The Unapologetic Mathematician

Mathematics for the interested outsider

The Riemann-Stieltjes Integral III

Last Friday we explained the change of variables formula for Riemann integrals by using Riemann-Stieltjes integrals. Today let’s push it a little further and prove a change of variables formula for Riemann-Stieltjes integrals.

We start with a function f:\left[a,b\right] which we assume to be Riemann-Stieltjes integrable by the function \alpha. Now, instead of the full generality we used before, let’s just let g:\left[c,d\right]\rightarrow\left[a,b\right] be a strictly increasing continuous function with g(c)=a and g(d)=b. Define h and \beta to be the composite functions h(x)=f(g(x)) and \beta(x)=\alpha(g(x)). Then h is Riemann-Stieltjes integrable by \beta on \left[c,d\right], and we have the equality

\displaystyle\int\limits_{\left[a,b\right]}fd\alpha=\int\limits_{\left[c,d\right]}hd\beta

For decreasing functions we get almost the exact same thing, so you should figure out the parallel statement and proof yourself.

Since g is strictly increasing, it must be one-to-one, and it’s onto by assumption. In fact, g is an explicit homeomorphism of the intervals \left[a,b\right] and \left[c,d\right], and its inverse g^{-1}:\left[a,b\right]\rightarrow\left[c,d\right] is also a strictly increasing continuous function. We can now use g and its inverse to set up a bijection between partitions of \left[a,b\right] and \left[c,d\right]: if a=x_0<x_1<...<x_n=b is a partition, then c=g^{-1}(x_0)<g^{-1}(x_1)<...<g^{-1}(x_n)=d is a partition, and vice versa. Further, refinements of partitions of one side correspond to refinements of partitions on the other side.

So if we’re given an \epsilon>0 then there’s some partition y_\epsilon of \left[a,b\right] so that for any finer partition y we have |f_{\alpha,y}-\int_{\left[a,b\right]}f,d\alpha|<\epsilon. Let x_\epsilon=g^{-1}(y_\epsilon) be the corresponding partition of \left[c,d\right], and let x be a partition of \left[c,d\right] finer than it. Then it’s easily verified that the Riemann-Stieltjes sum h_{x,\beta} is equal to the Riemann-Stieltjes sum f_{g(x),\alpha}. Everything else follows quickly from here.

March 3, 2008 Posted by | Analysis, Calculus | 12 Comments

   

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