Let $X$ be compact metric space and $\{\mu_n\}$ be a sequence of complex borel measures on $X$ which are pairwise mutually singular. Let $H_n=L^2(\mu_n)$ and $H=\bigoplus\limits_n H_n$, $\pi_n:C(X)\to B(H_n)$ be the representation of $C(X)$ defined as $$\pi_n(f)=M_f\ \text{ where }M_f(\phi)=f\phi$$ Define $\pi:=\oplus \pi_n$. Then $\pi:C(X)\to B(H)$ is a representation of $C(X)$. Define $\displaystyle{\mu:=\sum\limits_{n=1}^\infty \frac{\mu_n}{2^n\lVert \mu_n\rVert}}$. Let $\rho:C(X)\to B(L^2(\mu))$ be the representation $\rho(f)=M_f$. Show that, $\pi$ and $\rho$ are equivalent represenation.
This is from J.B. Conway A Course in Functional Analysis (Chapter VIII, example 5.8). So we have to define an isomorphism $U:L^2(\mu)\to H$ such that $U\circ\rho(f)=\pi(f)\circ U$.
I have defined $U(f)=\left\{\frac{f}{\sqrt{2^n\lVert\mu_n\rVert}}\right\}_n\in H$.
Then $\displaystyle{\lVert U(f)\rVert_2^2=\sum\limits_{n=1}^\infty \left\lVert\frac{f}{\sqrt{2^n\lVert\mu_n\rVert}}\right\rVert_2^2}=\lVert f\rVert_2^2$.
This $U$ also satisfies the condition $U\circ\rho(f)=\pi(f)\circ U$, but it is not onto. Can anyone help me to find out such $U$ which is prove the equivalence of the $C^*$-algebra $C(X)$.
Thanks for your help in advance.
Your $U$ is onto because your $\mu_n$'s are mutually singular: let $\{X_n\}$ be a partition of $X$ by Borel sets such that $\mu_n=0$ outside $X_n$. Then, for any $g=\{g_n\}\in H$, wlog (up to equality $\mu_n$-a.e.) $g_n=0$ outside $X_n$ and $$\sum_n\sqrt{2^n\|\mu_n\|}\;g_n=U^{-1}(g).$$