Notes on Chapter 6 (Complex Measures) of Walter Rudin's Real and Complex Analysis. Happy December!

Chapter 6

After having studied mostly positive measures (range in $[0,\infty]$), we move on to studying complex measures (which also cannot take on infinite values, by our convention). In particular, the sum of the measures of any countable partition of a set must converge absolutely, since the order of the sum should not matter.

Define the total variation $|\mu|$ of a complex measure $\mu$ as

$$|\mu|(E) = \sup\sum_{i=1}^\infty |\mu(E_i)|\qquad (E\in\mathfrak{M})$$

where the supremum is taken over all partitions of $E$.

The following is true:

Total variation: The total variation $|\mu|$ is a positive measure on $\mathfrak{M}$. □

Note that $|\mu|(E)\geq |\mu(E)|$, but in general they are not equal. By construction, this is the smallest positive measure that satisfies this property. In fact,

Total variation is bounded: If $\mu$ is a complex measure on $X$, then $|\mu|(X)<\infty$. □

The argument for this proceeds by splitting an offending set $E\in\mathfrak{M}$ with $|\mu|(E)=\infty$ into a bunch of sets such that the countable sum of measures of some of them don't converge, which is impossible (since a complex measure of a set takes on values in $\mathbb{C}$).

Now, with the total variation as norm as addition and multiplication defined in the usual ways, the complex measures form a normed linear space.

We now define a few more concepts in preparation:

For a real measure $\mu$, set

$$\mu^+ = \frac{1}{2}(|\mu| + \mu),\qquad \mu^- = \frac{1}{2}(|\mu| - \mu).$$

These are positive measures and give the positive and negative variations of $\mu$, together forming the Jordan decomposition of $\mu$.

Let $\mu$ be a positive measure, and let $\lambda$ be any measure (positive or complex) on the same $\sigma$-algebra. We say that $\lambda \ll \mu$ ($\lambda$ absolutely continuous with respect to $\mu$) if $\lambda(E) = 0$ whenever $\mu(E) = 0$.

The term 'absolutely continuous' can be explained by the following:

Absolutely continuous measures: Suppose $\mu$ is positive and $\lambda$ is complex. Then the following are equivalent:

$\lambda\ll\mu$.

For all $\varepsilon > 0$ there is a $\delta > 0$ s.t. $|\lambda(E)|<\varepsilon$ for all measurable $E$ with $\mu(E)<\delta$. □

If there is a set $A\in\mathfrak{M}$ s.t. $\lambda(E) = \lambda(A\cap E)$ for every measurable $E$, we say that $\lambda$ is concentrated on $A$. If $\lambda_1$ is concentrated on $A$ and $\lambda_2$ is concentrated on $B$ and $A\cap B = \emptyset$, then we write $\lambda_1\ \bot\ \lambda_2$ ($\lambda_1$ and $\lambda_2$ are mutually singular).

We now can state the (Lebesgue-)Radon-Nikodym theorem after a short useful lemma:

Lemma (6.9): If $\mu$ is a positive $\sigma$-finite measure, then there is a function $w\in L^1(\mu)$ s.t. $0 < w < 1$. □

This means that any such $\mu$ can be replaced by a finite measure $\tilde{\mu}$ (namely, $d\tilde{\mu} = w\ d\mu$) which has precisely the same sets of measure 0.

Lebesgue-Radon-Nikodym: Let $\mu$ be a positive $\sigma$-finite measure on a $\sigma$-algebra $\mathfrak{M}$ in a set $X$, and let $\lambda$ be a complex measure on $\mathfrak{M}$.

(Existence of Lebesgue decomposition of $\lambda$ relative to $\mu$) There is a unique pair of complex measures $\lambda_a$ and $\lambda_s$ on $\mathfrak{M}$ s.t. $$\lambda = \lambda_a + \lambda_s,\qquad \lambda_a\ll\mu,\qquad \lambda_s\ \bot\ \mu.$$ If $\lambda$ is positive and finite then so are $\lambda_a$ and $\lambda_s$.

(Existence of Radon-Nikodym derivative) There is a unique $h\in L^1(\mu)$ s.t. $$\lambda_a(E)=\int_E h\ d\mu$$ for every set $E\in\mathfrak{M}$. We write $d\ \lambda_a = h\ d\mu$. □

Note that together this means that every $\lambda\ll\mu$ has a derivative of this form.

