Q) Suppose that $\mu_1, \mu_2, \mu_3$ are $\sigma-$finite positive measures on X such that $\mu_1\ll\mu_2$ and $\mu_2\ll \mu_3$. Show that $\mu_1\ll\mu_3$ and its Radon-Nikodym derivative satisfies
$$\frac{d\mu_1}{d\mu_3}(x)=\frac{d\mu_1}{d\mu_2}(x)\frac{d\mu_2}{d\mu_3}(x) \text{ for } \mu_3-\text{a.e. }x\in X \tag{1}$$
Since $\mu_1\ll \mu_2$, there exists a unique $f$ upto $\mu_2-$a.e. and since $\mu_2\ll \mu_3$, there exists a unique $g$ upto $\mu_3-$a.e.. To show $(1)$, I need to show that $f(x)g(x)$ is unique upto $\mu_3-$a.e.. Since $\mu_2\ll \mu_3$, $\mu_3(A)=0\implies \mu_2(A)=0$ and so if $g$ is unique upto $\mu_3-$a.e., then $g$ is unique upto $\mu_2-$a.e.. So $f(x)g(x)$ is unique upto $\mu_2-$a.e. but how can I show that it is unique upto $\mu_3-$a.e.?
Best Answer
As $\mu_3(A)=0\Rightarrow\mu_2(A)=0\Rightarrow\mu_1(A)=0$, we have that $\mu_1\ll\mu_3$ and the Radon-Nikodym theorem says that there exists a unique $\mu_3-$a.e $h$ such that $\mu_1(A)=\int_Ah\,d\mu_3.$ We want to show that $h=fg$. For that let $f_n=\sum_{i=1}^na_i^n\chi_{A_i^n}$ and $g_k=\sum_{j=1}^kb_j^k\chi_{B_j^k}$ be simple functions such that $f_n\nearrow f$ and $g_k\nearrow g$. By the Monotone Convergence Theorem, we have
\begin{align*}\mu_1(A)&=\int_A f\,d\mu_2=\lim_{n\rightarrow \infty}\int_A f_n\,d\mu_2=\lim_{n\rightarrow \infty}\sum_{i=1}^na_i^n\mu_2(A\cap A_i^n)=\lim_{n\rightarrow \infty}\sum_{i=1}^na_i^n\int_{A\cap A_i^n}g\,d\mu_3\\&=\lim_{k,n\rightarrow \infty}\sum_{i=1}^n \sum_{j=1}^k a_i^nb_j^k\mu_3(A\cap A_i^n\cap B_j^k)=\lim_{k,n\rightarrow \infty}\int_A f_ng_k\, d\mu_3=\int_Afg\,d\mu_3.\end{align*}