Proof:Sincefis integrable on [a, b],fis bounded, i.e.,|f(x)| ≤Bfor allx∈[a, b] forsome constantB. Consider another functiong: [-B, B]→R,t7→tn, which is continuouson [-B, B]. By Theorem above, the composed functiong◦f=fnmust be integrable on[a, b].Appendix. Some problems discussed in the classA. Iff(x)is a continuous function on[a, b]withf(x)≥0andf6=constant. ProveRbaf(x)dx >0.Proof:Sincef≥0 andf6=constant, there exists a pointc∈(a, b) such thatf(c)>0.By the continuity off, for any>0, there is aδ >0 such that|f(x)-f(c)|<,∀|x-c|< δ.By taking=f(c)2, we obtainf(x) =|f(x)|=|f(x)-f(c)+f(c)| ≥ |f(c)|-|f(x)-f(c)|> f(c)-f(c)2=f(c)2,∀|x-c|< δ.Then by Theorem 0.8, Corollary 0.6,Zbaf(x) =Zc-δaf(x)dx+Zc+δc-δf(x)dx+Zbc+δf(x)dx≥0 +Zc+δc-δf(x)dx+ 0≥f(c)22δ >0.7.2.2.True or false:(a) Iffis continuous on[0,2]and0≤f(x)≤4for allx∈[0,2], thenI=R20fexists and0≤I≤8.(b) Iffis integrable on[a, b]andgis integrable on[c, d], wheref([a, b])⊆[c, d],theng◦fis integrable on[a, b].(c) Iffis integrable on[a, b], then|f|is integrable on[a, b]andRba|f| ≤ |Rbaf|.Solution:(a) True by the definition and by Corollay 0.6.(b) False. Letf(x) =(1q,ifx=pq∈Q∩[0,1], where(p, q) = 1;0,ifx∈[0,1]-Q.169

Letg(x) =(0,ifx= 0;1ifx >0.Then by Theorem 0.10,g◦f(x) =(1,if x∈Q∩[0,1];0,ifx∈[0,1]-Q.Herefis integrable on [0,1],gis integrable on [0,+∞), butg◦fis not integrable on [0,1].

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