Complex Measures

Definition 2.1   A complex measure on a measurable space $(X,\mathcal{S})$ is a function $\nu: \mathcal{S}\rightarrow \mathbb{C}$ such that

(a)

$\nu(\emptyset) = 0$ , and

(b)

If $\{ E_i \}_{i=1}^\infty$ are pairwise disjoint, $E_i \in \mathcal{S}$ , then

$$\nu(\bigcup_{i=1}^\infty E_i) = \sum_{i=1}^\infty \nu(E_i),$$

where the sum converges absolutely.

Remark: A positive measure is a complex measure when it is finite $(\mu(X) < \infty)$ .
For a complex measure, $\nu$ , write $\nu_{re}, \nu_{im}$ for its real and imaginary parts, respectively. Thus $\nu_{im}$ and $\nu_{re}$ are signed measures:

$$\nu = \nu_{re} + \nu_{im} = (\nu_{re}^+ - \nu_{re}^-) + i(\nu_{im}^+ - \nu_{re}^-)$$

Therefore, the range of $\nu$ is a bounded set.
For a complex measure, $\nu$ ,

$$L^1(\nu) := L^1(\nu_{re}) \cap L^1(\nu_{im}).$$

For $f \in L^1(\nu)$ , define

$$\int_X f d\nu = \int_X f d\nu_{re} + i \int_X f d\nu_{im}.$$

For $\nu$ a complex measure and $\mu$ a positive measure, $\nu \perp \mu$ means $\nu_{re} \perp \mu$ and $\nu_{im} \perp \mu$ .
Similarly, $\nu \ll \mu$ means $\nu_{re} \ll \mu$ and $\nu_{im} \ll \mu$ .

Theorem 3   Lebesgue-Radon-Nikodym
Let $(X,\mathcal{S})$ be a measurable space, $\mu$ a $\sigma$ -finite positive measure, and $\nu$ a complex measure. Then, there exists unique complex measures $\nu_0$ and $\nu_1$ such that $\nu = \nu_0 + \nu_1$ , $\nu_0 \perp \mu$ , and $\nu_1 \ll \mu$ . Moreover, there exists a $\mu$ -integrable function $f: X \rightarrow C$ such that $\nu_1(E) = \int_E f d\mu$ . Finally, $f$ is unique $\mu$ -almost everywhere.

Wednesday, 1-19-2005

Definition 2.2   Extended Integrable:
Let $\nu$ be a signed measure. Then, we say that $f: X \rightarrow \bar \mathbb{R}$ is extended integrable if $f$ is integrable of at least one of $\nu^+$ or $\nu^-$ . For such $f$ ,

$$\int f d\nu = \int f d\nu^+ - \int f d\nu^-$$

(Thus, we don't have a problem such as $\infty - \infty$ )

Because of this, we must slightly change the statement of the Radon-Nikodym Theorem:

Theorem 4   Radon-Nikodym
Let $\nu$ be a $\sigma$ -finite signed measure on $(X,\mathcal{S})$ and let $\mu$ be a $\sigma$ -finite positive measure on $(X,\mathcal{S})$ . If $\nu \ll \mu$ , then there exists an extended integrable function $f$ (i.e., either $f^+$ or $f^-$ is integrable) such that

$$\nu(E) = \int_E f d\mu$$

for all $E \in \mathcal{S}$ .

Let $\nu$ be a complex measure. We write

$$\mu = \nu_{Re}^+ + \nu_{Re}^- + \nu_{Im}^+ + \nu_{Im}^-$$

where $\mu$ is the total variation of $\nu$ . Note that $\mu$ is a positive measure and $\nu \ll \mu$ .

Lemma 2.1   Suppose $\lambda_1$ , $\lambda_2$ are positive measures such that $\nu \ll \lambda_1$ and $\nu \ll \lambda_2$ . Write $d \nu = f_i d \lambda_i$ . Then,

$$\vert f_1\vert d \lambda_1 = \vert f_2 \vert d \lambda_2$$

Proof. Let $\rho = \lambda_1 + \lambda_2$ , so $\rho$ is a positive measure with $\lambda_1 \ll \rho$ and $\lambda_2 \ll \rho$ . By the Theorem 1, there exists non-negative functions $h_1$ , $h_2$ such that $d \lambda_1 = h_1 d\rho$ and $d \lambda_2 = h_2 d\rho$ .
So, $d\nu = f_1 h_1 d\rho$ and similarly $d\nu = f_2 h_2 d\rho$ . By the uniqueness statement, we have $f_1 h_1 = f_2 h_2$ $\rho$ -almost everywhere. So, $\vert f_1\vert h_1 =\vert f_2 \vert h_2$ $\rho$ -almost everywhere. In other words, $\vert f_1\vert d\lambda_1 = \vert f_2\vert d\lambda_2$ . $\qedsymbol$

Definition 2.3   Total Variation:
The total variation $\vert \nu \vert$ of the complex measure $\nu$ is defined by $d\vert\nu\vert = \vert f\vert d\lambda$ where $\lambda$ is any positive measure with $\nu \ll \lambda$ and $d\nu = f d\lambda$ .

