From Radon-Nikodym to Riesz Representation

$L^{p^{\prime }}$ has a countable dense subset, say $\mathcal{G}$. [Recall an approximation result from Chapter 8.]

Append to $\mathcal{G}$ all finite rational linear combinations of its elements. We will refer to this augmented collection as $\mathcal{G}$; $\mathcal{G}$ [the augmented collection] is still countable. Next, let $\left\{ f_{n}\right\}_{n=1}^{\infty }$ denote a sequence in $B_{1}^{p}$.

For every $g \in L^{p^{\prime }}$, $\left\{ \int f_{n} g \, \mathrm{d} \mu \right\}_{n=1}^{\infty }$ is a bounded sequence of real numbers with bound $\left\| g\right\|_{p^{\prime }}$. From this we may deduce that $\left\{ \int f_{n} g \, \mathrm{d} \mu \right\}_{n=1}^{\infty }$ is a bounded sequence of real numbers with bound $\left\| g\right\|_{p^{\prime }}$ for every $g \in \mathcal{G}$.

Since $\mathcal{G}$ is countable, there is a subsequence $\left\{ n_{k}\right\}_{k=1}^{\infty }$ of the sequence $1, 2, \cdots$ such that $\lim_{k} \int f_{n_{k}} g \, \mathrm{d} \mu$ exists for every $g \in \mathcal{G}$. [This is a direct application of the diagonal method from advanced calculus: The array whose $r$th row is $\left\{ \int f_{r} g_{n} \, \mathrm{d} \mu \right\}_{n=1}^{\infty }$, with $\left\{ g_{n}\right\}_{n=1}^{\infty }$ an enumeration of $\mathcal{G}$, is bounded row by row, so Exercise 1 of this section applies.] Write $L_{\mathcal{G}}(g)=\lim_{k} \int f_{n_{k}} g \, \mathrm{d} \mu$ for every $g \in \mathcal{G}$; Hölder's Inequality gives $\left|L_{\mathcal{G}}(g)\right| \leqslant \left\| g\right\|_{p^{\prime }}$ for every $g \in \mathcal{G}$.

$L_{\mathcal{G}}$ is uniformly continuous on its domain $\mathcal{G}$. [We have

for every $g, g^{\prime } \in \mathcal{G}$.]

There exists a unique continuous function $L$ on $L^{p^{\prime }}$ such that $L=L_{\mathcal{G}}$ on $\mathcal{G}$. [From advanced calculus, we know that a uniformly continuous function $\psi$ defined on a dense subset $D$ of a metric space $(X, \rho )$ has a unique continuous extension to all of $X$.]

For every $g, g^{\prime } \in \mathcal{G}$ we have $L_{\mathcal{G}}(g+g^{\prime })=L_{\mathcal{G}}(g)+L_{\mathcal{G}}(g^{\prime })$. [Direct verification.]

For every $h, h^{\prime } \in L^{p^{\prime }}$ we have $L(h+h^{\prime })=L(h)+L(h^{\prime })$. [Let $\left\{ g_{n}\right\}_{n=1}^{\infty }$ denote a sequence in the dense collection $\mathcal{G}$ with $g_{n} \xrightarrow {p^{\prime }} h$; let $\left\{ g_{n}^{\prime }\right\}_{n=1}^{\infty }$ denote a sequence in $\mathcal{G}$ with $g_{n}^{\prime } \xrightarrow {p^{\prime }} h^{\prime }$. Thus $\left\{ g_{n}+g_{n}^{\prime }\right\}_{n=1}^{\infty }$ is a sequence in $\mathcal{G}$ with $g_{n}+g_{n}^{\prime } \xrightarrow {p^{\prime }} h+h^{\prime }$. Use the continuity of $L$ and (e) to verify the equation

]

Directly verify that $L_{\mathcal{G}}(cg)=cL_{\mathcal{G}}(g)$ for every $g \in \mathcal{G}$ and $c \in \mathbb {R}$. This in turn shows that $L(cg)=cL(g)$ for every $c \in \mathbb {R}$ and $g \in L^{p^{\prime }}$. [The first part is easy. The second part is obtained in the same spirit as was (f).]

By (g), $L$ is a linear functional on $L^{p^{\prime }}$. Show in addition that $L$ is bounded with $\left\| L\right\| \leqslant 1$. [We already know that $L$ is continuous from (d); use the relevant claim from Section 8.8 to achieve boundedness. Next, pick $h \in L^{p^{\prime }}$; let $\left\{ g_{n}\right\}_{n=1}^{\infty }$ denote a sequence in $\mathcal{G}$ with $g_{n} \xrightarrow {p^{\prime }} h$. Verify that

]

We now invoke the Riesz Representation Theorem for the bounded linear functional $L$ on $L^{p^{\prime }}$, obtaining a function $f \in L^{p}$ such that $L(g)=\int fg \, \mathrm{d} \mu$ for every $g \in L^{p^{\prime }}$. Since $\left\| L\right\| \leqslant 1$, we have $\left\| f\right\|_{p}=\left\| L\right\| \leqslant 1$ by Claim 1 of Section 8.9, hence $f \in B_{1}^{p}$. Recall the subsequence $\left\{ n_{k}\right\}_{k=1}^{\infty }$ given after part (b). Observe that for every $g \in \mathcal{G}$ we have

