# Function of bounded variation and Riemann Stieltjes integral.

1. 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?

1. 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.

• Since $dg = 0$, the integral is zero for every function $f$. So, yes, you are right. I don't know what you do in 2. though. Apr 3 '17 at 0:44
• @Fredrich Philipp, I guess $dg$ is a part of RS integral and not always a differential of function. And I have no idea what to do in 2, so.. Apr 3 '17 at 1:03
• dg can be seen as a measure. Here, it is the zero measure. Apr 3 '17 at 1:07

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 $$V(g)\geq 2\sum_{i=1}^\infty|g(x_i)|.$$

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.

• Maybe I wasn't clear in my question. $g(x)=0$ everywhere except countable set of points $I$. Apr 3 '17 at 14:26
• @EzWin : Your question as stated does not say that. Your statement is that $g$ is non-zero at a countable number of points, but there is no statement that it is non-zero only at a countable number of points. So you edit your question. You may mean the stronger statement that $g$ is zero except at .... . Apr 3 '17 at 14:39