Measure Lebesgue-Stieltjes of single point in $\mathbb{R}$

If $f$ is a nondecreasing function defined on $\mathbb{R}$, then the “length” of a half-open interval $(a, b]$, denoted by $\alpha_f((a;b])$ can be defined by $\alpha_f((a;b]) = f(b) - f(a)$

The Lebesgue-Stieltjes outer measure of an arbitrary set $E \subset \mathbb{R}$ is defined by $\lambda(E) = inf \displaystyle\left\{\sum_{h_k \in F}\alpha(h_k) | h_k = (a_k; b_k], E \subset\bigcup_{h_{k} \in F} h_k\right\}$* $F$ is a countable collection of half-open intervals $h_k$ of the form $(a_k; b_k]$. $\lambda$ is the Lebesgue- Stieltjes measure generated by $f$.(Modern real analysis, William P. Ziemer, pg. 96)

Question: Let $f : \mathbb{R}\rightarrow \mathbb{R}$ be a nondecreasing function and let $\lambda_f$ be the Lebesgue- Stieltjes measure generated by $f$. Prove that $\lambda_f(\{x_0\}) = 0$ if and only if f is left continuous at $x_0$.

My demonstration $(\Rightarrow)$ Let $\epsilon > 0$ there exist a countable collection $F$ of half-open intervals $h_k$ of the form $(a_k; b_k]$ such that $0 < \displaystyle\sum_{h_k \in F}\alpha_f(h_k) < \epsilon$ and $\displaystyle{x_0} \in \bigcup_{h_k \in F} h_k$. Then ${x_0} \in h_{k_i}$ for some $h_{k_i} \in F$, so $x_0 \in (a_{k_0}; b_{k_0}]$; take $\delta= |a_{k_0} - x_0|$, for any $x \in (x_0 - \delta; x_0]$ we have $f(x_0) - f(x)\leq \alpha_f((x_0 - \delta; x_0]) \leq \alpha_f(h_{k_i}) \leq \displaystyle\sum_{h_k \in F}\alpha_f(h_k) < \epsilon$ because $f$ is nondecreasing. Therefore $f$ is left continuous at $x_0$.

($\Leftarrow$) Let $\epsilon > 0$ there exist $\delta > 0$ such that for $x \in (x_0 - \delta; x_0]$ then $\alpha_f((x, x_0]) = f(x_0) - f(x) < \epsilon$. As ${(x, x_0]}$ is a enumerable collection of half-open intervals such that $\{x_0\} \subset (x ; x_0]$ then by the definition * $\lambda(\{x_0\}) = 0$ $\blacksquare$

Any mistake?

• It sounds fine to me. Jan 30 '17 at 19:15

I think you can do it faster: the sequence $${\left\{\left(x_0 - \frac{1}{n} , x_0\right]\right\}}_{n = 1}^{\infty}$$ is contractive, so, using properties about measures, $${\lambda}_f(\{x_0\}) = {\lambda}_f\left(\bigcap_{n = 1}^{\infty} \left(x_0 - \frac{1}{n} , x_0\right]\right) = \lim_{n \to \infty} {\lambda}_f\left(\left(x_0 - \frac{1}{n} , x_0\right]\right) =$$ $$= \lim_{n \to \infty} \left(f(x_0) - f\left(x_0 - \frac{1}{n}\right)\right) = f(x_0) - f(x_0^-)\mbox{.}$$ Now, it's trivial: if ${\lambda}_f(\{x_0\}) = 0$, then $f(x_0) = f(x_0^-)$, so $f$ is continuous from the left at $x_0$, but $f$ is also continuous from the right at $x_0$ because it's a distribution function (and $f$ is continuous from the right on $\mathbb{R}$ by definition of distribution function); definitely, $f$ is continous at $x_0$ (because it's continuous from the left and from the right at $x_0$). If $f$ is continuous at $x_0$, in particular it's continous from the left at $x_0$, so, by the equality ${\lambda}_f(\{x_0\}) = f(x_0) - f(x_0^-)$, it's clair that ${\lambda}_f(\{x_0\}) = 0$.