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

Question

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.

Answer 1

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 $|f(b_k) - f(a_k)| \le ||f'||_{\infty}(b_k - a_k)$.