Consider the sequence , where gives the fractional part of a real number. The equidistribution theorem states that is irrational iff for any sub-interval of the unit interval, for sufficiently large , roughly of the numbers fall inside . More precisely,

Let us try to understand why the sequence behaves differently depending on whether is rational or not. If for some integers and , then roughly of the first multiples of are integers, i.e., . In fact, it is not difficult to see that for , fraction of the sequence equals . So equidistribution does not occur when is rational.

Now for , consider the exponential function , which has period . As discussed above, roughly of the sequence is , so the exponential sum contains an infinite number of ones. For example if , the exponential sum evaluates to , which is Grandi’s series and diverges. For an arbitrary , the exponential sum is a repeating sum of all -th roots of unity, and one can use the same technique showing that Grandi’s series diverges to establish that this exponential sum diverges too. However, the case when is irrational differs: every nonzero integer multiple of is an non-integer, and thus for any . Writing , the exponential sum converges to the geometric series , since .

In fact, the above intuition can be summarized into the following criterion due to Weyl:

From the above discussion, Weyl’s criterion immediately implies the equidistribution theorem.

Suppose is rational, i.e., for some integers and . Without loss of generality, we assume . Then for all integer , by Euclid’s algorithm, is in . So the sequence is not equidistributed, e.g., no multiples of fall inside .

Now suppose is irrational. Since , we have

which tends to as .

Before proving Weyl’s criterion, it is instructive to state the notion of equidistribution analytically. For , let iff . Note that is a periodic function on all of .

1) For ,

2) For all integers ,

Written in this form, Weyl’s criterion is asserting that if the sum of every interval function at a sequence converges to its integral, then every exponential sum converges to its integral as well, and vice versa. The reason is that both the space of interval functions and the space of exponential functions can approximate one another. More precisely, an exponential function is clearly periodic and continuous (and thus uniformly continuous), and every uniformly continuous function can be approximated by a finite sum of interval functions. For the other direction, a periodic, interval function can be approximated by a periodic, continuous function, which by Fejér’s Theorem can also be approximated by a finite sum of exponential functions. Now we have all the machinery to write down a proof.

For the forward direction (1) => (2), fix and . Since the exponential function continuous on the closed interval , it is also uniformly continuous, i.e., there exists some such that for every , . Let be a step function where if . By construction, is a finite sum of interval functions with .

By linearity, we have

By taking sufficiently large and repeatedly applying triangle inequality, we have

implying that converges to as .

Now we consider the reverse direction (2) => (1). Fix and let . Define

Note that is continuous, periodic, and dominates . By Fejér’s Theorem, there exists such that . By linearity, we have

By taking sufficiently large and repeatedly applying triangle inequality, we have

implying that

Since dominates , we have

One can similarly define a periodic, continuous function that approximates from below, showing that is bounded below by , which together with the proceeding argument will show that the limit of must exist and converge to .