Problem 4

Exercise 3.4 (Radon-Nikodym) If $\mu(X)<\infty$ , $\{E_k\}_{k=1}^n$ are measurable, and $\{c_k\}_{k=1}^n$ are real, define a measure $\nu$ by

$\displaystyle \nu(E) := \sum_{k=1}^n c_k \mu(E\cap E_k)$

(3.26)

Proof. Verify that $\nu$ is a measure by checking countable additivity. Let $\{A_j\}_{j=1}^\infty$ be a sequence of disjoint measurable sets. Then

$\displaystyle \nu\left(\bigcup_{j=1}^\infty A_j\right)$	$\displaystyle = \sum_{k=1}^n c_k \mu\left(\left(\bigcup_{j=1}^\infty A_j\right) \cap E_k\right)$	(3.27)
	$\displaystyle = \sum_{k=1}^n c_k \mu\left(\bigcup_{j=1}^\infty (A_j\cap E_k)\right)$	(3.28)
	$\displaystyle = \sum_{k=1}^n c_k \sum_{j=1}^\infty \mu(A_j\cap E_k)$	(3.29)
	$\displaystyle = \sum_{j=1}^\infty \sum_{k=1}^n c_k\mu(A_j\cap E_k)$	(3.30)
	$\displaystyle = \sum_{j=1}^\infty \nu(A_j)$	(3.31)

To see that $\nu\ll\mu$ , suppose $\mu(A)=0$ . Then

$\displaystyle \nu(A) = \sum_{k=1}^n \mu(A\cap E_k) \leq \sum_{k=1}^n \mu(A) = 0$

(3.32)

Now we show the Radon-Nikodym derivative equals $\sum_k c_k \mathbf{1}_{E_k}$ . Observe:

$\displaystyle \int_A\frac{d\nu}{d\mu}d\mu$	$\displaystyle =\int_A \sum_{k=1}^n c_k \mathbf{1}_{E_k}$	(3.33)
	$\displaystyle = \sum_{k=1}^n \int_A c_k \mathbf{1}_{E_k}d\mu$	(3.34)
	$\displaystyle = \sum_{k=1}^n \int_{A\cap E_k} c_k d\mu$	(3.35)
	$\displaystyle = \sum_{k=1}^n c_k \mu(A\cap E_k)$	(3.36)
	$\displaystyle = \nu(A)$	(3.37)

showing that which was to be shown. $\qedsymbol$