I found this problem in a book I'm using to study (Curso de Análise - Vol 2, Elon Lages Lima).
"Let $\alpha:[a,b] \to \mathbb{R}$ be a bounded function. If $\displaystyle \int_{a}^{b}f(t)d\alpha$ exists for all continuous function $f:[a,b]\to\mathbb{R}$, then $\alpha$ is of bounded variation."
Since $\displaystyle \int_{a}^{b}f(t)d\alpha$ exists for all continuous function $f:[a,b]\to\mathbb{R}$, we have that $\displaystyle \int_{a}^{b}f(t)d\alpha$ exists when $f\equiv c$, $c \in [a,b]$.
By the definition of the Stieltjes integral, $$\int_{a}^{b}f(t)d\alpha = \lim_{|P|\to0} \sum(P^*),$$ where $\sum(P^*)=\sum_{k=1}^nf(\xi_i)[\alpha(t_i)-\alpha(t_{i-1})]$ , with $P^*=(P,\xi)$ , $P=\{a=t_0,t_1,...,t_n=b\}$ and $t_{i-1} \leq \xi_i \leq t_i$.
Since $f \equiv c$, $$\lim_{|P|\to0} \sum(P^*)=c\lim_{|P|\to 0}\sum_{t=1}^n[\alpha(t_i)-\alpha(t_{i-1})]=K \in \mathbb{R}$$
However I can't see a way of putting the |.| ($\left|[\alpha(t_i)-\alpha(t_{i-1})]\right|$) inside the limit and still be sure that it is finite. I've tried in other ways but I always get the same problem, that I don't know if $\sum_{t=1}^n|\alpha(t_i)-\alpha(t_{i-1})|$ is finite when $|P| \to 0$ (and hence $n \to \infty$).
I'd be grateful if someone could give a hint or a solution.
If $P = \{t_0,\ldots t_n\}$ is a partition of $[a,b]$ and $\alpha : [a,b] \to \mathbb R$ let us define $$\sum(P) = \sum_{i=1}^n |\alpha(t_i) - \alpha(t_{i-1})|.$$
Suppose that $\alpha$ does not have bounded variation. Then, for any $M > 0$, there exists a partition $P_0$ with the property that $\displaystyle \sum(P_0) > M$. Moreover, if $P_1$ is any refinement of $P_0$ then necessarily we have $\displaystyle \sum(P_1) \ge \sum(P_0) > M$. If follows that $\displaystyle \lim_{|P| \to 0} \sum (P)$, if it exists, must exceed $M$. Since $M$ is arbitrary, we conclude that $\displaystyle \lim_{|P| \to 0} \sum (P)$ does not converge to a finite value.
Now let $f = 1$. Then $\displaystyle \sum (P*) = \sum(P)$ for any partition $P$. If $\displaystyle \int_a^b f(t) \, d\alpha$ exists, then $$\int_a^b f(t) \, d\alpha = \int_a^b d\alpha = \lim_{|P| \to 0} \sum(P^*) = \lim_{|P| \to 0} \sum(P)$$ which is a contradiction.