0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ This happens when you study too much for your mind. Thanks @guest. $\endgroup$ – Filburt Feb 11 '17 at 13:33
  • $\begingroup$ Please take a look at my other related question. I need a hand: math.stackexchange.com/questions/2139423/… $\endgroup$ – Filburt Feb 11 '17 at 13:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.