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.
to show a function is RS integrable even when lim S(P,f,a) does not exist
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0@Arturo Magidin: You are probably right. – 2010-10-29
1 Answers
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".
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0Well, anyone who thinks that simply copying a question and posting it is the appropriate way to go is clearly in need of more than a gentle hint... – 2010-10-30