Isomorphic Hilbert spaces and mutual singularity Let $(X, \Omega)$ a measurable space, $\mu_1, \mu_2$ a $\sigma-$finite measures on $X$, and $\mu = \mu_1 + \mu_2$. Define
$$V : L^2(\mu) \longrightarrow L^2(\mu_1) \oplus L^2(\mu_2) $$ defined by $Vf = (f, f)$. We easily see that it is well-defined. Furthermore, a moment's consideration shows that $V$ is linear, injective mapping. What I want to show is if $V$ is isomorphism of Hilbert spaces, then $\mu_1$ and $\mu_2$ are mutually singular.
What I tried is that if $\mu_1$ is concentrated on $A \subseteq X$ and $X = A \cup B$ where $A, B$ disjoint, then by the surjectivity, there exist $f \in L^2(\mu)$ such that $f = 1_B$, with $\mu_1 - a.e$, and $f = 1_A$, with $\mu_2 - a.e$. And Im stucked. Can anyone help this problem?
It is from Conway's functional analysis, page 25 #2.
Thanks.
 A: Here is another try which crucially uses Radon-Nikodym and that densities of $\sigma$-finite measures $v\ll \mu$ are $\mu$-a.e. unique. Let $\mu=\mu_1+\mu_2$ and $\varphi_k$ be $\mu$-densities of $\mu_k$. They are both positive with $\varphi_1+\varphi_2=1$
$\mu$-a.e. (because the sum is a $\mu$-density of $\mu$), in particular, $0\le\varphi_k\le 1$ $\mu$-a.e.
Assuming that $V:L^2(\mu)\to L^2(\mu_1)\oplus L^2(\mu_2)$, $f\mapsto (f,f)$ is surjective we want to show that $\mu_2(\{\varphi_1>0\})=0$ because then $\mu_2$ is consentrated on $\{\varphi_1=0\}$ and $\mu_1$ is concentrated on the complement.
In general, $\varphi_1$ does not need to be $L^2(\mu_1)$ so that we cannot use $(\varphi_1,0)$. Therefore, we fix any measurable set with $\mu(A)<\infty$. Then $\varphi_1 1_A \in L^2(\mu_1)$ so that there is $f\in L^2(\mu)$ with $(\varphi_1 1_A,0)=V(f)=(f,f)$. This means $f=\varphi_1 1_A$ $\mu_1$-a.e. and $f=0$ $\mu_2$-a.e. Therefore, for every measurable set $B$, we have $$\int_B f1_A d\mu = \int_B f 1_Ad\mu_1 +\int_Bf1_A d\mu_2 = \int_B \varphi_1 1_A d\mu_1 = \int_B \varphi_1^2 1_A d\mu.$$
The uniqueness of densities implies $f1_A=\varphi_1^2 1_A$ $\mu$-a.e. and since $A$ was arbitrary also $f=\varphi_1^2$ $\mu$-a.e. Since $f=0$ $\mu_2$-a.e. this implies $\varphi_1=0$ $\mu_2$-a.e. as required (and since $f=\varphi_1$ $\mu_1$-a.e. we also get $\varphi_1=\varphi_1^2$ $\mu_1$-a.e. so that $\varphi_1$ is in fact an indicator function).

This is a strange argument and it may very well be that I miss something more elegant.
