Riemann-Lebesgue Lemma

December 20, 2006

The Riemann-Lebesgue lemma is quite general, but since we only know Riemann integration, Ill stateit in that form.

Theorem 1. Let f be Riemann integrable on [a, b]. Then

lim

baf(t) cos(t)dt = 0 (1)

lim

baf(t) sin(t)dt = 0 (2)

lim

baf(t)eitdt = 0 (3)

Proof. I will prove only the first statement. Since f is integrable, given > 0, there is a partition

{a = x0, x1, . . . , xn = b}, so that /2 > baf

n1

mixi 0, where mi is the minimum of f on [xi1, xi].

But the sum can be written asn

1 mixi = ba g, where g =

mi[xi1,xi], and the inequality takes the

form

/2 > ba(f g) 0.

Now we use the fact that f g 0 to get baf(t) cos(t)dt

ba(f(t) g(t)) cos(t)dt

+ bag(t) cos(t)dt

(4) ba(f g) +

(1/)mi (sin(xi) sin(xi1)) . (5)The function g has been fixed. Take large enough that(1/)mi (sin(xi) sin(xi1)) < /2,and we are done.

This proof works nearly verbatim for Lebesgue integration and non-compact intervals.

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