Show that $\lim_{n\to \infty }\sum_{i=0}^{m_n-1}f(t_i^{(n)})(g(t_{i+1}^{(n)}-g(t_i^{(n)})$ exist. Let $g:[0,1]\to \mathbb R$ with bounded variation and $f:[0,1]\to \mathbb R$ continuous. The aim of the exercise is to construct Stiljes integral.
Let $\pi_n=\{t_0^n,...,t_{m_n}^{n}\} $ a partition of $[0,1]$ s.t. $$|\pi_n|=\max_{i=0,...,m_n-1}|t_{i+1}^n-t_i^n|\underset{n\to \infty }{\longrightarrow } 0.$$
I would like to show that $$\lim_{n\to \infty }\sum_{i=0}^{m_n-1}f(t_i^n)\big(g(t_{i+1}^n)-g(t_i^n)\big)$$
exist. How can I do ? 


*

*The only think I could conclude is that $\sum_{i=0}^{m_n-1}f(t_i^n)\big(g(t_{i+1}^n)-g(t_i^n)\big)$ is uniformly bounded in $n$, but this is not conclusive. 

*Also, I tried tried to prove that it's a Cauchy sequence, but it doesn't really hase sense here, because when you change $n$, you change the partition, and thus the $t_i$. Any idea ?
 A: There is maybe a better/easier way, but it looks to be a bit more technical than it looks.
Let $\sigma =\{t_0,...,t_k\}$ a subdivision of $[0,1]$ and $$S^\sigma (f)=\sum_{i=0}^{k-1}M_i\Delta g_i\quad \text{and}\quad S_\sigma (f)=\sum_{i=0}^{k-1}m_i\Delta g_i,$$
where $\Delta g_i:=g(t_{i+1})-g(t_i).$ Since $g$ has bounded variation on $[0,1]$ and $f$ is continuous on $[0,1]$, then $S^\sigma $ and $S_\sigma $ are uniformly bounded. Therefore, 
$$\overline{S}:=\inf\{S^\sigma \mid \sigma \text{ subdivision of }[0,1]\}$$
and $$\underline{S} =\sup\{S_\sigma \mid \sigma \text{ subdivision of }[0,1]\},$$
are well defined. Since $f$ is continuous and $g$ has bounded variation, one can prove that $\overline{S}=\underline{S}.$
Now, as a classical exercise, you can prove that $$\lim_{n\to \infty }S_{\pi_n}(f)=\lim_{n\to \infty }S^{\pi_n}(f).$$
Indeed, since $$ S^{\pi_n}(f)-S_{\pi_n}(f)\underset{n\to \infty }{\longrightarrow }0,$$
and $$S_{\pi_n}(f)\leq \underline{S}=\overline{S}\leq S^{\pi_n}(f),$$
the result follow.
Now, $$S_{\pi_n}(f)\leq \sum_{i=0}^{m_n-1}f(t_i^n)(g(t_{i+1}^n)-g(t_i^n))\leq S^{\pi_n}(f),$$
and thus, the claim follow.
I'm not sure if there is an easier way.
