1549901369-Elements_of_Real_Analysis__Denlinger_

7.9 *Lebesgue's Criterion for Riemann Integrability 447

The collection {Ux : x E [a, b]} is an open cover of [a, b], which is compact.

Thus, it has a finite subcover, say

Let 6 be the minimum radius of these neighborhoods; i.e.,

6= ~min{6x 1 ,6x 2 ,··· ,6xn}.

Now, suppose I is any closed subinterval of [a, b] such that l(I) < 6. Since

the open sets Uxi cover [a, b], I has nonempty intersection with at least one of

them; say 3 xo EI n Ux,·

(

f X; Ex! ~ ]

XI. -..!. 2 U "x i x ,· + _21 Dx,·

)

X;+Dx;

Figure 7.12

Note that 6 ~ ~6x,. Thus, I is a closed interval of length l(I) < ~6x,
containing a point xo of the open interval Ux,. So I ~ N 0 x (xi) (since Ux, has

radius ~6xJ· Therefore, '

w1(I) ~ WJ (Nox, (x i ) n [a, bl) by Remark 7.9.3 (b)

< E:. by (48) •

Theorem 7.9.7 (Lebesgue's Criterion Is Sufficient for Integrability)
Suppose f : [a, b] ~ JR satisfies Lebesgue's criterion: the set of discontinuities
off in [a, b] has measure 0. Then f is Riemann integrable on [a, b].

Proof. Suppose f : [a, b] ~ JR and the set of discontinuities off in [a, b]
has measure 0. If f is constant on [a, b], then f is integrable there, so suppose
f is not constant on [a, b]. Let
m =inf f[a, b] and M = sup f[a, b].
Then m < M, so M - m > 0.
Let S = {x E [a, b]: f is not continuous at x}. Recall that

S = {x E [a,b] : w1(x) > O} = U So(!)
o>O
where S 0 (f) = {x E [a, b] : w1(x) 2'. 5}. (See Remarks 7.9.3.)