Let $\{\mu_n\}$ be measures on $(X,M)$ with $\sup\{\mu_n\}(X) \leq M < \infty$.
Let $\mu = \sum_{n \geq 1 } 2^{-n} \mu_n$
Let $\nu$ be a $\sigma-$finite measure.
Suppose $\mu_n= \mu_{n,0} + \mu_{n,1}$ is the Lebesgue decomposition of $\mu_n$ with respect to $\nu$.
Find the Lebesgue decomposition of $\mu = \mu_0 +\mu_1$ and $\frac{d\mu_0}{d\nu}$.
My attempt:
By Lebesgue's decomposition theorem, we know that $\mu_{n,0} \ll \nu$ and $\mu_{n,1} \perp \nu$.
By breaking down the definitions, this means that
if $\nu(E) = 0$ ,then $\mu_{n,0}(E)=0$. So $\mu_0(E) := \sum_{n \geq 1 } 2^{-n} \mu_{n,0}(E)=0.$
There exists disjoint $A,B \in M$ such that $A \cup B=X$ and $\mu_{n,1}(A)=0$ and $\nu(B)=0$. So $\mu_1(A):= \sum_{n \geq 1 } 2^{-n} \mu_{n,1}(A)=0.$ And $\mu_1 \perp \nu$
Are we done finding the Lebesgue Decomposition here? I feel like just wrote down the definitions, not sure if I missed anything.
Now to find $\frac{d\mu_0}{d\nu}$ (the part I'm stuck on), I showed that $\mu$ is a finite measure. I don't know if Radon Nikodym Theorem is relevant here. By definition, we have $d \mu_0 = \frac{d\mu_0}{d\nu} d\nu$. Not sure how to proceed from here.
Thank you in advance.
You have an issue in 2. The $A,B$ actually depend on $n$, so your proof does not work. Fix this by taking $A_n$ and $B_n$ and noting that $\nu(\cup_n B_n) = 0$ and $\mu_1(\cap_n A_n) = 0$.
As for $\frac{d\mu_0}{d\nu}$, for any positive measurable $f$, $$\int f d\mu_0 = \sum_n2^{-n}\int f d\mu_{n,0} =\sum_n2^{-n}\int f \frac{d\mu_{n,0}}{d\nu} d\nu $$ $$ = \int f \left(\sum_n 2^{-n} \frac{d\mu_{n,0}}{d\nu}\right)d\nu. $$ Thus $\frac{d\mu_0}{d\nu} = \sum_n 2^{-n} \frac{d\mu_{n,0}}{d\nu}$. In general Radon Nikodym derivatives add and multiply by constants the same way that measures do.