I'm trying to do the exercise below.
18. If $\mu$ is a $\sigma$-finite Radon measure on $X$ and $\nu \in \mathcal{M}(X)$, let $\nu=\nu_1+\nu_2$ be the Lebesgue decomposition of $\nu$ with respect to $\mu$. Then $\nu_1$ and $\nu_2$ are Radon. (Use Exercise 8.)
Also, can we say that $\| v\| = \|v_1\| + \| v_2\|$?
Here $M(X)$ is the space of complex Radon measures on $X$, and if $\mu \in M(X)$, we define $\|\mu \| = |\mu|(X)$, where $|\mu|$ is the total variation of $\mu$.
I know that by Lebesgue decomposition we have $v_1 \ll \mu$ and $v_2 \perp \mu$. By the Radon- Nikodym theorem, there exists some $f \in L^+$ such that $v_1(E) = \int_E f d\mu$ for $E \in B_X.$ By exercise 8, we have $v_1$ is Radon.
But from here, I'm not sure how to proceed.
Any help will be appreciated!
Thank you.
At this point you have that $\nu, \nu_1$ are both complex Radon measures (i.e. they are in $M(X)$) and that $\nu= \nu_1 + \nu_2$, which is to say $\nu_2 = \nu - \nu_1$. Now Proposition 7.16 says that $M(X)$ is a vector space, so it follows that $\nu_2 \in M(X)$ as well, i.e. $\nu_2$ is Radon.
To see that $\|\nu\| = \|\nu_1\| + \|\nu_2\|$, note we have disjoint measurable sets $E_1, E_2$ such that $|\nu_1|(E_2) = |\nu_2|(E_1) = 0$. Let $f_i = d\nu_i/d|\nu|$ for $i=1,2$. Show that on the one hand, $|f_1 + f_2|= 1$, $|\nu|$-a.e. On the other hand, show that the set $\{f_1 \ne 0\} \cap \{f_2 \ne 0\}$ is $|\nu|$-null, and conclude that $|f_1 + f_2| = |f_1| + |f_2|$, $|\nu|$-a.e. Now note that $\|\nu_i\| = \int |f_i|\,d|\nu|$ and so on.