Given a finite set of real numbers $x_1\leq x_2\leq ... \leq x_n$, define the function:
$F(x):= \frac{1}{n} \underset{k=1}{\overset{n}{\sum}} \chi_{(-\infty, x_k]}(x)$
Considering it's induced Stieltjes measure $\mu_F$, I would like to verify (or eliminate) a few properties:
(1) Is $\mu_F\big( (-\infty,x_1) \big)=0$ and $\mu_F\big( (x_n,\infty) \big)=0$?
(2) Is $\mu_F$ supported on $\{ x_k \}_{k=1}^n$, i.e $\mu_F(\mathbb{R})= \underset{k=1}{\overset{n}{\sum}} \mu_F(\{ x_k \})$ ?
(3) Is $\mu_F \big( [x_k,x_{k+1}] \big)= \frac{1}{n} $ for all $k\in [n-1]$?
(4) Is $\int_{-\infty}^tf(x)dF(x)= \frac{1}{n} \underset{k=1}{\overset{n}{\sum}} \chi_{(-\infty, x_k]}(x)\cdot f(x_k)$?
I am trying to understand integration against this Stieljes measure and I am pretty weak on the subject of Stieltjes integral, so I would appreciate pointers on how to relate to this.
You have $\mathcal{B}(\mathbb{R})= \sigma(\{[c,d):c,d\in\mathbb{R}, c<d\})$. You should make yourself clear, that it holds:
\begin{align} \mu_F([c,d))&=F(d-)-F(c-)\\ \mu_F((c,d])&=F(d+)-F(c+)\\ \mu_F((c,d))&=F(d-)-F(c+)\\ \mu_F([c,d])&=F(d+)-F(c-) \end{align}
(1) is true. You should try to solve this on your own.
(2) is also true, actually it holds
$$\mu_F=-\frac{1}{n}\sum_{i=1}^n\delta_{x_k}$$
(3) Is not true, since
$$\mu_F([x_k,x_{k+1}])=F(x_{k+1}+)-F(x_k-)=\frac{1}{n}\big[(n-(k+1))-(n-(k-1))]=-\frac{2}{n}$$
(4) Does not make sense. $x$ has to be integrated, so the right hand side should not depend on $x$. Just try to solve the integral, since you know the measure from (2). Note, that
$$\int_{(-\infty, t]}f(x)dF(x)$$ and $$\int_{(-\infty, t)}f(x)dF(x)$$ can have different values. (What happens, if $t=x_k$ for some $k\in\{1,\ldots, n\}$?)
Please consider checking my answer. Thanks! :-)