We have that $\left\{ f_{n_{k}}\right\}_{k=1}^{\infty }$ converges weakly to $f$, completing the proof. [Pick $g^{\prime }$ in $L^{p^{\prime }}$ and $\varepsilon >0$; find $g \in \mathcal{G}$ with $\left\| g-g^{\prime }\right\|_{p^{\prime }}<\frac{\varepsilon }{3}$. Define $\alpha_{k}=\left|\int f_{n_{k}} g \, \mathrm{d} \mu -\int fg \, \mathrm{d} \mu \right|$ for each $k \in \mathbb {N}$, and observe that the above shows $\lim_{k} \alpha_{k}=0$. Verify that

For large enough $k$ we thus have $\left|\int fg^{\prime } \, \mathrm{d} \mu -\int f_{n_{k}} g^{\prime } \, \mathrm{d} \mu \right|<\varepsilon$. Since $g^{\prime } \in L^{p^{\prime }}$ is arbitrary, the entire proof is complete.]

Quickly show that $\int fg \, \mathrm{d} \mu =\int f^{\prime } g \, \mathrm{d} \mu$ for every $g \in L^{p^{\prime }}$. [Both quantities are equal to $\lim_{n} \int f_{n} g \, \mathrm{d} \mu$.]

Show that $f=f^{\prime }$ $\mu$-a.e. on $\Omega$. [Define $L_{f}(g)=\int fg \, \mathrm{d} \mu$ for every $g \in L^{p^{\prime }}$; define $L_{f^{\prime }}(g)=\int f^{\prime } g \, \mathrm{d} \mu$ for every $g \in L^{p^{\prime }}$. Both $L_{f}$ and $L_{f^{\prime }}$ are bounded linear functionals on $L^{p^{\prime }}$ that agree on $L^{p^{\prime }}$ by (a). Use the uniqueness claim of the Riesz Representation Theorem.]

Both $\mu$ and $\nu$ are measures with domain $\mathcal{F}$, and $\nu \leqslant \mu$. [It should be straightforward to verify that $\nu$ is a measure. From this and the interchange of the roles of $x$ and $y$ one can show that $\mu$ is a measure. That $\nu (E) \leqslant \mu (E)$ for all $E \in \mathcal{F}$ is trivial.]

Define $L$ on $L^{1}(\Omega , \mathcal{F}, \mu )$ by writing $L(f)=\int f \, \mathrm{d} \nu$ for every $f \in L^{1}(\Omega , \mathcal{F}, \mu )$. Verify that $L^{1}(\Omega , \mathcal{F}, \mu ) \subseteq L^{1}(\Omega , \mathcal{F}, \nu )$, so that the definition makes sense. Next, verify that $L$ is a bounded linear functional on $L^{1}(\Omega , \mathcal{F}, \mu )$ with $\left\| L\right\| \leqslant 1$. [Again, this should be straightforward.]

Now assume that the conclusion of the Riesz Representation Theorem holds: there is $g \in L^{\infty }(\Omega , \mathcal{F}, \mu )$ such that $L=L_{g}$ for this $g$, where the notation is as in Exercise 4. That is, assume that $L(f)=\int fg \, \mathrm{d} \mu$ for every $f \in L^{1}(\Omega , \mathcal{F}, \mu )$.

Accept without proof the assertion the existence of some $x_{0} \in [0,1]$ such that $\int_{[0,1]}\left|g(x_{0},y)\right| \, \mathrm{d} \lambda (y)=0$. [The proof of this fact requires the machinery of the next chapter.]

Let $V=\left\{ x_{0}\right\} \times [0,1]$ and $f=\; \mathbb {1}_{V}$. Show that $\nu (V)=1$ and $f \in L^{1}(\Omega , \mathcal{F}, \mu )$.

Verify that

so that we achieve a contradiction. [The inequality may be accepted on faith, as it depends on some concepts yet to be defined.] This gives a semirigorous counterexample to the Riesz Representation Theorem in the case $p=1$. The problem here is that $(\Omega , \mathcal{F}, \mu )$ is not a $\sigma$-finite measure space. [Actually, $(\Omega , \mathcal{F}, \mu )$ fails even to be decomposable.]

Let $P=\left\{ \omega \in \Omega : f(\omega ) \geqslant 0\right\}$. Then $(P, P^{c})$ is a Hahn decomposition of $\Omega$ with respect to $\phi$. [Obvious.]

