I'm finding some difficulties with an exercise from Conway and I ask for some help in understanding it:
"Let X be a compact space and let $\{\mu_n\}$ be a sequence of measures in X. For each $n$ let $\pi_n:C(X) \rightarrow \mathbb{B}(L^2(\mu_n))$ be defined by $\pi_n(f) = M_f$ on $L^2(\mu_n)$ --$M_f$ is the multiplication operator--. Then $\pi=\oplus_n \pi_n$ is a representation. If the measures $\{\mu_n\}$ are pairwise mutually singular, prove that $\pi$ is unitarily equivalent to the representation $f \rightarrow M_f$ of $C(X) \rightarrow \mathbb{B}(L^2(\mu))$ where $\mu=\sum_{n=1}^\infty \frac{\mu_n}{2^n \|\mu_n\|}$."
Some personal considerations:
For simplicity, let the $\mu_n$ be real positive borel measures, so $\|\mu_n\|=\mu_n(X)$.
The measure $\mu$ depends from the order of the measures $\{\mu_n\}$.
It exists an injection from $L^2(\mu)$ to $\oplus_n(L^2(\mu_n))$ simply by replicating the function to the differents $L^2(\mu_n)$. It exists an injection from $L^2(\mu_n)$ to $L^2(\mu)$ for some reasons due to the mutually singularity of the measures (I have to formalize better this argument). By Cantor-Bernstein it exists a bijection between the two spaces. I have to find an Isometrical bijection.
I believe that the coefficients of the measure $\mu$ i.e. $\frac{1}{2^n \|\mu_n\|}$ have some role in letting the bijection be isometrical but I can't figure out how to prove it.
Can you give me a hand? Thank you