Let $\mu_1,\mu_2$ be two finite Radon measures on $\mathbb{R}^n$ such that $\mu_1$ is absolutely continuous w.r.t. $\mu_2$, i.e. $\mu_1\ll\mu_2$, and denote by $g_{\mu_2}(\mu_1)$ the Radon-Nikodym derivative of $\mu_1$ w.r.t. $\mu_2$. Take $f\in L^1\left(\mathbb{R}^n,\mathcal{B}\left(\mathbb{R}^n\right), \mu_1\right)$ (where $\mathcal{B}\left(\mathbb{R}^n\right)$ is the Borel $\sigma$-algebra over $\mathbb{R}^n$) and suppose that $A\subset\mathbb{R}^n$ is a Borel set. I am trying to show that $\int_Afd\mu_1=\int_Afg_{\mu_2}(\mu_1)d\mu_2$ with the use of the standard machinery, i.e. first for indicator functions, then for simple functions etc. My problem is that I am stuck with the base case of indicator functions.
Concretely, suppose that $B\subset \mathbb{R}^n$ is a $\mu_1$-measurable subset. Take $f(x) = 1_B(x)$. Then $\int_Afd\mu_1 = \mu_1(A\cap B)$. If $A\cap B$ were to be a $\mu_2$-measurable set, then by the Radon-Nikodym derivative $\mu_1(A\cap B)=\int_{A\cap B}g_{\mu_2}(\mu_1)d\mu_2=\int_{A}1_Bg_{\mu_2}(\mu_1)d\mu_2$. However, I can't recall any approximation/Radon measure property which would allow me to conclude that $B$ is also $\mu_2$-measurable. If this were the opposite case, that $f = 1_B$ for some $\mu_2$-measurable subset $B$, then we could use the fact that $B = E\cup N$ with $\mu_2(N) = 0$ and $E$ a Borel set, implying that $B$ is also $\mu_1$-measurable.
For the record, as of now the only way I know how to prove claims like these are with the monotone/dominated convergence theorems, which are easy to use when working with the limits of simple functions. Therefore I am trying to crack the base case as is.