Function of bounded variation and Riemann Stieltjes integral. 
*

*Let $g(x) = C$ everywhere. Is true that for any function $f$ $\int_a^b f dg$ exists?


My idea was to consider the definition, and since for any points $x_k, x_{k+1}: g(x_k)-g(x_{k-1}) = 0$ RS-sum is always zero for any function f. Am I wrong somewhere? 


*Let $I= ${$x_i$}$_1^\infty$ - countable set of points where and only where $g(x)\neq0$


$$ g_n(x)=
\begin{cases}
0, x\in I\setminus (x_1,x_2,x_3,\dots,x_n) \\
g(x), \text{elsewhere}
\end{cases}$$
Then $V([a,b],g_n) \leq V([a,b], g)$.
Not sure if I have any idea how to use that I is countable.
 A: (a) has already been answered so i'll concentrate on (b).
The first thing to say is that $V(g_n)=2\sum_{i=1}^n|g_n(x_i)|=2\sum_{i=1}^n|g(x_i)|$. Indeed for every partition $(t_i)$ of $[a;b]$ one has
$$\sum_{i=0}^k|g_n(t_i)-g_n(t_{i+1})|\leq \sum_{i=0}^k|g_n(t_i)|+|g_n(t_{i+1})|\leq 2\sum_{i=0}^k|g_n(t_i)|\leq 2\sum_{i=1}^n|g_n(x_i)|$$
so $V(g_n)\leq 2\sum_{i=1}^n|g(x_i)|$. On the other hand it's not hard to find a partition $(t_i)$ that reach this bound : choose $(t_i)$ such that each $x_j$ is a $t_i$ and such that between every $x_j$ and $x_l$ there is also a $t_i$.
Using the same argument for $g$ we get $$V(g)\leq 2\sum_{i=1}^\infty|g(x_i)|$$ and once again the other bound is easy to obtain, for every integer $k$ choose a partition $(t_i)$ such that $x_j$ is a $t_i$ for every $j\leq k$ and choose also $(t_i)$ such that between every $x_j$ and $x_l$ with $j,l\leq k$ there is a $t_i$ such that $g(t_i)=0$. For such a partition one has 
$$\sum_{i=0}^p|g(t_i)-g(t_{i+1})|=2\sum_{i=1}^k|x_i|$$
And so taking the limit $k\to \infty$ yields 
$$V(g)\geq 2\sum_{i=1}^\infty|g(x_i)|.$$
A: For (a), the answer is yes, provided your integral definition does not require some requirement such as "$f$ is bounded," which is a common requirement.
For (b), let $g\equiv 1$. The $V_a^b(g)=0$. However, if you modify $g$ to be non-zero at a finite number of points, then the variation of the modified function is no longer $0$. That suggests your problem as stated is false, unless I'm misreading.
