The Unapologetic Mathematician

Mathematics for the interested outsider

Integrability over subintervals

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 f and the variation of the integrator \alpha, 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 \alpha is a function of bounded variation, f is integrable with respect to \alpha on \left[a,b\right], and c is a point between a and b, then f is integrable with respect to \alpha on \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

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], then this theorem states that f is integrable over \left[c,b\right], and another invocation of the theorem shows that f is integrable over \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 \epsilon>0 there is a partition x_\epsilon of \left[a,b\right] so that U_{\alpha,x}(f)-L_{\alpha,x}(f)<\epsilon for any partition x finer than x_\epsilon.

We may as well assume that c is a partition point of x_\epsilon, since we can just throw it in if it isn’t already. Then the partition points up to c form a partition x_\epsilon' of \left[a,c\right]. Further, any refinement x' of x_\epsilon' is similarly part of a refinement x of x_\epsilon. So by assumption we know that U_{\alpha,x}(f)-L_{\alpha,x}(f)<\epsilon, and we get down to x' by throwing away terms in the sum for partition points above c. Each of these terms is nonnegative, and so we see that

U_{\alpha,x'}(f)-L_{\alpha,x'}(f)<U_{\alpha,x}(f)-L_{\alpha,x}(f)<\epsilon

That is, f satisfies Riemann’s condition with respect to \alpha on \left[a,c\right], and so it’s integrable.

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March 21, 2008 - Posted by | Analysis, Calculus

2 Comments »

  1. [...] variation (in fact it’s decreasing), and so it’s integrable with respect to . Then it’s integrable over the subinterval . Why not just start by saying it’s integrable over ? Because now we [...]

    Pingback by Improper Integrals I « The Unapologetic Mathematician | April 18, 2008 | Reply

  2. [...] and a function that’s Riemann-Stieltjes integrable with respect to over that interval. Then we know that is also integrable with respect to over the subinterval . Let’s use this to define a [...]

    Pingback by The Integral as a Function of the Interval « The Unapologetic Mathematician | March 14, 2009 | Reply


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