Let $f$ be a real function defined on $[a, b]$. Assume that the Riemann-Stieltjes integral $\int_a^b f(t) d \alpha(t)$ exists for every increasing function $\alpha$. Show that $f$ is continuous on $[a, b]$.
How to do this question? Any hint is helpful.
Proof: Let $u(t)$ be the "unit step function" given by
$$u(t) = \begin{cases}1 , & t \geq 0 \\0 ,& \ t <0\end{cases}$$
Take any number $c \in (a,b]$. Consider partitions $P$ of $[a,b]$ with $c \in P$. Then $c$ is the right endpoint of a subinterval $[x_{i-1},x_i]$ for some consecutive points $x_{i-1}$ and $x_i$ of $P$. Let
$$\alpha(t)= u(t-c) = \begin{cases}1 , & t \geq c \\0 ,& \ t <c\end{cases}$$.
The existence of $\int_a^b f(t)d\alpha(t)$ means that $\lim_{m(P) \rightarrow 0} R(f, \alpha, P)$ exists. Write $P=\{x_0, x_1,...,x_n\}$. Note that $\alpha(x_j) - \alpha(x_{j-1})=0$ for all $j \ne i$. Thus $R(f, \alpha, P)= \Sigma_{j=1}^n f(s_j)[\alpha(x_j) - \alpha(x_{j-1})]=f(s_i)$, where $s_i \in [x_{i-1}, x_i].$ The fact that $\lim_{m(P) \rightarrow 0} R(f, \alpha, P)$ exists means that as $s_i \to c-$, $lim_{x \to c-} f(x)$ exists and is equal to $f(c)$, as one can take $s_i$ to be $x_i=c$. Thus $f$ is continuous from the left at every number $c \in (a,b]$.
A similar argument using $$\alpha(t)= \begin{cases}1 , & t > c \\0 ,& \ t \le c\end{cases}$$ shows that $f$ is continuous from the right at every $c \in [a,b)$.
Therefore, $f$ is continuous on $[a,b]$