Kenneth Kuttler

Corollary 3.0.11 Let f : [a,b]→ R be piecewise continuous. Then f is Riemann inte-grable. Also ∫ b

af dt =

∑i=1

∫ zi

zi−1

gidt (3.4)

Proof: Let Pi be a partition for [zi−1,zi] . Since there are only finitely many of theseintervals, there exists δ > 0 such that if ∥Pi∥< δ , then for each i,∣∣∣∣∣∑Pi

gi−∫ zi

zi−1

gidt

∣∣∣∣∣< ε

Let M f be an upper bound for | f | on [a,b], Mg an upper bound for all |gi|. Now let ∥P∥<δ < ε where P is a partition of [a,b] , these points denoted as x j. Let P̂i be those points ofP which are in (zi−1,zi] and let Pi consist of P̂i along with zi−1 and zi. Thus ∥Pi∥< δ . Thenfor yi ∈ [xi−1,xi] ,∣∣∣∣∣ n

∑i=1

∫ zi

zi−1

gidt−∑P

∣∣∣∣∣≤ n

∑i=1

∣∣∣∣∣∣∫ zi

zi−1

gidt− ∑x j∈P̂i

f (y j)(x j− x j−1

)∣∣∣∣∣∣Now for x j ∈ P̂i, f (y j) = gi (y j) except maybe at end points where these differ by no morethan 2

(M f +Mg

)≡ 2M. Thus the above is no more than

≤n

∑i=1

∣∣∣∣∣∫ zi

zi−1

gidt− ∑x j∈Pi

gi (y j)(x j− x j−1

)∣∣∣∣∣+ n

∑i=1

4(M f +Mg

)δ

< nε +4Mnδ < ε (n+4Mn)

Since ε is arbitrary, this shows that f is indeed Riemann integrable and equals 3.4. ■Note that what has actually been shown is that if a bounded function f satisfies f = gi

on [zi−1,zi] except for possibly the end points, and gi is Riemann integrable on [zi−1,zi] ,then f is Riemann integrable on [a,b]. The above proof applies with no change.

It is important to notice that the integral is linear. That is, for α,β numbers and f ,gpiecewise continuous functions,∫ b

a(α f +βg)dx = α

∫ b

af dx+β

∫ b

agdx

This is easy to see because such linearity holds for sums. Thus∫ b

a(α f +βg)dx ≡ lim

∥P∥→0∑P

α f +βg

= lim∥P∥→0

α ∑P

f +β ∑P

g = α

∫ b

af dx+β

∫ b

agdx

I leave the details to you. Actually, this works under the assumption that f ,g are Riemannintegrable but in this case, you have to show that a linear combination of Riemann inte-grable functions is Riemann integrable. This is not hard but I don’t want to waste time onit.

The above is the Riemann integral. There is another integral which can be proved to beequivalent to the above. It is called the Darboux integral.

37Corollary 3.0.11 Let f : [a,b] + R be piecewise continuous. Then f is Riemann inte-grable. Alsob n Zi| fat=¥y° gidt (3.4)a i=17%i-1Proof: Let P; be a partition for [z;_1,z;]. Since there are only finitely many of theseintervals, there exists 6 > 0 such that if ||P,|| < 6, then for each i,-[ gidtcami—<_€ELet My, be an upper bound for |f| on [a,b], Mg an upper bound for all |g;|. Now let ||P|| <6 < € where P is a partition of [a,b], these points denoted as x;. Let P; be those points ofP which are in (z;_1,z;] and let P; consist of P; along with z;_; and z;. Thus ||P;|| < 6. Thenfor y; € [xj-1 xl ;“edt Dy <yif gidt — Y fos) Xj —Xj- 1)<i- <i- xjeP;Now for x; € P,, f (yj) = gi (y;) except maybe at end points where these differ by no morethan 2 (M. f+M 9) = 2M. Thus the above is no more thanyf gidt — y gi (yj) (Xj —xj-1)<i xjEP;+) 4(Myp+M,) 6i=1< he 4Mn5 < €(n+4Mn)Since € is arbitrary, this shows that f is indeed Riemann integrable and equals 3.4.Note that what has actually been shown is that if a bounded function f satisfies f = g;on [z;-1,zj] except for possibly the end points, and g; is Riemann integrable on [z;-1, zi],then f is Riemann integrable on [a,b]. The above proof applies with no change.It is important to notice that the integral is linear. That is, for a, numbers and f,gpiecewise continuous functions,[Carr Bejar=a [sap [easaThis is easy to see because such linearity holds for sums. Thusb[ (ar+Bs)ax = lim Yaf+Bg||P +0°p= lim ot oy s+B Y= a’ jap [ gdxIPII leave the details to you. Actually, this works under the assumption that f,g are Riemannintegrable but in this case, you have to show that a linear combination of Riemann inte-grable functions is Riemann integrable. This is not hard but I don’t want to waste time onit.The above is the Riemann integral. There is another integral which can be proved to beequivalent to the above. It is called the Darboux integral.