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?
 A: 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
$$
\left|L - \sum_{i=1}^n f(t_i)\;\big(g(t_i)-g(t_{i-1})\big)\right| < \epsilon .  
$$  
[There are other equivalent definitions, of course, so you may have to adapt this to your definition.  For example, something with upper sums and lower sums, or with refinement of partitions.]
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.
