Differentiation of a measure Let  $ v$ a measure, $A$ a set and $g$ defined as :
$$v(A)=\int_{A}^{}g(x)dx$$
Is there a "formal" way for writing the following ?: 
$$dv(A)=d\int_{A}g(x)dx=d\int_{}1_{A}g(x)dx=1_{A}g(x)dx$$
 A: Note first that for a measure $\nu$ on $\mathbf R$, there does not always exists a $g \in L^0(\mathbf R)$ such that 
$$ \nu(A) = \int_A g(x) \, dx $$
If it does, $g$ is called a density of $\nu$. The Radon-Nikodym-theorem gives a neccesary and sufficient condition on $\nu$ - in case $\nu$ is $\sigma$-finite - for such a $g$ to exists, we must have $\nu(A) = 0$ for Lebesgue-nullsets $A \subseteq \mathbf R$. 
If such a $g$ exists, we will show 
$$ \int_{\mathbf R} fg \, dx = \int_{\mathbf R} f\, d\nu \tag + $$
- short $d\nu = g\, dx$ - for $f \in L^1(\nu)$. If $f = \chi_A$ with finite measure $\nu(A) < \infty$, then $(+)$ holds true by definition of $\nu$, as 
$$ \int_{\mathbf R} \chi_A g \, dx = \int_A g\,dx = \nu(A) = \int_{\mathbf R} \chi_A \, d\nu $$
If $f$ is a simple functions, $(+)$ follows by the linearity of integration. If $f \in L^0(\nu)^+$, there is a sequence of integrable simple functions $f_n$ with $f_n \nearrow f$. Beppo-Levi gives 
$$  \int_{\mathbf R} f g \, dx = \lim_n \int_{\mathbf R} f_n g\, dx = \lim_n  \int_{\mathbf R} f_n \, d\nu  = \int_{\mathbf R} f\, d\nu $$
If finally $f \in L^1(\nu)$, write $f = f^+ - f^-$ and use linearity again.
