316 CHAPTER 13. THE GENERALIZED RIEMANN INTEGRAL

The following example, is very significant. It exposes an unpleasant property of thegeneralized Riemann integral. You can’t multiply two generalized Riemann integrablefunctions together and expect to get one which is generalized Riemann integrable. This islike the case of summation. You know that if you multiply the terms of two conditionallyconvergent series, the resulting series is not necessarily convergent. Also, just becausef is generalized Riemann integrable, you cannot conclude | f | is. Again, this is like thesituation with summation. This is very different than the case of the Riemann integral. It isunpleasant from the point of view of pushing symbols. The reason for this unpleasantnessis that there are so many functions which can be integrated by the generalized Riemannintegral.

Example 13.4.5 Consider the function

f (x) =

{x2 sin

(1x2

)if x ̸= 0

0 if x = 0

Then f ′ (x) exists for all x ∈ R and equals

f ′ (x) =

{2xsin

(1x2

)− 2

x cos(

1x2

)if x ̸= 0

0 if x = 0

Then f ′ is generalized Riemann integrable on [0,1] because it is a derivative. Now let ψ (x)denote the sign of f ′ (x) . Thus

ψ (x)≡

 1 if f ′ (x)> 0−1 if f ′ (x)< 00 if f ′ (x) = 0

Then ψ is a bounded function and you can argue it is Riemann integrable on [0,1] . How-ever, ψ (x) f ′ (x) = | f ′ (x)| and this is not generalized Riemann integrable .

Although you can’t in general multiply two generalized integrable functions and getone which is also a generalized integrable function as shown in the example, sometimesyou can. The following computation will show what the conditions should be.

Let h∈R∗ [a,b] with F an increasing integrator function which we assume is continuoushere. This last assumption could be generalized. What are conditions for g such that hg isalso R∗ [a,b]? Let εn→ 0 and let δ n be a gauge function such that if Pn is a tagged divisionwhich is δ n fine, then ∣∣∣∣∫ b

ahdF−S (P,h)

∣∣∣∣< εn

Without loss of generality, δ n (x)≤ εn. Say Pn consists of the division points a = xn0 < · · ·<

xnmn = b. By Henstock’s lemma, for tm

i the tags,∣∣∣∣∣ k

∑i=1

h(tni )(F (xn

i )−F(xn

i−1))−∫ xn

k

ahdF

∣∣∣∣∣< εn

Let Hnk ≡ ∑

ki=1 h(tn

i )(F (xn

i )−F(xn

i−1))

,Hn0 ≡ 0. Thus, from the above, if k ≥ 1∣∣∣∣Hn

k −∫ xn

k

ahdF

∣∣∣∣= ∣∣enk

∣∣< εn (13.16)