Given $\Omega\subset\mathbb{R}^n$, the measure space $(\Omega,\mathcal{B}(\Omega),\lambda)$, some Borel measure $\mu:\mathcal{B}(\Omega)\to[0,\infty]$ and a Lebesgue measurable function $f:\Omega\to\mathbb{R}$, und what conditions is it possible to calculate $$\int_\Omega f(x)\mathrm{d}\mu(x)=\int_\Omega f(x)\mu(x)\mathrm{d}x\ ?$$
Can you give me some reference on where I can find a result like this (perhaps in a more general setting)? I have the feeling that many authors implicitly use this "transformation", but I would like to see a rigorous proof of it.
In many cases $\mu(\{x\})$ will be zero, since Borel measures on $\mathbb{R}^n$ are Lebesgue-Stieltjes measures. This post explores this case.
Therefore I suppose that $\mu(x)$ probably does not mean $\mu(\{x\})$, but rather the density of $\mu$ w.r.t. the Lebesgue measure $\lambda$. To avoid confusion, denote $g$ as this density, i.e. a measurable nonnegative function such that $\mu = g\ d\lambda$. If $f$ is also nonnegative and measurable, i.e. a density w.r.t. $\mu$, then we will have $$ \int f(x)\ d\mu(x) = \int f(x)\cdot g(x)\ d\lambda(x) $$
This can be proven by letting $f=\chi_A$ for $A\in\mathcal{B}$, then using linearity to show this for simple functions. The monotone convergence theorem yields this result for nonnegative measurable $f$.
Note that if actually $\mu(x)=\mu(\{x\})$ you would be suggesting that all measures $\nu = f\ d\mu$ are absolutely continuous to the Lebesgue measure with this very specific density $f\cdot \mu(\{\cdot\})$.