Let F be a Borel probability measure on $\mathbb{R}$. And let u be an F integrable function. Consider the integral:
$$\int u(a)aF(da)$$
Now consider a Borel measure $\mu$
\begin{equation} \mu(B) = \int \mathbb{1}_{B}(u(a))a F(da) \qquad (\star) \end{equation}
Where B is a Borel set of $\mathbb{R}$.
How can we show that $\int u(a)aF(da)= \int a \mu(da)$? If this is true? I feel intuitively only that the equality is correct, but have not been able to show analytically.
This is my attempt to show the result, assuming F has a density f. We have
$$\int u(a)a F(da) = \int u(a) af(da)da$$
let $\phi(a)=af(a)$ and note
$$ \int u(a)a F(da) = \int u(a) \phi(a)da = \int u(a) \Phi(da)$$
where $\Phi$ is the measure given by $\Phi(B) = \int \mathbb{1}_{B}(a)aF(da)$. Now letting $\mu = \Phi\circ u^{-1}$ denote the image measure of $\Phi$ under $u$, we have
$$ \int u(a)a F(da) = \int a \Phi\circ u^{-1}(da) = \int a \mu(da)$$
We just need to confirm $(\star)$ is met. We have
$$ \mu(B) = \Phi\{\omega \vert u(\omega) \in B\} = \int \mathbb{1}_{\{\omega \vert u(\omega) \in B\}}(a)aF(da) = \int \mathbb{1}_{B}(u(a))aF(da) $$
I would appreciate any help to extend the argument, if correct, to general measures! WIthout the density assumption.