If $\int_A f d\nu = \int_A f d\nu'$ for some $f>0$ and all measurable $A$, does $\nu$ equal $\nu'$? Let $\nu,\nu'$ be finite measures (same total mass) on $\mathbb{R}^d$ (with Borel sigma-algebra), and $f: \mathbb{R}^d \rightarrow (0,1)$ be a  measurable function.
Note: $f$ is fixed (this the the point of difference to similar questions that have been asked). 
Suppose that for all Borel sets $A$
$$\int_A f d\nu = \int_A f d\nu'.$$
Is it then true that $\nu=\nu'$? Is this a simple consequence of a particular well known theorem? 
A reference would be very useful. Thanks!

This is true.
Use a monotone class method to prove that for every positive Borel function $g$, we have 
$$\int fg d\nu = \int fg d\nu'$$
Take any Borel set $A$ and define $g = \frac{1}{f}\mathbf{1}_A$. This yelds $\int \mathbf{1}_A d\nu = \int \mathbf{1}_A d\nu'$, that is $\nu(A) = \nu'(A)$.
