R is the reals. f, g are continuous functions R → R with period 1. Show that lim_{n→∞} ∫_{0}^{1} f(x) g(nx) dx = (∫_{0}^{1} f(x) dx) (∫_{0}^{1} g(x) dx).

**Solution**

The idea is to split the integration range into n equal parts. Thus we get ∫_{0}^{1} f(x) g(nx) dx = ∑ ∫_{r/n}^{r/n+1/n} f(x) g(nx) dx. For large n, f is roughly constant over the range, so we get S f(r/n) ∫_{r/n}^{r/n+1/n} g(nx) dx. Changing the integration variable to t = nx, gives ∑ f(r/n) 1/n ∫_{0}^{1} g(t) dt since g is periodic. But lim ∑ f(r/n) 1/n is just ∫_{0}^{1} f(x) dx, so we get the required (∫_{0}^{1} f(x) dx) (∫_{0}^{1} g(x) dx).

It remains to look at the error involved in approximating f. The function f is continuous and [0, 1] is compact, so it must be uniformly continuous on [0, 1]. Thus we given any ε > 0, we can find N such that for n > N, we have |f(x) - f(r/n)| < ε on [r/n, r/n + 1/n] for each of r = 0, 1, 2, ... , n-1. So the error is at most ∑ ∫_{r/n}^{r/n+1/n} |f(x) - f(r/n)| |g(nx)| dx < ∑ ε 1/n ∫_{0}^{1} |g(t)| dt = ε ∫_{0}^{1} |g(t)| dt, which can be made arbitrarily small.

© John Scholes

jscholes@kalva.demon.co.uk

14 Jan 2002