1549901369-Elements_of_Real_Analysis__Denlinger_

7.9 *Lebesgue's Criterion for Riemann Integrability 445

Remarks 7.9.3 Let a > 0.

(a) \:/nonempty A~ V(f), w1(A) 2'. 0.

(b) \:/nonempty A, B ~ V(f), A~ B::::} w1(A)::::; w1(B).

(c) \:/x E [a, b], w1(x) 2'. 0.

(d) f is continuous at x 0 if and only if w1(x) = 0.

(e) \:/8 > 0, the set S 0 (!) = {x E V(f): w1(x) 2'. 8} is closed, hence compact.

(f) If S denotes the set of points of [a, b] where f is discontinuous, then

00

S = {x E V(f): w1(x) > O} = U So(!)= U S1;n(f).

o>O n=l

Theorem 7.9.4 Suppose f is Riemann integrable on [a, b], and let 8 > 0.
Then, Ve > 0, S 0 (!) can be covered by a finite coll ection of open intervals of
total length less than€. [Hence, S 0 (!) has measure O.]

Proof. Suppose f is Riemann integrable on [a, b], and let 8 > 0 b e fixed.
Let c > 0. Since f is integrable on [a, b], there is a partition P = { xo, x 1 , · · · , Xn}
of [a, b] such that

• 8c
  S(f, P) - S..(f, P) <
  2

.

Since P is a finite set, we can easily cover it by a finite collection of open
intervals of total length less than c/2; just use { Nc/sn(xi): i = 1, 2, · · · , n }. So,
the proof will be complete if we can show that S 0 (!) -P can be covered by a
finite collection of open intervals of total length less than c/2.

Let N = {i: (Xi-1,Xi) n So(!) -=I-0}. Then {(xi-1,Xi): i EN} is a finite
collection of open intervals covering S 0 (!) -P with total length L 6i, where
iEJV
6i = Xi -Xi-l · The proof will be complete if we can show that L 6i < c /2.
iEJV
Now, i EN::::} 3x E (xi-1,Xi) n So(!)

::::} w1(x) 2'. 8

::::} Mi(!) - mi(!) 2'. 8

where Mi(!) =sup f[xi-1, xi] and mi(!) =inf f[xi-1, xi].

Thus, L [Mi(!) - mi(f)]6i 2'. 8 L 6i. (46)

iEJV iEJV

• 8c
  On the other h and, L [Mi(!) - mi(f)]6i::::; S(f, P) - S..(f, P) < -
  2
  . (47)
  iEJV