Given $\mu$ finite Borel measure and $f:\mathbb{R}\to \mathbb{R}$ by $f(x)=\mu((-\infty,x])$. Prove that for all $c\in \mathbb{R}$: $$\int_{\mathbb{R}}[f(x+c)-f(x)]dx= c\mu(\mathbb{R}).$$
Actually, I've tried a lot of things, one of them is trying a lemma that says $\mu \times \mu(E)=\int \mu(E)d\mu(x)$ to get some inequality. I really don't know how to get something good in this exercise. Types?
Since $f(x)=\int_{\mathbb{R}}\chi_{(-\infty,x]}(y)d\mu(y)$, let $g(x,y)=\chi_{(-\infty,x+c]}(y)-\chi_{(-\infty,x]}(y)$ is measurable and $$\int_{\mathbb{R}\times\mathbb{R}}|g(x,y)|dxd\mu(y)=c\mu(\mathbb{R})<\infty,$$ by use of Fubini's theorem we conclude that \begin{align*}\int_{\mathbb{R}}[f(x+c)-f(x)]dx &=\int_{\mathbb{R}}(\int_{\mathbb{R}}g(x,y)d\mu(y))dx=\int_{\mathbb{R}}(\int_{\mathbb{R}}g(x,y)dx)d\mu(y)\\ &=\int_{\mathbb{R}}cd\mu(y)=c\mu(\mathbb{R}). \end{align*}