Suppse that an increasing function f: [a, infinity)-> R is Riemann integrable on

Question

Suppse that an increasing function f: [a, infinity)-> R is Riemann integrable on [a, b] for all b>= a E R, and f(x)>=0 for all x>=a. Prove that f is improper Riemann integrable if and only if f(x)=0 for all x>=a

Explanation / Answer

Suppse first that I = [a, b] is a compact interval. Then, for every x ? I, |x| = M = max(|a|, |b|). If n > M, then, for every x? I, |x/n| < 1 and, therefore, (1 + x/n) > 0. It follows that, for n > M, f_n is increasing on I. So, discarding, if necessary, a finite number of functions (what doesn't affect convergence), we can assume (f_n) is a sequence of increasing functions that converges on the compact interval I to the continuous f(x) = e^x. By Polya's theorem, the convergence is uniform on I. If I is any bounded interval, then its closure is compact and the above conclusion holds. Since I is subset of its closure, the desired conclusion follows. Source(s): Polya's theorem: if (f_n) is a sequence of monotone real valued functions that converges on a compact interval I to a continuous function f, then the convergence is uniform.

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