# Absolutely continuous and Lebesgue-Stieltjes measure

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.

• 'having trouble with the order'? The order in which the intervals are arranged is of no consquence at all in this. You just need disjoint intervals with total length less than $\delta$. So you can get this result from Theorem 3.5. – Kavi Rama Murthy May 9 '19 at 8:03
• What is your definition of $\mu_g$? Isn't $\mu_g (a,b]=g(b)-g(a)$? – Kavi Rama Murthy May 9 '19 at 8:13
• @KaviRamaMurthy Ok,I think I made a mistake on the definition of $\mu_g$, but to apply the theorem we need $\mu_g$ a finite measure, does the definition of $g$ implies that? – 6666 May 9 '19 at 8:18
• @KaviRamaMurthy if you check that theorem from Folland's book, you will see the proof needs the $\mu_g$ to be finite, I have attached the poof. – 6666 May 9 '19 at 8:23

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.
• When $g:X\to\mathbb{R}$ is a non-decreasing continuous function, then for the $\mu_g$'s continuity from above, is $\mu_g(X)<\infty$ still required? – 6666 May 9 '19 at 9:08
• In the case of $x^{3}$ the function is continuous and increasing but the conclusion is false. To show that it is false consider intervals of the type $(n,n+\frac 1 n)$. – Kavi Rama Murthy May 9 '19 at 9:12