# A function which is not Riemann-Stieltjes integrable on [0,2]

Question - Give examples for two increasing functions g and f on [0,2] such that f is not Riemann-Stieltjes integrable on [0,2] with respect to g.

I cant figure out a way to find two such functions.Can you tell me the way to think of two such functions?

• I guess you know some theorems about when $\int_0^2 f\;dg$ exists. So you will have to choose $f,g$ to falsify the hypotheses of those theorems. – GEdgar Nov 27 '15 at 13:56
• By the way, I think this is a nice question! You have to think about the definition of Riemann-Stieltjes integral. – GEdgar Nov 27 '15 at 14:05
• @GEdgar well it seems like i have no clue :/ .Could you please tell me where to look – Razor1692 Nov 28 '15 at 9:46

So you probably have some theorems about existence of R-S integral $\int_a^b f\;dg$, where there is a condition that $f$ and $g$ have no discontinuity in common. So for our example we must let $f$ and $g$ have a discontinuity in common. (Since the problem has integral from $0$ to $2$, it is convenient to do this at the point $1$.)
Let's take this definition. Let $a < b$ be reals, let $f,g : [a,b] \to \mathbb R$ be two functions, and let $L \in \mathbb R$. Then $L = \int_a^b f\;dg$ means: for every $\epsilon > 0$ there is a partition $$s_0=a < s_1 < s_2 < \dots < s_n = b$$ of $[a,b]$ such that, for every choice of tags $(t_i)_{i=1}^n$ [ that is $s_{i-1}\le t_i \le s_i$ for $i=1,2,\dots,n$ ], we have $$\left|L - \sum_{i=1}^n f(t_i)\;\big(g(t_i)-g(t_{i-1})\big)\right| < \epsilon .$$
OK here is our example. $a=0, b=2$. $f,g$ are both equal to the function: $f(x)=0$ for $0 \le x < 1$, $f(x) = 1$ for $1 \le x \le 2$. Take $\epsilon < 1/2$. I claim there is no $L$ that satisfies the definition. Let $$s_0=0 < s_1 < s_2 < \dots < s_n = 2$$ be any partition of $[0,2]$. There is an index $j$ so that $s_{j-1} < 1 \le s_j$. Note that $g(s_i)-g(s_{i-1}) = 0$ for all $i$ except $i=j$, and $g(s_j)-g(s_{j-1}) = 1$. Now there is a choice of tags with $t_j=1$, and then $$\sum_{i=1}^n f(t_i)\;\big(g(t_i)-g(t_{i-1})\big) = f(t_j)\;\big(g(t_j)-g(t_{j-1})\big) = 1 \cdot 1 = 1 . \tag{A}$$ But there is also a choice of tags with $s_{j-1}< t_j < 1$, and then $$\sum_{i=1}^n f(t_i)\;\big(g(t_i)-g(t_{i-1})\big) = f(t_j)\;\big(g(t_j)-g(t_{j-1})\big) = 0 \cdot 1 = 0 . \tag{B}$$ There is no $L \in \mathbb R$ within distand $\epsilon$ of both (A) and (B). So this R-S integral does not exist.