First, assume $\lambda$ is a positive bounded measure. We use Lemma 6.9 to associate $w$ to $\mu$. Then $d\varphi = d\lambda + w\ d\mu$ defines a positive bounded measure $\varphi$. We can then show that $$f\mapsto\int_X f\ d\lambda$$ is a bounded linear functional on $L^2(\varphi)$. Since $L^2(\varphi)$ is a Hilbert space, bounded linear functionals arise uniquely from inner products, so there is a $g\in L^2(\varphi)$ s.t. $$\int_X f\ d\lambda = \int_X fg\ d\varphi$$ for every $f\in L^2(\varphi)$. We can show that $g(x)\in[0,1]$ a.e. with respect to $\varphi$, so we can redefine it to have $0 < g < 1$, and rewrite the previous relation as $$\int_X (1-g)f\ d\lambda = \int_X fgw\ d\mu.$$ Using $g$ and the above integral relation (substituting appropriate values for f), one can construct both the decomposition and the Radon-Nikodym derivative $h$ and show they have the desired properties.

Finally, for complex measures, split them into the real and complex parts and apply the previous case to the positive and negative variations of them both.

If $\lambda$ is not complex but positive and $\sigma$-finite, most of the theorem is true but $h$ is only 'locally' in $L^1$, i.e. the integral of $h$ over each of the sets in the a decomposition of the space into sets of finite measure. However, if we move beyond $\sigma$-finiteness, the theorems fail.

Some useful consequences are the following:

Polar representation of $\mu$: Let $\mu$ be a complex measure. Then there is a measurable $h$ such that $|h(x)| = 1$ for all $x\in X$ and s.t. $d\mu = h\ d|\mu|$. □

This is obtained by using the Radon-Nikodym theorem with the fact that $\mu\ll |\mu|$ and tweaking the resulting derivative a bit on a set of measure zero.

Derivative of total variation: Suppose $\mu$ is a positive measure, $g\in L^1(\mu)$ and $$\lambda(E) = \int_E g\ d\mu\qquad(E\in\mathfrak{M}).$$ Then $$|\lambda|(E) = \int_E |g|\ d\mu\qquad(E\in\mathfrak{M}).$$ □

This follows from the polar representation of $\lambda$ and some manipulation.

Hahn Decomposition: Let $\mu$ be a real measure on $X$. Then there exist measurable sets $A$ and $B$ that are disjoint and partition $X$, s.t. $$\mu^+(E)=\mu(A\cap E),\qquad\mu^-(E)=-\mu(B\cap E)\qquad(E\in\mathfrak{M}).$$ □

In other words, $\mu$ induces a decomposition of $X$ into two disjoint sets, one which 'carries the positive weight' of $\mu$ and another which 'carries the negative weight'. This once again follows from the polar representation and noting that $h = \pm 1$ since $\mu$ is real.

Corollary: If $\mu = \lambda_1 - \lambda_2$ with $\lambda_1$ and $\lambda_2$ positive measures, then $\lambda_1\geq \mu^+$ and $\lambda_2\geq \mu^-$. □

This corollary gives a minimality property for the Jordan decomposition.

With this in hand, we can finally characterize the bounded linear functionals on $L^p$:

$L^p$ duality: Suppose $1\leq p < \infty$, $p, q$ are conjugate exponents, $\mu$ is a $\sigma$-finite measure on $X$, and $\Phi$ is a bounded linear functional on $L^p(\mu)$. Then there is a unique $g\in L^q(\mu)$, s.t. $$\Phi(f)=\int_X fg\ d\mu\qquad (f\in L^p(\mu)).$$ Furthermore, $\|\Phi\| = \|g\|_q$. □

Hence, $L^q(\mu)$ is isometrically isomorphic to the dual of $L^p(\mu)$ under these conditions. Note that this does not hold for $p=\infty$.

We can assume $\|\Phi\| > 0$ as the other case is trivial, and consider the case $\mu(X)<\infty$.