Let $\phi =\phi^{+}-\phi^{-}$ denote the Jordan decomposition of $\phi$ into finite measures $\phi^{+}$ and $\phi^{-}$; recall that Jordan decompositions are unique by Exercise 4 of Section 9.1. Show that $\phi^{+}(E)=\int_{E} f^{+} \, \mathrm{d} \mu$ and $\phi^{-}(E)=\int_{E} f^{-} \, \mathrm{d} \mu$ for every $E \in \mathcal{F}$. [We have $\phi^{+}(E)=\phi (E \cap P)=\int_{E \cap P} f \, \mathrm{d} \mu =\int_{E \cap P} f^{+} \, \mathrm{d} \mu$ for every $E \in \mathcal{F}$, where $P$ is as in part (a).]

Define $\left|\phi \right|(E)=\phi^{+}(E)+\phi^{-}(E)=\int_{E}\left|f\right| \, \mathrm{d} \mu$. $\left|\phi \right|$ is a finite measure, and for every $E \in \mathcal{F}$ we have $\left|\phi \right|(E)=\sup \sum_{i=1}^{n}\left|\phi (E_{i})\right|$, the supremum being over all finite decompositions of $E$ into disjoint $\mathcal{F}$-sets $E_{1}, \cdots , E_{n}$. [Let $E \in \mathcal{F}$, and let $\beta$ denote the supremum. If $E_{1}, \cdots , E_{n}$ are disjoint $\mathcal{F}$-sets, then

which gives $\beta \leqslant \left|\phi \right|(E)$. For the reverse inequality, consider the disjoint $\mathcal{F}$-sets $E \cap P$ and $E \cap P^{c}$, where $P$ is as in (a). Then

This gives $\beta =\left|\phi \right|(E)$.]

Define $\nu (E)=L(\; \mathbb {1}_{E})$ for every $E \in \mathcal{F}$. Then $\nu$ is an additive set function on $\mathcal{F}$. [Note that part (i) of the proof in this section works for $p=1$.]

There exists a unique [up to a $\mu$-null set] function $g \in L^{1}(\Omega , \mathcal{F}, \mu )$ such that $\nu (E)=\int_{E} g \, \mathrm{d} \mu$. [Note that part (ii) of the proof in this section works for $p=1$.] By the definition of $\left|\nu \right|$, we have $\left|\nu \right|(E)=\int_{E}\left|g\right| \, \mathrm{d} \mu$ for every $E \in \mathcal{F}$.

We may without any loss of generality assume that $\left|g(\omega )\right| \leqslant \left\| L\right\|$ for every $\omega \in \Omega$. Therefore, we have $g \in L^{\infty }(\Omega , \mathcal{F}, \mu )$ as well. [Let $A=\left\{ \omega \in \Omega : \left|g(\omega )\right|>\left\| L\right\| \right\}$. If $\mu (A)>0$, then $\left|\nu \right|(A)=\int_{A}\left|g\right| \, \mathrm{d} \mu >\int_{A}\left\| L\right\| \, \mathrm{d} \mu =\left\| L\right\| \mu (A)$. By part (c) of the previous exercise, there exists a finite collection $A_{1}, \cdots , A_{n}$ of disjoint $\mathcal{F}$-sets with union $A$ such that $\sum_{i=1}^{n}\left|\nu (A_{i})\right|>\left\| L\right\| \mu (A)$. But then

a contradiction. Thus $\left|g\right| \leqslant \left\| L\right\|$ $\mu$-a.e. on $\Omega$.]

Show that $L(s)=\int sg \, \mathrm{d} \mu$ for every finite-valued $\mathcal{F}/\mathcal{B}^{*}$-measurable simple function $s$ defined on $\Omega$. [Follow the proof of (iii) given in the proof of the section.]

$L(f)=\int fg \, \mathrm{d} \mu$ for every $f \in L^{1}(\Omega , \mathcal{F}, \mu )$. [Follow the proof of (v) given in the proof of the section, or do the usual thing of using MCT to show the result is true for nonnegative functions in $L^{1}(\Omega , \mathcal{F}, \mu )$ and then using positive and negative parts to give the result in general.]

The function $g \in L^{\infty }(\Omega , \mathcal{F}, \mu )$ is unique up to a $\mu$-null set. [This should be straightforward. Suppose that $\psi \in L^{\infty }(\Omega , \mathcal{F}, \mu )$ is such that $\int f\psi \, \mathrm{d} \mu =0$ for every $f \in L^{1}(\Omega , \mathcal{F}, \mu )$. If $\left\{ \omega \in \Omega : \psi (\omega ) \neq 0\right\}$ has positive $\mu$-measure, then, without loss of generality, $A=\left\{ \omega \in \Omega : \psi (\omega )>0\right\}$ has positive and finite $\mu$-measure. Let $f=\; \mathbb {1}_{A}$. Then $\int f\psi \, \mathrm{d} \mu =\int_{A} \psi \, \mathrm{d} \mu >0$, a contradiction. Use this to get the general uniqueness claim.]

Show that $\left|\phi \right| \in \mathfrak {F}(\Omega , \mathcal{F}, \mu )$.

Show that $\mathfrak {F}(\Omega , \mathcal{F}, \mu )$ is a normed linear space [with setwise linear operations].