If we have a finite signed measure $\mu$ on $\mathbb R$ and we define the following function $$ F(x)=\mu([x,\infty)) $$ can we conclude that this function is of bounded variation? I tried to use the Jordan decomposition to get $\mu=\mu^+-\mu^-$ which then is $$F (x)=\mu^+([x,\infty))-\mu^-([x,\infty)) $$ but the problem is, that this is the difference of two decreasing (non-increasing) functions. For the bounded variation we would need that it is the difference of two non-decreasing functions, so this somehow does not work. Anything I can do about it?
For the context: I wanted to show that the convolution of the finite signed measure with an indicator function is of bounded variation, so $$ \mu\ast1_{(-\infty,t]}(x)=\int_{\mathbb R}1_{(-\infty,t]}(x-y)\mu(dy)=\mu(x-t,\infty) $$
Idea:
$F$ is of bounded variation if $-F$ is of bounded variation (is this correct?). If so, we could re-write the above and get $$ G(x)=-F(x)=-\Big(\mu^+([x,\infty))-\mu^-([x,\infty))\Big)=-\mu^+([x,\infty))-\Big(-\mu^-([x,\infty))\Big) $$ now $-G$ would be the difference of two non-decreasing functions and hence of bounded variation. Does this argument work?