As I noted when I first motivated bounded variation, we’re often trying to hold down Riemann-Stieltjes sums to help them converge. In a sense, we’re sampling both the integrand 
and the variation of the integrator 
, and together they’re not big enough to make the Riemann-Stieltjes sums blow up as we take more and more samples. And it seems reasonable that if these sums don’t blow up over the whole interval, then they’ll not blow up over a subinterval.
More specifically, I assert that if is a function of bounded variation, is integrable with respect to on ![\left[a,b\right]](http://s0.wp.com/latex.php?latex=%5Cleft%5Ba%2Cb%5Cright%5D&bg=e6e6e6&fg=333333&s=0 "\left[a,b\right]")
, and 
is a point between 
and 
, then is integrable with respect to on ![\left[a,c\right]](http://s0.wp.com/latex.php?latex=%5Cleft%5Ba%2Cc%5Cright%5D&bg=e6e6e6&fg=333333&s=0 "\left[a,c\right]")
.
Then, in the equation expressing “linearity” in the interval
![\displaystyle\int\limits_{\left[a,b\right]}fd\alpha=\int\limits_{\left[a,c\right]}fd\alpha+\int\limits_{\left[c,b\right]}fd\alpha](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cint%5Climits_%7B%5Cleft%5Ba%2Cb%5Cright%5D%7Dfd%5Calpha%3D%5Cint%5Climits_%7B%5Cleft%5Ba%2Cc%5Cright%5D%7Dfd%5Calpha%2B%5Cint%5Climits_%7B%5Cleft%5Bc%2Cb%5Cright%5D%7Dfd%5Calpha&bg=e6e6e6&fg=333333&s=0 "\displaystyle\int\limits_{\left[a,b\right]}fd\alpha=\int\limits_{\left[a,c\right]}fd\alpha+\int\limits_{\left[c,b\right]}fd\alpha")
we have two of these integrals known to exist. Therefore the third one does as well, and the equation is true$. If we have a subinterval ![\left[c,d\right]\subseteq\left[a,b\right]](http://s0.wp.com/latex.php?latex=%5Cleft%5Bc%2Cd%5Cright%5D%5Csubseteq%5Cleft%5Ba%2Cb%5Cright%5D&bg=e6e6e6&fg=333333&s=0 "\left[c,d\right]\subseteq\left[a,b\right]")
, then this theorem states that is integrable over ![\left[c,b\right]](http://s0.wp.com/latex.php?latex=%5Cleft%5Bc%2Cb%5Cright%5D&bg=e6e6e6&fg=333333&s=0 "\left[c,b\right]")
, and another invocation of the theorem shows that is integrable over ![\left[c,d\right]](http://s0.wp.com/latex.php?latex=%5Cleft%5Bc%2Cd%5Cright%5D&bg=e6e6e6&fg=333333&s=0 "\left[c,d\right]")
. So being integrable over an interval implies that the function is integrable over any subinterval.
As we said earlier, we can handle all integrators of bounded variation by just considering increasing integrators. And then we can use Riemann’s condition. So, given an 
there is a partition 
of so that 
for any partition 
finer than .
We may as well assume that is a partition point of , since we can just throw it in if it isn’t already. Then the partition points up to form a partition 
of . Further, any refinement 
of is similarly part of a refinement of . So by assumption we know that , and we get down to by throwing away terms in the sum for partition points above . Each of these terms is nonnegative, and so we see that
That is, satisfies Riemann’s condition with respect to on , and so it’s integrable.
March 21, 2008
Posted by John Armstrong |
Analysis, Calculus |
2 Comments
