37.3. FOURIER TRANSFORM 755

Then for f ∈ L1, let fR (t)≡X[−R,R] (t) f (t). Then for R large,∫∞

−∞

| f (t)− fR (t)|dt =∫

∞

R| f (t)|dt +

∫ −R

−∞

| f (t)|dt < ε

Now fR is Riemann integrable and so there is a step function s(t) = ∑ni=1 aiXIi (t) such that

|s(t)| ≤ | fR (t)| and ∫ R

−R| fR (t)− s(t)|dt =

∫∞

−∞

| fR (t)− s(t)|dt < ε

This follows from the definition of the Riemann integral as a limit of integrals of stepfunctions, details are left for you. Therefore,∫

∞

−∞

|s(t)− f (t)|dt < 2ε

Now ∣∣∣∣∫ ∞

−∞

f (t)sin(rt)dt∣∣∣∣ ≤ ∫

∞

−∞

|( f (t)− s(t))sin(rt)|dt +∣∣∣∣∫ ∞

−∞

s(t)sin(rt)dt∣∣∣∣

≤ 2ε +

∣∣∣∣∫ ∞

−∞

s(t)sin(rt)dt∣∣∣∣ (37.1)

It remains to verify that limr→∞

∫∞

−∞s(t)sin(rt)dt = 0. Since s(t) is a sum of scalars times

XI for I an interval, it suffices to verify that limr→∞

∫∞

−∞X[a,b] (t)sin(rt)dt = 0. However,

this integral is just ∫ b

asin(rt)dt =

−1r

cos(rb)+1r

cos(ra)

which clearly converges to 0 as r→ ∞. Therefore, for r large enough, 37.1 implies∣∣∣∣∫ ∞

−∞

f (t)sin(rt)dt∣∣∣∣< 3ε

Since ε is arbitrary, this shows that 3. holds. ■

Definition 37.3.3 The following notation will be used assuming the limits exist.

limr→0+

g(x+ r)≡ g(x+) , limr→0+

g(x− r)≡ g(x−)

Theorem 37.3.4 Suppose that g∈ L1 (R) and that at some x, g is locally Holder continuousfrom the right and from the left. This means there exist constants K,δ > 0 and r ∈ (0,1]such that for |x− y|< δ ,

|g(x+)−g(y)|< K |x− y|r (37.2)

for y > x and|g(x−)−g(y)|< K |x− y|r (37.3)

for y < x. Then

limr→∞

2π

∫∞

0

sin(ur)u

(g(x−u)+g(x+u)

2

)du

=g(x+)+g(x−)

2.