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 $f$.
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 $$\mu_F(\{1\})=F(1)-F(1-)=F(1)-\lim_{x\nearrow 1}F(x)=1-0=1>0.$$