THE ADDITIVITY OF THE INTEGRALFOKKO VAN DE BULTIn this note we prove the following Theorem.Theorem 1. Let f and g be integrable functions on the interval [a, b]. Then f + gis also integrable on [a, b] and we haveZba(f + g)(x)dx =Zbaf(x)dx +Zbag(x)dx.Proof. The proof given here is almost the same as the one described in the bookon pages 85 and 86, and probably written down better there.Suppose f and g are integrable on [a, b]. Choose an arbitrary n ∈ N. Thenwe see that there exists a step function sn≤ f such thatRbasn(x)dx ≥ I(f ) −14n=Rbaf(x)dx −14n, as otherwise I(f ) −14nwould be an upperbound to S ={Rbas(x)dx | s ≤ f, s a step function} less than I(f), in violation of the definitionI(f) = sup(S). Likewise we find tn, ˜snand˜tnwithRbatn(x)dx ≤Rbaf(x)dx +14n,Rba˜sn(x)dx ≥Rbag(x)dx −14nandRba˜tn(x)dx ≤Rbag(x)dx +14n.Combining these equations, and using the additivity for step functions gives usthatZba(sn+ ˜sn)(x)dx ≥Zbaf(x)dx +Zbag(x)dx −12n,Zba(tn+˜tn)(x)dx ≤Zbaf(x)dx +Zbag(x)dx +12n.Moreover we know that sn+ ˜sn≤ f + g ≤ tn+˜tn(for all n ∈ N), so sn+ ˜snis astep function bounding f + g from below, and tn+˜tnis a step function boundingf + g from above. Thus we find that (still, for all n ∈ N) we haveZbaf(x)dx +Zbag(x)dx +12n≥¯I(f + g) ≥ I(f +g) ≥Zbaf(x)dx +Zbag(x)dx −12n,for all n ∈ N. [From here I modified the proof slightly from that given on theblackboard] Now we use a slight modification of Theorem I.31, proven as Theorem2 below, to see that this implies that in fact¯I(f + g) =Rbaf(x)dx +Rbag(x)dx andthat¯I(f + g) =Rbaf(x)dx +Rbag(x)dx as well. As the lower integral and the upperintegral are now seen to be equal we find that f + g is integrable, and its integralequals the value of the lower integral, which isZba(f + g)(x)dx =¯I(f + g) =Zbaf(x)dx +Zbag(x)dx,as desired. 12 FOKKO VAN DE BULTFor completeness we prove the little extension to Theorem I.31 in the book, thatwe used above.Theorem 2. If three numbers a, x, and y satisfya −yn≤ x ≤ a +yn,for all n ∈ N then x = a.Proof. Either x ≥ a, or x ≤= a. Suppose x ≥ a, then clearly we have the inequali-tiesa ≤ x ≤ a +yn,so by Theorem I.31 we see x = a. If x ≤ a, then we have the inequalitiesx ≤ a ≤ x +yn,so, again by Theorem I.31 we have a = x.We conclude that in all cases we find that x and a are equal.