Proving equivalence of operators imply equivalence of measures

Question

Let $A:L^2([0,1],\mu)\to L^2([0,1],\nu)$ an unitary operator. Prove that $$d\mu=\rho(x) d\nu$$ for some $L^1(\mu) \ni \rho(x) >0 (\mu\text{ a.e})$

I thought maybe saying $$\int_{[0,1]}|f(x)|^2d\mu=\int_{[0,1]}|Af(x)|^2d\nu=\int_{[0,1]}|f|^2\cdot \frac{|A(f)|^2}{|f|^2}d\nu.$$ But for $f\in L^2([0,1],\nu)$ which is zero on a set with positive measure, it cannot be established.

How can I find that $\nu\ll \mu$ with positive derivative?

Answer 1

The current statement is also false. Consider uniform probability measures $\mu$ and $\nu$ on the intervals $[0,\frac13]$ and $[\frac23,1]$ respectively. Consider the operator $A$ that shifts every function class in $L^2[0,\frac13]$ by adding $\frac23$ to its argument. It is easy to verify that this shift operation preserves the inner product, i.e., $A$ is unitary. Yet there cannot exist a function $\rho$ with the required properties, because $\rho(x)d\nu$ remains concentrated on the interval $[\frac23,1]$ whatever values $\rho$ takes.

Answer 2

If absolute continuity is not given then the statement is false. As an example, take finite measures $\mu$ concentrated in ${0}$ and $\nu$ concentrated in ${1}.$ The $L^2$ spaces are one-dimensional and a unitary operator is easily constructed.