# Measure for continuous non-decreasing function

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 $$\mu_f((a,b))=\lim_{\epsilon\rightarrow 0^+}(f(b-\epsilon)-f(a+\epsilon))$$

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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 $\mu_f$.

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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