320 CHAPTER 13. THE GENERALIZED RIEMANN INTEGRAL

and let δ (α) = δ (β ) = ε > 0. If you have any δ fine division P of [a,b] , note thatif t is a tag such that t /∈ {α,β} , then (t−δ (t) , t +δ (t)) cannot contain α and itcannot contain β . Therefore, both α,β are tags. Furthermore, explain why each ofα and β is an interior point of the interval for which they are tags. Now explain whythe division points are

a = x0 < · · ·< xk < α < xk+2 < · · ·< xm < β < xm+2 < · · ·< xn = b

where (xk+2− xk) < 2ε,(xm+2− xm) < 2ε . Now explain why if I = (α,β ) , thenS (P,XI)=F (xm)−F (xk+2) and if I =(α,β ], S (P,XI)=F (xm+2)−F (xk+2) and ifI = [α,β ), then S (P, f )=F (xm)−F (xk) and if I = [α,β ] , then S (P, f )=F (xm+2)−F (xk). Now as ε → 0, show these converge respectively to F (β−)− F (α+),F (β+)−F (α+), F (β−)−F (α−) , and F (β+)−F (α−). Thus indicator func-tions of intervals are generalized Riemann integrable and we can even compute them.Consider a case where (α,β ) is not contained in [a,b] and use the same method tocompute

∫ ba X(α,β )dF . Also consider intervals of the form (α,∞).

17. The gamma function is defined for x > 0 as

Γ(x)≡∫

∞

0e−ttx−1dt ≡ lim

R→∞

∫ R

0e−ttx−1dt

Show the integral from 0 to R exists as a generalized Riemann integral directly fromthe definition. Also show that

Γ(x+1) = xΓ(x) , Γ(1) = 1.

How does Γ(n) for n an integer compare with (n−1)!? This was all done earlier forthe improper Riemann integrals. Why is there no change with generalized Riemannintegrals?