$f:\mathbb R\to\mathbb R, x\mapsto f(x)$. $\mu$ is a Borel measure. Is it possible to simplify $$\frac{d}{dx}\int fd\mu$$ to $f\frac{d\mu}{dx}$?
I only learnt the definition of symmetric differentiation of $\mu$ as $$D\mu=\frac{d\mu}{dm}$$ where $m$ is Lebesgue measure.
Could the notation $\frac{d\mu}{dx}$ be properly defined in any ways??
If $\mu$ is a $\sigma$-finite Borel measure on $\mathbb{R}$, then $\mu = \mu_{ac} + \mu_{s}$, where $\mu_{ac}$ is absolutely continuous with respect to Lebesgue measure on $\mathbb{R}$. Let's assume that $\mu(K) < \infty$ whenever $K$ is a compact subset of $\mathbb{R}$, though it's not clear to me I need to make this assumption in what follows. This will enable us to differentiate $\mu$.
Since $\mu_{ac}$ is absolutely continuous with respect to Lebesgue measure and $\sigma$-finite, it follows by the Radon-Nikodym Theorem that there is a Borel measureable function $f : \mathbb{R} \to \mathbb{R}_{\geq 0}$ such that $\int_{-R}^{R} f(x) \, dx < \infty$ whenever $R > 0$ and $\mu_{ac}(A) = \int_{A} f(x) \, dx$ whenever $A \subseteq \mathbb{R}$ is Borel measurable.
The following is proved in textbooks on harmonic analysis: $$f(x) = \lim_{\epsilon \to 0} \frac{\mu([x - \frac{\epsilon}{2},x + \frac{\epsilon}{2}])}{\epsilon}$$ for each $x \in E$, where the Lebesgue measure of $\mathbb{R} \setminus E$ is zero.
Moreover, if $g : \mathbb{R} \to \mathbb{R}$ is Borel measurable, then $$g(x) f(x) = \lim_{\epsilon \to 0} \epsilon^{-1} \int_{x - \frac{\epsilon}{2}}^{x + \frac{\epsilon}{2}} g(y) \mu(dy)$$ for Lebesgue almost every $x$.
Edit: Suppose $\mu$ and $\nu$ are locally finite Borel measures on $\mathbb{R}^{d}$ with $d \geq 1$ (locally finite meaning compact sets get finite measure). Write $d \mu = f \, d \nu + d \mu_{s}$, where $\mu_{s}$ is singular with respect to $\nu$. Then there is a $E_{s} \subseteq \mathbb{R}^{d}$ such that $\mu_{s}(\mathbb{R}^{d} \setminus E_{s}) = 0$ and $$\forall x \in E_{s} \quad \lim_{\epsilon \to 0^{+}} \frac{\mu_{s}(B(x,\epsilon))}{\nu(B(x,\epsilon)} = \infty.$$ Similarly, there is a $E_{c} \subseteq \mathbb{R}^{d}$ such that $\nu(\mathbb{R}^{d} \setminus E_{c}) = 0$ and $$\forall x \in E_{c} \quad \lim_{\epsilon \to 0^{+}} \frac{\mu_{s}(B(x,\epsilon))}{\nu(B(x,\epsilon)} = 0.$$
Moreover, $$\forall x \in E_{s} \quad \lim_{\epsilon \to 0^{+}} \frac{\mu(B(x,\epsilon))}{\nu(B(x,\epsilon))} = f(x).$$
The proofs of these statements (at least, the ones I've seen) use covering theorems (Besicovitch's Covering Theorem and a mild version of Vitali's Covering Theorem when $\nu$ is nice).