Let $\mu$ be a finite Lebesgue-Stieltjes measure on $\mathscr{B}(\mathbb{R})$ such that $\mu(\{x\})=0$ for all $x$. Show that any distribution function F corresponding to $\mu$ is uniformly continuous on $\mathbb{R}$.
I want to show that for all $\epsilon>0$ there exits $\delta>0$ such that $\vert x-y\vert<\delta$ implies $\vert F(x)-F(y) \vert<\epsilon$.
Take $\delta=\epsilon.$ I know that since F is assumed to correspond to $\mu$ we have:
$\mu(a,b]=F(b)-F(a)$ for all $a,b\in\mathbb{R}$. I Also know that $\mu(\{x\})=0$ implies $\mu(a,b] = \mu(a,b) + \mu(\{b\}) = \mu(a,b).$ How can I show that $\mu(a,b) = \vert a-b\vert?$ Is this trivially true?
Hint: If not, then for some $\epsilon>0,$ no $\delta$ works. Thus for $\delta = 1/n, n=1,2,\dots $ there exist $a_n<b_n,$ with $b_n-a_n<1/n,$ such that $|\mu(a_n,b_n]| \ge \epsilon.$ Either $|b_n|\to \infty$ or some subsequence of $b_n$ converges ...