# Lebesgue-Stieltjes measure absolutely continous with respect to Lebesgue measure.

I aim to show the following result:

Let $$f:\mathbb{R}\to\mathbb{R}$$ be a nondecreasing, continuously differentiable function and let $$\lambda_f$$ be the corresponding Lebesgue-Stieltjes measure generated by $$f$$. Prove:

$$\lambda_f <<\lambda$$ (that is, $$\lambda_f$$ is absolutely continuous with respect to Lebesgue-measure $$\lambda$$).

I tried consider the case when all the things happen in a compact interval $$[a,b]$$ (the general case easily follow from this one). So $$f$$ is uniformly continuous in $$[a,b]$$. Let $$E\subset[a,b]$$ with $$\lambda(E)=0$$ and let $$\epsilon>0$$ arbitrary. We want to find a cover $$\{(a_k, b_k]\}_{k=1}^\infty$$ of $$E$$ such that $$\sum_{k=1}^\infty [f(b_k)-f(a_k)]<\epsilon.\tag{1}$$ If we find $$\{(a_k, b_k]\}_{k=1}^\infty$$ sufficiently fine so that $$f(b_k)-f(a_k)<\epsilon/2^k$$ for all $$k\in \mathbb{N}$$, then (1) follows.

Here we have a problem: $$\epsilon/2^k$$ depends on $$k$$.

Since $$\lambda(E)=0$$, there exists a cover $$\{(a_k, b_k]\}_{k=1}^\infty$$ of $$E$$ such that $$\sum_{k=1}^\infty(b_k-a_k) < \frac{\epsilon}{\star},$$ where $$\star\in\mathbb{R}^+$$ I can control. But the number $$\star$$ cannot depends on $$k$$.

Now, what should I do?

Right now I will try to use the Mean Value Theorem in each interval and see what happen.

By the mean value theorem, for each $k$ we have there's a point $c_k \in [a_k, b_k]$ so that $|f(b_k) - f(a_k)| = |f'(c_k) (b_k - a_k)|$. Since $f$ is $C^1$ we have that $||f'||_{\infty} < \infty$.
So $|f(b_k) - f(a_k)| \le ||f'||_{\infty}(b_k - a_k)$.