# Lebesgue-Stieltjes measure of Singletons - when is it zero?

Let $f: \mathbb{R} \to \mathbb{R}$ be increasing, right-continuous and nonnegative. Define the Lebesgue-Stieltjes measure $\mu_f$ by $$\mu_f(a,b] = f(b) - f(a).$$

When does $\mu_f\{b\} = 0$?

I think it is true for linear such $f$, since $f(0) = 0$. Proof: Let $f$ be linear and let $\epsilon > 0$. Then $$\mu_f\{b\} \leq \mu_f(b-\epsilon, b] = f(b) - f(b - \epsilon) = f(\epsilon).$$ Since this holds $\forall \epsilon > 0$, $$\mu\{b\} = \lim_{\epsilon \to 0^+} \mu_f\{b\} \leq \lim_{\epsilon \to 0^+} f(\epsilon) = f(0) = 0,$$ where I used right-continuity of $f$ in the second-to-last equality and linearity of $f$ for the last.

Hence $\mu_f\{b\} = 0$ for linear such $f$

It looks like it's also true for affine such $f$. Is there a more general result?

Update Thanks to Cain's comment, we also get $\mu_f\{b\} = 0$ for continuous such $f$.

• If you'll inspect your claim you will notice you've only used the fact that a linear function is continous. Try to generalize this idea. – Cain Sep 18 '15 at 20:03
• @Cain Ah, good eye. I see it. Thanks for the insight! – bcf Sep 18 '15 at 20:09
• even more, notice that you just need right continuity. – L.F. Cavenaghi Sep 18 '15 at 20:17
• @LeonardoFranciscoCavenaghi hmm, okay. Since every Lebesgue-Stieltjes function is right-continuous, we get $\mu_f\{b\} = 0$ for any such $f$?? I seemed to have thought it was possible that $\mu_f\{b\} > 0$ for some $f$. – bcf Sep 18 '15 at 20:22
• @L.F.Cavenaghi is mistaken here. Points can have positive measure even when $f$ is right continuous; consider $f = 1_{[0,\infty)}$ which is a point mass at 0. They will have zero measure iff $f$ is also left continuous. Measurability is not relevant; indeed, every Lebesgue-Stieltjes function is measurable. – Nate Eldredge Oct 31 '17 at 15:53

Hopefully someone finds this later answer useful. What is actually true (see exercise 1.28 in Folland's Real Analysis) is that $$\mu_F(\{a\})=F(a)-F(a-),$$ where $F$ is increasing and right-continuous. It need not, of course, be true for such $F$ that $F(a-)=F(a)$. Consider $$F(x)=\begin{cases} 0&\text{if}~x<1\\ 1&\text{if}~x\geq 1 \end{cases}.$$ Then $F$ is increasing and right continuous, but $$\mu_F(\{1\})=F(1)-F(1-)=F(1)-\lim_{x\nearrow 1}F(x)=1-0=1>0.$$