Friday, April 10, 2009


It clears up the mystery of the popcorn function.

Suppose a function is zero at every irrational point. What is its integral? Well, certainly, the points left out are insignificant. Not merely a set of measure zero but even countable. Such a set plays no role in determining the integral so your function can have any values on the rationals, or even remain undefined on the rationals. The simple answer is that your function is integrable on every interval with a zero integral.

Oh, what? Sorry? You have only learned the Riemann integral. Alas, the answer now is entirely different. Now the function must be defined at these missing rational points and the answer depends on how you define them. If the resulting function is integrable then certainly the value of the integral is zero, but it may or may not be integrable. Let xn be a listing of all the rationals and let your function be defined to be f(xn)=cn and with f(x)=0 at x irrational. What is a necessary and sufficient condition for f to be integrable [i.e., integrable in the dumb Riemann sense]? That's a tough question, but one that is not particularly important.

In 1875, K. J. Thomae discovered the now-famous example of a function of this kind that is continuous at all the irrationals and discontinuous at the rationals. This function has many names: the modified Dirichlet function, Thomae function, Riemann function, raindrop function, ruler function, and popcorn function.

His example is a nice curiosity in the study of continuous functions. But it is usually presented to students of integration theory as example of a seriously discontinuous function that is integrable. The student gets the impression that it is important to have continuity, that discontinuities must be controlled, that without proper configuration of the values of a function the integral is badly affected, and that integration theory has its mysteries. That's good teaching?

If we drop the Riemann integral then the popcorn function would not be mentioned in the context of integration theory and can return to its proper place in the study of continuity.

Is there a function continuous at every rational and discontinuous at every irrational?
Is there a function discontinuous at every rational and continuous at every irrational?

The answer to the first question is "no" and the answer to the second question is "popcorn."

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