# The Unapologetic Mathematician

## 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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Posted by John Armstrong | 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 | 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 | March 14, 2009 | Reply

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