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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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When you post a homework problem, try to give precise information about what you have done and why you are confused, rather than copying it as if you were the teacher and were giving an assignement to the group. Many of us have already taken this course, so we have already turned in our assigned homework. If it is indeed a homework problem, then add the homework tag as well. –  Arturo Magidin Oct 29 '10 at 16:32
    
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

1 Answer 1

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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Some times it is hard to give a hint without spoiling the "Ahhh" :) –  AD. Oct 29 '10 at 19:48
    
Well, 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... –  Arturo Magidin Oct 30 '10 at 2:42

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