I'm asked to show that
$\lim_{n\to\infty}\frac{b-a}{n}\sum_{k=1}^n f(a+k\frac{b-a}{n}) = \int_{a}^{b}f(x)dx$.
I tried to come up with a step function that is constant on each of the open intervals $(a+k\frac{b-a}{n},a+(k+1)\frac{b-a}{n})$ and has a jump size of $\frac{b-a}{n}$ at $a+k\frac{b-a}{n}$. I think $\alpha_n = [floor(\frac{x-a}{\frac{b-a}{n}})](\frac{b-a}{n})$ works, so I want to show that $\alpha_n$ converges pointwise to $x$ and then invoke the Helly's second theorem.
I don't know how to show that function converges to $x$, any help will be appreciated.
You're overthinking this problem. The definition of a Riemann-Stieltjes integral $\int_a^b{f(x)\,d\alpha(x)}$ is $$\int_a^b{f(x)\,d\alpha(x)}=\lim_{\substack{P\,\vdash[a,b]\\ P_{j-1}\leq x_j\leq P_j}}{f(x_j)(\alpha(P_j)-\alpha(P_{j-1}))}$$ where the limit is along the net of partitions of $[a,b]$ with arbitrary $x_j$ interleaved, partially ordered by subdivision of partitions. (If those last words sound like nonsense to you, I claim that they are a fancy way of saying roughly "take the mesh $\mu(P)\to0$" that avoids dealing with Darboux sums.)
So choose $P$ and $\{x_j\}_j$ wisely: $$P_n=\left\{a,a+\frac{b-a}{n},a+2\cdot\frac{b-a}{n},\dots,b\right\}$$ $$x(n)_k=a+k\cdot\frac{b-a}{n}\quad\forall k\in[0,n]$$