
“real: chapter_13” — 2011/5/22 — 23:35 — page 47 — #47
Lebesgue Measure and Integration 13-47
The following example shows that the converse of the above
theorem does not hold.
Example 13.5.15 Let f :
R → R defined by
f (x) =
0ifx is rational
1ifx is irrational.
Then f ∈ L
1
[0, 1] (indeed, f = 1 a.e.onR and so
1
0
f (x)dm(x) = 1).
However for any partition P of [0, 1], U (P, f ) = 1 and L(P, f ) = 0
and hence f is not Riemann integrable.
Theorem 13.5.16 Suppose f : (a, b) →
R (−∞ ≤ a < b ≤∞) is
such that f is Riemann integrable over every closed interval [c, d]⊂
(a, b) and that the improper integral
b
a
|f (t)|dt = lim
c→a+
lim
d→b−
d
c
|f (t)|dt
exists. Then the improper ...