For any measurable set $E\subset X$, define $\lambda(E) = \Phi(\chi_E)$. It can be shown that this is a complex measure, and that $\lambda\ll\mu$. Applying Radon-Nikodym, we get that for every measurable $E$,

$$\Phi(\chi_E) = \lambda(E) = \int_E g\ d\mu = \int_X \chi_E g\ d\mu.$$

By linearity, it follows that

$$\Phi(f) = \int_X fg\ d\mu$$

for every simple measurable $f$, so for $f\in L^\infty f(\mu)$ since every such $f$ is a uniform limit of simple functions $f_i$.

With some effort, we can show that $g\in L^q(\mu)$ and that $\|g\|_q \leq \|\Phi\|$. Hölder's inequality implies that $\|\Phi\|\leq\|g\|_q$. Now since the integral equality holds on $L^\infty(\mu)$ which is dense in $L^p(\mu)$, and both sides of the equality are continuous functions on $L^p(\mu)$, they coincide on all of $L^p(\mu)$ and so we are done in the case of $\mu(X)<\infty$.

In the case that $\mu(X)=\infty$ but $\mu$ is $\sigma$-finite, choose $w\in L^1(\mu)$ as in Lemma 6.9, to get a finite measure $d\tilde{\mu} = w\ d\mu$. We have that $F\rightarrow w^{1/p}F$ is a linear isometry of $L^p(\tilde{\mu})$ onto $L^p(\mu)$, so $$\Psi(F) = \Phi(w^{1/p}F)$$ is a bounded linear functional on $L^p(\tilde{\mu})$ with $\|\Psi\|=\|\Phi\|$.

We can now apply the first part of the proof and tweak the result to get the required $g$.

We can now discuss the complex version of the Riesz representation theorem. The previous version characterized the positive linear functionals on $C_c(X)$; with this we will be able to characterize the bounded linear functionals $\Phi$ on $C_c(X)$. Since $C_c(X)$ is dense in $C_0(X)$ under the sup norm, every such $\Phi$ uniquely extends to a bounded linear functional on $C_0(X)$, so we can assume we are dealing with the Banach space $C_0(X)$.

If $\mu$ is a complex Borel measure, the polar representation asserts there is a complex Borel function $h$ with $|h|=1$ such that $d\mu = h\ d|\mu|$, so we can define integration w.r.t a complex measure as $$\int f\ d\mu = \int fh\ d|\mu|.$$ This can be seen to be a bounded linear functional on $C_0(X)$, with norm at most $|\mu|(X)$.

Finally, call a complex Borel measure $\mu$ regular if $|\mu|$ is regular. We have the following:

Riesz Representation Theorem: 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 $f\in C_0(X)$. Furthermore, $\|\Phi\|=|\mu|(X)$. □

The key idea is to explicitly contruct a positive linear functional $\Lambda$ on $C_c(X)$, such that $$|\Phi(f)| \leq \Lambda(|f|) \leq \|f\| \qquad (f\in C_c(X)),$$ where $\|f\|$ denotes the supremum norm.

We can now associate this $\Lambda$ with a positive Borel measure $\lambda$ using the other version of the Riesz representation theorem. Since $$|\Phi(f)| \leq \Lambda(|f|) \leq \int_X |f|\ d\lambda\qquad(f\in C_c(X)),$$ $\Phi$ is a linear functional $C_c(X)$ of norm at most 1 with respect to the $L^1(\lambda)$ norm on $C_c(X)$. So we have a norm preserving extension of $\Phi$ to $L^1(\lambda)$, and by the duality theorem on $L^1(\lambda)$ there is a Borel function $|g|$ with $|g| \leq 1$ s.t. $$\Phi(f) = \int_X fg\ d\lambda\qquad(f\in C_c(X)).$$

Both sides of this equality are continous on $C_0(X)$, and $C_c(X)$ is dense in $C_0(X)$, so the equality holds on $C_0(X)$ and we get our desired representation with $d\mu = g\ d\lambda$.

With some further work we can show that $\|\Phi\| = |\mu|(X)$, and that the desired $\Lambda$ can be constructed as $$\Lambda f = \sup\{|\Phi(h)|: h\in C_c(X), |h|\leq f\},$$ for $f\in C_c^+(X)$ (nonnegative real members of $C_c(X)$), and then extended to all real functions in $C_c(X)$ and then all of $C_c(X)$.

Papa Rudin Chapter 6

⤷jarsp (2022-12-03)

Chapter 6