The Unapologetic Mathematician

Mathematics for the interested outsider

The Integral as a Function of the Interval

Let’s say we take an integrator \alpha of bounded variation on an interval \left[a,b\right] and a function f that’s Riemann-Stieltjes integrable with respect to \alpha over that interval. Then we know that f is also integrable with respect to \alpha over the subinterval \left[a,x\right]\subseteq\left[a,b\right]. Let’s use this to define a function F on \left[a,b\right] by

\displaystyle F(x)=\int\limits_{\left[a,x\right]}fd\alpha

We can immediately say some interesting things about this function. First of all, F is, like \alpha, of bounded variation. Next, wherever \alpha is continuous, so is F. Finally, if \alpha is increasing, then F is differentiable wherever \alpha is differentiable and f is continuous. At such points, we have F'(x)=f(x)\alpha'(x). Notice that, as usual, the first two results will hold if we show that they hold for increasing integrators.

These results are similar to those we get from the Fundamental Theorem of Calculus, and we can use some of the same techniques to prove them. In particular, we call on the Integral Mean-Value Theorem for Riemann-Stieltjes integrals. If we take points {x} and y in \left[a,b\right], this tells us that

\displaystyle F(y)-F(x)=\int\limits_{\left[a,y\right]}fd\alpha-\int\limits_{\left[a,x\right]}fd\alpha=\int\limits_{\left[x,y\right]}fd\alpha=f(x_0)\left(\alpha(y)-\alpha(x)\right)

where x_0 is some point between x and y.

Now we let M be the supremum of |F| on \left[a,b\right]. For any partition of \left[a,b\right] we calculate the variation

\displaystyle\sum\limits_{i=1}^n|F(x_i)-F(x_{i-1})|\leq M\sum\limits_{i=1}^n|\alpha(x_i)-\alpha(x_{i-1})|\leq MV_\alpha(a,b)

thus giving an upper bound to the variation of F.

Similarly, let {x} be a point where \alpha is continuous. Then given an \epsilon>0 we can find a \delta so that |y-x|<\delta implies that |\alpha(y)-\alpha(x)|<\frac{\epsilon}{M}. Thus |F(y)-F(x)|<\epsilon, and we see that F is continuous at {x} as well.

Finally, we can divide by y-x to find


Then as y gets closer to {x}, x_0 gets squeezed in towards {x} as well. If f is continuous at {x} and \alpha is differentiable there, then the limit of this difference quotient exists, and has the value stated.


March 31, 2008 - Posted by | Analysis, Calculus


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  2. […] over the subinterval . Why not just start by saying it’s integrable over ? Because now we have a function on defined […]

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