Let $f\in L^\infty(\mathbb{R})$ and $m$ a nonnegative, bounded measure on $\mathbb{R}$ that almost everywhere has no atoms. Let $g\in L^\infty(\mathbb{R})$ and we have \begin{align} f(t)-f(s) = \int_s^t g(r)m(r)\,dr. \end{align}
I want to show now that for every test function $\varphi \in C^\infty_c(\mathbb{R})$ we have \begin{align} \int_0^T \partial_t\varphi(r)f(r) \,dr = \int_0^T \varphi g m\, dr \end{align}
First Try: I get it for step functions in the sense of distributions. But I'm not sure it is possible to prove the convergence against the derivative of $\varphi$ on the left side, because it is just a weak convergence and $f$ is not smooth.(I also do not know how to show the weak convergence)
Second Try: if $m$ was the Lebesgue measure it would give me that $f$ is AC, right? That means the second equation holds? (differentiable a.e. $\Rightarrow$ weak?)
Does someone has an idea how to prove that? or perhaps a link where I could find a solution and cite? Thanks in advance !
The result as stated is false. For example, if $\phi\in C^\infty_c(\Bbb R)$ is equal to $1$ at every point of $[0,T]$ then the identity you want to prove is $\int_0^T gm(dr)=0$, which doesn't follow from the hypotheses. Assuming the notation $\partial_t\phi(r)$ means just $\phi'(r)$, it's true for $\phi\in C^\infty_c((0,T))$ (except of course that you're missing a minus sign that clearly "should" be there, since this is a version of integration by parts):
In that case we have $$\int_0^T\phi(r)g(r)\,dm(r)=\int_0^T\int_0^r\phi'(s)g(r)\,ds\,dm(r);$$now apply Fubini's theorem:
$$\begin{align}\int_0^T\phi(r)g(r)\,dm(r)&=\int_0^T\int_s^T\phi'(s)g(r)\,dm(r)\,ds \\&=\int_0^T\phi'(s)(f(T)-f(s))\,ds \\&=-\int_0^T\phi'(s)f(s)\,ds,\end{align}$$since $\int_0^T\phi'=0$.