & = \frac{f(x)h(x)}{f(x) + g(x)}, \quad h_2(x) = \frac{g(x)h(x)}{f(x) + g(x)} \quad (x \in V), \tag {12} \\ h_1(x) & = h_2(x) = 0 \quad (x \notin V). \end{align}

It is clear that \(h_1\) is continuous at every point of \(V\). If \(x_0 \notin V\), then \(h(x_0) = 0\); since \(h\) is continuous and since \(|h_1(x)| \leq |h(x)|\) for all \(x \in X\), it follows that \(x_0\) is a point of continuity of \(h_1\). Thus \(h_1 \in C_c(X)\), and the same holds for \(h_2\).

Since \(h_1 + h_2 = h\) and \(|h_1| \leq f\), \(|h_2| \leq g\), we have

\begin{equation} |\Phi (h)| = |\Phi (h_1) + \Phi (h_2)| \leq |\Phi (h_1)| + |\Phi (h_2)| \leq \Lambda f + \Lambda g. \end{equation}

Hence \(\Lambda (f + g) \leq \Lambda f + \Lambda g\), and we have proved (10).

If \(f\) is now a real function, \(f \in C_c(X)\), then \(2f^+ = |f| + f\), so that \(f^+ \in C_c^+(X)\); likewise, \(f^- \in C_c^+(X)\); and since \(f = f^+ - f^-\), it is natural to define

\begin{equation} \Lambda f = \Lambda f^+ - \Lambda f^- \quad (f \in C_c(X), f \text{ real}) \tag {13} \end{equation}

and

\begin{equation} \Lambda (u + iv) = \Lambda u + i\Lambda v. \tag {14} \end{equation}

Simple algebraic manipulations, just like those which occur in the proof of Theorem 1.32, show now that our extended functional \(\Lambda \) is linear on \(C_c(X)\).

Theorem 9 Rudin 6.19

If \(X\) is a locally compact Hausdorff space, then every bounded linear functional \(\Phi \) on \(C_0(X)\) is represented by a regular complex Borel measure \(\mu \), in the sense that

\begin{equation} \Phi f = \int _X f \, d\mu \tag {1} \end{equation}

for every \(f \in C_0(X)\).

Proof ▶

Once we have the \(\Lambda \) from Lemma 8, we associate with it a positive Borel measure \(\lambda \), as in Theorem 2.14. The conclusion of Theorem 2.14 shows that \(\lambda \) is regular if \(\lambda (X) {\lt} \infty \). Since

\begin{equation} \lambda (X) = \sup \{ \Lambda f : 0 \leq f \leq 1, f \in C_c(X)\} \end{equation}

and since \(|\Lambda f| \leq 1\) if \(\| f\| \leq 1\), we see that actually \(\lambda (X) \leq 1\).

We also deduce from (4) that

\begin{equation} |\Phi (f)| \leq \Lambda (|f|) = \int _X |f| \, d\lambda = \| f\| _1 \quad (f \in C_c(X)). \tag {5} \end{equation}

The last norm refers to the space \(L^1(\lambda )\). Thus \(\Phi \) is a linear functional on \(C_c(X)\) of norm at most 1, with respect to the \(L^1(\lambda )\)-norm on \(C_c(X)\). There is a norm-preserving extension of \(\Phi \) to a linear functional on \(L^1(\lambda )\), and therefore Theorem 6 (the case \(p = 1\)) gives a Borel function \(g\), with \(|g| \leq 1\), such that

\begin{equation} \Phi (f) = \int _X fg \, d\lambda \quad (f \in C_c(X)). \tag {6} \end{equation}

Each side of (6) is a continuous functional on \(C_0(X)\), and \(C_c(X)\) is dense in \(C_0(X)\). Hence (6) holds for all \(f \in C_0(X)\), and we obtain the representation (1) with \(d\mu = g \, d\lambda \).

Lemma 10 Rudin 6.19

Moreover, the norm of \(\Phi \) is the total variation of \(\mu \):

\begin{equation} \| \Phi \| = |\mu |(X). \tag {2} \end{equation}

Proof ▶

Since \(\| \Phi \| = 1\), (6) shows that

\begin{equation} \int _X |g| \, d\lambda \geq \sup \{ |\Phi (f)| : f \in C_0(X), \| f\| \leq 1\} = 1. \tag {7} \end{equation}

We also know that \(\lambda (X) \leq 1\) and \(|g| \leq 1\). These facts are compatible only if \(\lambda (X) = 1\) and \(|g| = 1\) a.e. \([\lambda ]\). Thus \(d|\mu | = |g| \, d\lambda = d\lambda \), by Theorem 5, and

\begin{equation} |\mu |(X) = \lambda (X) = 1 = \| \Phi \| , \tag {8} \end{equation}

which proves (2).

Theorem 11

Placeholder to combine the three results which make up The Reisz Theorem.

Proof ▶