7.3. THE RIEMANN DARBOUX INTEGRAL∗ 179
Proof: This is verified by adding in one point at a time. Thus let P = {x0, · · · ,xn} andlet Q = {x0, · · · ,xk,y,xk+1, · · · ,xn}. Thus exactly one point y, is added between xk and xk+1.Now the term in the upper sum which corresponds to the interval [xk,xk+1] in U ( f ,P) is
sup{ f (x) : x ∈ [xk,xk+1]}(xk+1 − xk) (7.1)
and the terms which corresponds to the interval [xk,xk+1] in U ( f ,Q) are
sup{ f (x) : x ∈ [xk,y]}(y− xk)+ sup{ f (x) : x ∈ [y,xk+1]}(xk+1 − y) (7.2)≡ M1 (y− xk)+M2 (xk+1 − y) (7.3)
All the other terms in the two sums coincide. Now
sup{ f (x) : x ∈ [xk,xk+1]} ≥ max(M1,M2)
and so the expression in 7.2 is no larger than
sup{ f (x) : x ∈ [xk,xk+1]}(xk+1 − y)+ sup{ f (x) : x ∈ [xk,xk+1]}(y− xk)
= sup{ f (x) : x ∈ [xk,xk+1]}(xk+1 − xk) ,
the term corresponding to the interval [xk,xk+1] and U ( f ,P) . This proves the first part ofthe lemma pertaining to upper sums because if Q ⊇ P, one can obtain Q from P by addingin one point at a time and each time a point is added, the corresponding upper sum eithergets smaller or stays the same and similarly, the resulting lower sum is no smaller.
Lemma 7.3.3 If P and Q are two partitions, then
L( f ,P)≤U ( f ,Q) .
Proof: By Lemma 7.3.2,
L( f ,P)≤ L( f ,P∪Q)≤U ( f ,P∪Q)≤U ( f ,Q) .
Definition 7.3.4I ≡ inf{U ( f ,Q) where Q is a partition}
I ≡ sup{L( f ,P) where P is a partition}.
Note that I and I are well defined real numbers.
Theorem 7.3.5 I ≤ I.
Proof: From Lemma 7.3.3,
I = sup{L( f ,P) where P is a partition} ≤U ( f ,Q)
because U ( f ,Q) is an upper bound to the set of all lower sums and so it is no smaller thanthe least upper bound. Therefore, since Q is arbitrary,
I = sup{L( f ,P) where P is a partition}≤ inf{U ( f ,Q) where Q is a partition} ≡ I
where the inequality holds because it was just shown that I is a lower bound to the set ofall upper sums and so it is no larger than the greatest lower bound of this set.
Now here is the definition of the Darboux integral based on the observation that I ≥ I.