I am working in $L^2$ and I want to change from some measure, $\rho$, to the usual measure. This measure is continuous w.r.t. the Lebesgue measure, and so, I can write it as $\rho(dt) = \rho(t)dt$. I am trying to prove the following thing:
$W_{\rho} = W_{dt}\sqrt{\rho}$, where W is some function in L².
Indeed, I have computed $\langle W,W \rangle_{\rho} = \int W² (\rho dt) = \int (W\cdot \sqrt{\rho})^2 dt = \langle W\cdot \sqrt{\rho}, W\cdot \sqrt{\rho} \rangle_{dt}$, and I somehow need to get the following equality:
$$ \mathcal{F}(W_{\rho}) = \mathcal{F}(W_{dt}\sqrt{\rho}) $$
where $\mathcal{F}$ is the Fourier Transform. But I am confused, because I am getting:
$$ \mathcal{F}(W_{\rho}) = \int e^{-2\pi i x \cdot t} W(t) \rho(t) dt = \mathcal{F}(W_{dt}\cdot \rho) $$
Where am I going wrong?