Properties of $\vert \nu \vert$ :

(a)

$\forall E$ measurable, $\vert \nu(E) \vert \leq \vert\nu\vert (E)$

(b)

$\nu \ll \vert \nu \vert$ and $\left\vert\left[\frac{d\nu}{d\vert\nu\vert}\right] \right\vert = 1$ almost everywhere.

(c)

If $\nu_1, \nu_2$ are complex measures, then $\vert \nu_1 + \nu_2\vert \leq \vert\nu_1\vert + \vert\nu_2\vert$ .

Proof. Let $\lambda$ be a positive measure such that $\nu \ll \mu$ and write $d\nu = f d\lambda$ . Then,

(a)

$\vert\nu(E)\vert = \vert \int_E f d\lambda \vert \leq \int_E \vert f\vert d\lambda = \vert \nu \vert (E).$

(b)

Part (a) give $\nu \ll \vert \nu \vert$ . Write

$$\nu(E) = \int_E \left[ \frac{d\nu}{d\vert\nu\vert} \right] d\vert\nu\vert$$

Then,

$$\int_E 1 d\vert\nu\vert = \vert\nu\vert(E) = \left\vert \left[ \frac{d\nu}{d\vert\nu\vert} \right] \right\vert =$$

Therefore, by uniqueness of Theorem 1,

$$\left\vert \left[ \frac{d\nu}{d\vert\nu\vert} \right] \right\vert = 1,$$    $\vert \nu \vert$ -almost everywhere$$$$

(c)

Let $\lambda_1, \lambda_2$ be positive measures such that $\nu_1 \ll \lambda_1$ and $\nu_2 \ll \lambda_2$ ; write $d\nu_1 = f_1 d\lambda_1$ . Put $\rho = \lambda_1 + \lambda_2$ . Then,

$$(\nu_1 + \nu_2)(E) = \int_E f_1 d\lambda_1 + \int_E f_2 d\lambda_2$$

$$= \int_E f_1 \left[ \frac{d \lambda_1}{d\rho} \right] d\rho + \int_E f_2 \left[ \frac{d \lambda_2}{d \rho} \right] d \rho$$

$$= \int_E \left( f_1 \left[ \frac{d\lambda_1}{d\rho} \right] + f_2 \left[ \frac{d\lambda_2}{d \rho}\right] \right) d\rho$$

So,

$$\vert \nu_1 + \nu_2\vert (E) = \int_E \left\vert f_1 \left[ \frac... ...ho} \right] + f_2 \left[ \frac{ d\lambda_2 }{d \rho} \right] \right\vert d\rho$$

$$\leq \int_E \vert f_1\vert \left\vert \left[ \frac{d\lambda_1}{d\... ...ght\vert = \int_E \vert f_1\vert d\lambda_1 + \int_E \vert f_2\vert d\lambda_2$$

$$= \vert\nu_1\vert (E) + \vert\nu_2\vert (E)$$

$\qedsymbol$

Theorem 5   Let $\nu$ be a complex measure. Then for any measurable $E$ ,

$$\vert\nu\vert(E) = \sup \left\{ \sum_{i=1}^\infty \vert \nu(E_i)\... ...\cap E_j = \emptyset (i \neq j), \text{ and } E_i \text{ measurable } \right\}$$

Proof. Found in Rudin, Real and Complex Analysis. $\qedsymbol$

An application to Harmonic Analysis:
Let $X = \mathbb{R}$ , $\mathcal{L}$ be Lebesgue-measurable sets. For complex measures $\nu_1, \nu_2$ , define

$$(\nu_1 * \nu_2)(E) = \int_\mathbb{R}\left[ \int_\mathbb{R}\chi_E( x + y) d\nu_1(x) \right] d\nu_2(y)$$

This is the convolution of $\nu_1$ and $\nu_2$ . Note that by using the Radon-Nikodym derivatives, we can convert this to the product of two positive measures.

This makes the set of complex measures into an algebra over $\mathcal{C}$ , with vector space under $+$ and usual associative multiplication.

The measures which are absolutely continuous with respect to the Lebesgue measure $m$ form an ideal.

Let $\nu_t$ be the point mass at $t$ . Then, $\nu_t * \nu_s = \nu_{(t+s)}$ . Hence, the real line is embedded in this algebra in a natural sort of way.