Let $g:\mathbb{R}\to\mathbb{R}$ be a non-decreasing continuous function, and denote by $\mu_g$ be the corresponding Lebesgue-Stieltjes measure. Let $\mu$ denote the Lebesgue measure. Assume $\mu(F)=0$ implies $\mu_g(F)=0$. Then how to show that to any $\epsilon>0$ there exists $\delta > 0$ so that any $t$ and $x_1 < y_1 < x_2 < y_2 <… < x_t < y_t$, we have $\sum(y_j-x_j)<\delta$ implies $\sum(g(y_j)-g(x_j))<\epsilon$.
and I am thinking about using the following theorem:
But which requires $\mu_g$ to be finite.
Best Answer
It is necessary to assume that $g$ is bounded on $\mathbb R$. Otherwise there are simple counterexamples: $g(x)=x^{3}$ is one such. In this case $\mu_g << m$ but $g$ is not even uniformly continuous. The stated property of $g$, which is called absolute continuity, is much stronger than uniform continuity. When $g$ is bounded Theorem 3.5 gives the result.