Lebesgue’s Criterion for Riemann integrability
Here we give Henri Lebesgue’s characterization of those functions which are Riemann integrable.
Recall the example of the he Dirichlet function, defined on [0,1] by
f (x) =
1
q, if x = pq is rational in lowest terms
0, otherwise .
This function is continuous at all irrational numbers and discontinuous at the ratio- nal numbers. It is also Riemann-integrable (with integral 0). It turns out that there is a connection here. It is the nature of the set of discontinuities that determines integrability.
For a real-valued function f defined on a set X, and I ⊂ X, let ωf (I) = sups,t∈I|f (s) − f (t)|, the oscillation of f on I, as usual. The oscillation of f at a point x is defined as
ωf(x) = inf{ωf (B(x, δ)) : δ > 0}.
It is easy to prove that f is continuous at x if and only if ωf(x) = 0.
Lemma. Let f : [a, b] → R. Then, for every α > 0, {x : ωf(x) < α} is open in [a, b] and {x : ωf(x) ≥ α} is a closed set (in R).
Proof. Let G = {x ∈ [a, b] : ωf(x) < α}. Let c ∈ G. Then, ωf(c) < α and by definition, there is a δ > 0 such that ωf (B(c, δ) ∩ [a, b]) < α. If x ∈ B(c, δ) ∩ [a, b], and U is a neighbourhood of x contained in B(c, δ), then ωf (U ) < α, so ωf(x) ≤ ωf (U ) < α, also. Thus, G is open in [a, b].
Since [a, b] is closed and G is open in [a, b], {x : ωf(x) ≥ α} = [a, b] \ G, is closed in [a, b] and in R.
Let `(I) denote the length of the interval I. A subset N of R is said to have measure 0, if for each ε > 0, there exists countable family H = {I1, I2, . . . } of open intervals covering N , with total lengthP
k`(Ik) < ε.
Lemma.
(1) Every countable set of reals has measure 0.
(2) If B has measure 0 and A ⊂ B, then A also has measure 0.
(3) If Ak has measure 0, for all k ∈ N, thenS
k∈NAk also has measure 0.
.
Proof. (1) Let A = {a1, a2, . . . } be countable, ε > 0, and for every k, let Ik be the interval (a − ε/2k+1, a + ε/2k+1). Then, A ⊂S
kIk — that is these intervals cover A. For each k, the length of Ik is ε/2k, and the total length isP
k`(Ik) ≤ P∞
k=1ε/2k = ε. Thus, A has measure 0.
(2) is obvious, because a family of intervals that covers B also covers A.
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2 LEBESGUE’S CRITERION FOR RIEMANN INTEGRABILITY
To prove (3), one uses a modification of the proof of (1). Let ε > 0. For each k, let Hk be a countable family of intervals whose total length is less than ε/2k. Then, S
kHk is still a countable family of intervals, and their total length is less thanP
kε/2k = ε.
Theorem. (Lebesgue’s Criterion for integrablility) Let f : [a, b] → R. Then, f is Riemann integrable if and only if f is bounded and the set of discontinuities of f has measure 0.
Notice that the Dirichlet function satisfies this criterion, since the set of dis- continuities is the set of rationals in [0, 1], which is countable.
Proof. Let f be Riemann integrable on [a, b]. Then, f is certainly bounded. Let D be the set of points of discontinuity of F . Then D = {x : ωf(x) > 0}. We are to show that D has measure 0. For each α > 0, let N (α) = {x ∈ [a, b] : ωf(x) ≥ α}.
Then, D =S∞
k=1N (1/k). Thus, we need only prove that each N (α) has measure 0.
Fix such an α and let ε > 0. By the Basic Integrability Criterion, we can choose a partition P = {x0, x1, . . . , xn} of [a, b] with
Xn i=1
ωf ([xi−1, xi])(xi− xi−1) < αε/2.
Assume, as we may, that the xi are distinct. Let F be the set of all i for which (xi−1, xi) intersects N (α). Then for each i ∈ F , ωf ([xi−1, xi]) ≥ α. Thus,
αX
i∈F
∆xi≤X
i∈F
ωf ([xi−1, xi])∆xi< αε/2,
so that the sum of the lengths of the intervals (xi−1, xi) is less than ε/2. These cover N (α) except for the elements of {x0, x1, . . . , xn}. But these can be covered by intervals whose lengths total less than ε/2, so that N (α) can be covered with open intervals of total length less than ε, as required.
For the converse, let f be bounded and suppose that the set D of discontinuities of f is of measure 0.
Fix ε > 0 and let E = {x : ωf(x) ≥ ε}. Since E ⊂ D, E has measure 0.
Thus, E can be covered by a countable family of open intervals, whose total length is less than ε. Since E is closed and bounded, it is compact, so a finite family of such intervals will do, say E ⊂Sm
i=1Ui. For each i, let Ii be the closure of Ui. For simplicity, by replacing pairs that intersect, we may assume that no two Iiintersect.
Let D = {I1, . . . , Im}.
The set K = [a, b] \Sm
i=1Uiis compact (in fact, is the union of a finite number of disjoint closed intervals) and consists of points where ωf(x) < ε. For each x ∈ K, there is a closed interval J with x ∈ int J and ωf ([J ]) < ε. By compactness, a finite number of such intervals covers K. By intersecting with K, we can assume that
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LEBESGUE’S CRITERION FOR RIEMANN INTEGRABILITY 3
they are all subsets of K. Thus, let C = {J1, . . . , Jk}, be closed intervals whose union is K and such that ωf ([Jj]) < ε, for all j. We can (and do) assume that the intervals Jk do not overlap.
The family D ∪ C = {[x0, x1], [x1, x2], . . . , [xn−1, xn]} partitions [a, b] and Xn
i=1
ωf ([xi−1, xi])(xi− xi−1) = Xm i=1
ωf (Ii)`(Ii) + Xk j=1
ωf (Jj)`(Jj)
≤X
i
2kf k`(Ii) + Xk j=1
ε`(Jj)
= 2kf kX
i
`(Ii) + ε(b − a)
≤ 2kf kε + ε(b − a),
which is arbitrarily small. Thus, the Basic Integrablity Criterion is satisfied and f is integrable.
You may have noticed that part of this argument is similar to that in the proof that the composition g ◦ f of a continuous function g with an integrable function f is integrable. We see now that the composition result is an immediate consequence of Lebesgue’s criterion.
Lemma. Let f : [a, b] → [c, d] be integrable and g : [c, d] → R be continuous.
Then, g ◦ f is integrable.
Proof. The set of points of discontinuity of f has measure 0, since f is integrable.
But g ◦f is continuous wherever f is, so the set of discontinuities of g ◦f is contained in that of f , so has measure 0 also.
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