I'm currently studying the proof of Theorem 6.19 from Real and Complex Analysis by Walter Rudin, which concerns the Riesz-Markov-Kakutani Theorem. My question is about understanding the proof of the uniqueness of the measure $\mu$. $\vert \mu \vert$ is the total variation of the measure $\mu$.
Theorem 6.19
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
$$
\Phi f=\int_X f d \mu
$$
for every $f \in C_0(X)$. Moreover, the norm of $\Phi$ is the total variation of $\mu$ :
$$
\|\Phi\|=|\mu|(X) .
$$
The section of the proof dealing with the uniqueness of $\mu$ is as follows:
Proof(Uniqueness Part)
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 6.12 there is a Borel function $h$, with $|h|=1$, such that $d \mu=h d|\mu|$. For any sequence $\left\{f_n\right\}$ in $C_0(X)$ we then have
$$
|\mu|(X)=\int_X\left(\bar{h}-f_n\right) h d|\mu| \leq \int_X\left|\bar{h}-f_n\right| d|\mu|,
$$
and since $C_c(X)$ is dense in $L^1(|\mu|)$ (Theorem 3.14), $\left\{f_n\right\}$ can be so chosen that the last expression in (3) tends to 0 as $n \rightarrow \infty$. 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 $\Phi$.
Questions:
- How is the inequality $$
|\mu|(X)=\int_X\left(\bar{h}-f_n\right) h d|\mu| \leq \int_X\left|\bar{h}-f_n\right| d|\mu|,
$$ derived? - For the argument about density to be valid, doesn't it require $\overline{h} \in L^{1}(\vert \mu \vert)$? By Theorem 6.12 (used to derive the inequality) $h$ is only measurable.
- How does this part of the proof establish the uniqueness of $\mu$?
My Thoughts:
-
$\int_{X} f d \mu =0$ for all $f\in C_{0}(X)$, hence the first equal sign is derived as follows from:
$$
\begin{align}
|\mu|(X)
&=
\int_{X}1 d\vert \mu \vert
=
\int_{X}1 d\vert \mu \vert -\int_{X} f_{n} d \mu \\
&=
\int_{X}1 d\vert \mu \vert -\int_{X} f_{n}h d \vert \mu \vert
=
\int_{X}1 – f_{n}h d \vert \mu \vert \\
&=
\int_{X} \vert h \vert – f_{n}h d \vert \mu \vert
=
\int_{X}\left(\bar{h}-f_{n}\right)h d \vert \mu \vert
\end{align}
$$
where I've used that $\int_{X} f d \mu=\int_{X} fh d \vert \mu \vert$ by Theorem 6.12. The inequality
$\leq $ follows since $\vert h \vert=1$ and $\vert\mu\vert (X) \in [0,\infty)$ and hence $ \int_{X} \left(\bar{h}-f_n\right) h d|\mu| = \left \vert \int_X\left(\bar{h}-f_n\right) h d|\mu| \right \vert $ thus
$$
\begin{align}
\int_{X} \left( \bar{h}-f_n \right) h d|\mu|
&=
\left\vert \int_X\left(\bar{h}-f_n\right) h d|\mu| \right\vert \\
& \leq
\int_X \left\vert \left(\bar{h}-f_n\right) h \right \vert d|\mu| \\
&=
\int_X \left\vert \left(\bar{h}-f_n\right) \right \vert \vert h\vert d|\mu|
=
\int_X \left\vert \left(\bar{h}-f_n\right) \right \vert d|\mu|
\end{align}
$$ -
I have no current ideas for proving this.
-
The set of all complex measures is a Banach space w.r.t. to the total variation norm $\vert\vert \cdot \vert \vert$ given by $\vert\vert \mu \vert \vert=\vert \mu \vert (X)$ . Hence I would need to show that for two regular complex Borel measures $\mu$ and $\nu$ we have $\vert \vert \nu_{1} – \nu_{2} \vert \vert=0$ which is equivalent to $\nu_{1}=\nu_{2}$, i.e. the measure is unique. I assume this is somehow hidden in the proof (maybe $\mu = \nu_{1}-\nu_{2}$), but I do not see where.
Any help would be greatly appreciated!
Theorem 6.12
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
$$
d \mu=h d|\mu| .
$$
Theorem 6.13
Suppose $\mu$ is a positive measure on $\mathfrak{M}, g \in L^1(\mu)$, and
$$
\lambda(E)=\int_E g d \mu \quad(E \in \mathfrak{M}) .
$$
Then
$$
|\lambda|(E)=\int_E|g| d \mu \quad(E \in \mathfrak{M})
$$
Best Answer
To 1. : Seems good to me (apart from minor typographic errors (the sudden appearence of $f_n$ in the integrand after the last equality sign in the first block and the missing $\mu|$ in the second line of the same block)).
To 2. : Since $|\mu|(X) < \infty$ and $|\bar{h}|= | h| =1$ it follows that $\bar h \in L^1 (|\mu | )$.
To 3. : Assume that there are two regular complex Borel measures $\mu_1, \mu_2$ that satisfy the assumption of the proposition. Then $\mu := \mu_1 - \mu_2 $ is also a regular complex Borel measure, which satisfies $\int f \mu =0$ for all $f \in C_0 (X)$. Therefore (by what has been shown) $\mu=0$ and so $\mu_1 = \mu_2$.