# to show a function is RS integrable even when lim S(P,f,a) does not exist

Define $f$ on $[-1,1]$ by $$f(x) = \left\{ \begin{array}{ll} 0 & \mbox{if x\lt 0;}\\ 0 & \mbox{if x=0;}\\ 1 & \mbox{if x\gt 0.} \end{array}\right.$$ Let the integrator $a$ be defined by $$a(x) = \left\{\begin{array}{ll} 0 & \mbox{if x\lt 0;}\\ 0 & \mbox{if x=0;}\\ 1 & \mbox{if x\gt 0.} \end{array}\right.$$ Show that $f$ is Riemann-Stieltjes integrable on $[-1,1]$ even though $$\lim_{||P||\to 0} S(P,f,a)$$ does not exist.

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So... what is your definition of $S(p,f,a)$, exactly? I assume you are refering to the integral $\int f\,da$. – Arturo Magidin Oct 29 '10 at 16:33
@Arturo Magidin: I guess $S$ is for some sum and $P$ is a partition of $[-1,1]$? It is not clear at all. – AD. Oct 29 '10 at 17:52
@AD. Yes, it's some kind of Riemman sum associated to the partition P and the functions f and a. At a guess, pick an arbitrary $x_i^*$ in the interval $[x_i,x_{i+1}]$, and take $\sum f(x_i^*)(a(x_{i+1})-a(x))$. But it could means omething else. It could be specific partitions, or specific choices of $x_i^*$, or even something else. – Arturo Magidin Oct 29 '10 at 18:43
@Arturo Magidin: You are probably right. – AD. Oct 29 '10 at 19:51

HINT: Take a sequence of partitions which alternate between including $0$ and not including $0$ while having their size go to $0$ (the most natural choice will do, since $0$ is the midpoint of the interval); pick your "tag" appropriately in the ones that do not include $0$ so that the sum gives you one value, and see what you get when the partition does include $0$. With some appropriate choices, you'll get a sequence that does not converge, showing that the limit does not exist.
To prove the function is Riemman-Stieltjes integrable relative to $a$, break up the integral into two interals, possibly redefining $f$ and $a$ at a single point in one or both of them, so that you can use the standard formula for Riemman-Stieltjes when the functions are "nice".