Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f$ be a non-decreasing function. We know that $f$ has a countable number of discontinuities; call these $x_j$. How can we find a continuous non-decreasing function $g$ such that $$\mu_f=\mu_g+\sum c_j\delta(x-x_j)?$$ Here, $c_j=f(x_j^+)-f(x_j^-)$, and $\mu_f$ is defined on intervals as $$\mu_f((a,b))=\lim_{\epsilon\rightarrow 0^+}(f(b-\epsilon)-f(a+\epsilon))$$

share|cite|improve this question
up vote 1 down vote accepted

There is no loss of generality in assuming that $f$ is continuous from the right. Then we have $\mu_f(a,b] = f(b)-f(a)$.

Define $\sigma(x) = f(x)-\lim_{y \uparrow x} f(y)$, and let $\Omega = \{ x | \sigma(x) >0 \}$. $\Omega$ is countable.

To show that $\sigma$ is summable in some sense, let $a<b$ and let $\delta_k$ be a finite subset of $\Omega \cap (a,b]$ (the $\delta_k$ can be in increasing order without loss of generality). Then we have (for $n$ large enough): \begin{eqnarray} f(b)-f(a) &=& f(b)-f(\delta_m) + f(\delta_m) - f(\delta_m-{1\over n}) + \cdots f(\delta_1) - f(\delta_1-{1\over n}) +f(\delta_1-{1\over n}) -f(a) \\ &\ge & f(\delta_m) - f(\delta_m-{1\over n}) + \cdots f(\delta_1) - f(\delta_1-{1\over n}) \\ &\ge& \sigma(\delta_m)+\cdots \sigma(\delta_1) \end{eqnarray} Since this is true for all finite subsets, we have $f(b)-f(a) \ge \sum_{\omega \in \Omega \cap (a,b]} \sigma(\omega)$.

We need to define a 'jump' function (terminology from Kolmogorov & Fomin) Let $\gamma(x) = \begin{cases}\sum_{\omega \in \Omega\cap (0,x]} \sigma(\omega) , & x > 0 \\ 0, & x=0 \\ -\sum_{\omega \in \Omega\cap [x,0)} \sigma(\omega), & x < 0 \end{cases}$. A little work shows that $\gamma(b)-\gamma(a) = \sum_{\omega \in \Omega\cap (a,b]} \sigma(\omega) $, which simplifies proofs by avoiding special cases. It also shows that $\gamma$ is non-decreasing.

The function $\gamma$ is right continuous. To see this, pick some $x$ and $x<y$. Then we can write $\gamma(y) = \gamma(x) + \sum_{\omega \in \Omega\cap (x,y]} \sigma(\omega)$. Then we have $0 \le \gamma(y)-\gamma(x) = \sum_{\omega \in \Omega\cap (x,y]} \sigma(\omega) \le f(y)-f(x)$. Right continuity of $\gamma$ then follows from right continuity of $f$.

Now suppose $y<x$, then, as above, $f(x)-f(y) \ge \gamma(x)-\gamma(y) = \sum_{\omega \in \Omega\cap (y,x]} \sigma(\omega) \ge \sigma(x)$. Now, taking limits as $y \uparrow x$, we see that $\lim_{y \uparrow x} (\gamma(x)-\gamma(y)) = \sigma(x)$.

Now let $g = f-\gamma$. Then $g$ is right continuous because both $f,\gamma$ are. Furthermore, $g$ is left continuous because $\lim_{y \uparrow x} (g(x)-g(y)) = \lim_{y \uparrow x} (f(x)-f(y) - (\gamma(x)-\gamma(y))) = \sigma(x)-\sigma(x) = 0$. Hence $g$ is continuous.

To finish, we need to show that $\mu_\gamma$ has the desired form. Let $x_n$ be an enumeration of $\Omega$, and let $\delta_p$ be the Dirac measure concentrated at $p$. Then $\mu_\gamma (a,b] = \gamma(b)-\gamma(a) = \sum_{\omega \in \Omega\cap (a,b]} \sigma(\omega) = \sum_n \sigma(x_n) \delta_{x_n} (a,b]$. Since $c_n = \sigma(x_n)$, we have the desired form $\mu_\gamma = \sum_n c_n \delta_{x_n}$.

Then we have $\mu_f = \mu_g + \sum_n c_n \delta_{x_n}$, the Lebesque decomposition of $\mu_f$.

share|cite|improve this answer
Thanks for your answer, copper.hat! I'm still confused about your reasoning after $f(b)-f(a)\geq\ldots\geq\sigma(\delta_m)+\ldots+\sigma(\delta_1)$. You only chose a finite subset of $(a,b]$, so how can you conclude for $\Omega\cap(a,b]$, which maybe countably infinite? (I'm referring to your conclusion "Since this is true for all finite subsets, we have ...") – PJ Miller Nov 30 '13 at 5:43
Well, if $a_n$ are non-negative, then if $B \ge \sum_{n \in I} a_n$ for all $I$ finite, then $B \ge \sum_{n \in \mathbb{N}} a_n$. This is because if $B < \sum_{n \in \mathbb{N}} a_n$, then it must be true for a finite subset. – copper.hat Nov 30 '13 at 5:48
Notice that in the question, only the limiting values of $f$ are used for both the measure and for the $c_j$. So, we could do the proof like that, but everywhere that $f(x)$ appears, it needs to be replaced by $\lim_{y\downarrow x} f(y)$. So, it is a convenience to just assume it in the first place. If you prefer, you could define $\tilde{f}(x) = \lim_{y\downarrow x} f(y)$, note that $\tilde{f}$ is right continuous, use $\tilde{f}$ in the proof, then 'translate' back to $f$ at the end. (You also need to show $\lim_{y \uparrow x} f(y) = \lim_{y \uparrow x} \tilde{f}(y)$.) – copper.hat Nov 30 '13 at 6:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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