The Unapologetic Mathematician

Mathematics for the interested outsider

Necessary Conditions for Integrability

We’ve talked about sufficient conditions for integrability, which will tell us that a given integral does exist. Now we’ll consider the situation where we know that a given integral exists, and see what we can tell about its integrand or integrator.

First, let’s consider an increasing integrator \alpha over an interval \left[a,b\right], and an interior point c\in\left(a,b\right). Let’s assume that both \alpha and the integrand f are discontinuous from the right at c. Then the integral \int_{\left[a,b\right]} f\,d\alpha cannot exist.

Consider what the assumption of discontinuity means: there is some \epsilon so that for every \delta>0, no matter how small, there are points x,y\in\left(c,c+\delta\right) so that |f(x)-f(c)|\geq\epsilon and |\alpha(y)-\alpha(c)|\geq\epsilon. That is, as we approach c from the right we can always find a sample point that differs from the value we want by at least \epsilon, both in the integrand and in the integrator. So let’s see what happens when we throw c in as a partition point.

Suddenly we’re stuck! Riemann’s condition tells us to look at the sum

\displaystyle U_{\alpha,x}(f)-L_{\alpha,x}(f)=\sum\limits_{i=1}^n\left(M_i(f)-m_i(f)\right)\left(\alpha(x_i)-\alpha(x_{i-1})\right)

where each term is positive. So to make this sum go down to zero in Riemann’s condition we need to make each term go down to zero. But for the subinterval whose left endpoint is c, we can’t make this happen! The change in \alpha will always be at least \epsilon for some smaller subinterval, and the difference in sample values of f must always exceed \epsilon as well. Thus the difference between the upper and lower sum will always be at least \epsilon^2>0, violating Riemann’s condition, and thus integrability.

Of course, we could run through the same argument approaching c from the left. We could also use an integrator \alpha of bounded variation, as usual.

What this tells us is that while we may allow integrands or integrators to be discontinuous, they can’t be discontinuous from the same side at the same point. For example, any discontinuity in an integrator of bounded variation is fine as long as we’re pairing it off with a continuous integrand, as we did last time.

March 25, 2008 Posted by | Analysis, Calculus | 1 Comment

   

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