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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real-analysis
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0So... what is your definition of $S(p,f,a)$, exactly? I assume you are refering to the integral $\int f\,da$. – 2010-10-29
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0@Arturo Magidin: I guess $S$ is for some sum and $P$ is a partition of $[-1,1]$? It is not clear at all. – 2010-10-29
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0@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. – 2010-10-29
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0@Arturo Magidin: You are probably right. – 2010-10-29