3
$\begingroup$

Assume that $f\colon \Bbb R \rightarrow\Bbb R$ is left-continuous nondecreasing and let $\mu$ be a Borel measure in $\Bbb R$ such that $\mu([a,b))=f(b)-f(a)$ for $a, $a,b \in\Bbb R$.

I would like to prove that $\int \phi \,d\mu=-\int\phi'(x)f(x)\,dx$ for each $\phi\colon \Bbb R\rightarrow\Bbb R$ smooth with compact support.

How to show that LHS and RHS in the above equality are equal $\int \int_{ \{(x,y)\in\Bbb R\times\Bbb R:x

1 Answers 1

2

First, we show it when $\mu$ is a finite measure. We can write $\iint_{\{x because $\int_{(-\infty,y)}\phi'(x)dx=\lim_{A\to -\infty}\int_A^y\phi'(t)dt=\lim_{A\to -\infty}\phi(y)-\phi(A)=\phi(y),$ $\phi$ having a compact support.

We can also write $\iint_{\{x where $l:=\lim_{t\to +\infty}f(t)$. Since $\int_{\Bbb R}\phi'(x)dx=0$, we are done in this case.

To deal with the general case, we truncate $f$: denote $f_n(x):=\begin{cases} -n&\mbox{ if }f(x)<-n,\\ f(x)&\mbox{ if } -n\leq f(x)\leq n,\\ n&\mbox{ if }f(x)>n. \end{cases}$ Let $\mu_n$ the Borel measure associated. We need to show that $\lim_{n\to +\infty}\int_{\Bbb R}\phi(x)d\mu_n(x)=\int_{\Bbb R}\phi(x)d\mu(x),$ and the same for the other integral. But it is true because $\phi$ have a compact support and $\mu_n$ restricted to the support of $\phi$ doesn't change.