Riemann Integral and Bounded function. Ng Tze Beng

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1 Riem Itegrl d Bouded fuctio. Ng Tze Beg I geerlistio of re uder grph of fuctio, it is ormlly ssumed tht the fuctio uder cosidertio e ouded. For ouded fuctio, the rge of the fuctio is ouded d hece y suset of the rge is lso ouded. A cosequece of this d lso of the completeess of the rel umers is tht supremum d ifimum exist for y suset of the rge. This mes upper Riem sum d lower Riem sum with respect to y prtitio c e defied. Aother cosequece of the oudedess of the fuctio is tht the set of ll upper Riem sum is ouded elow d so we hve the the existece of the ifimum of tht set d this is the Upper Riem Itegrl of the fuctio. It is lso cosequece of the oudedess of the fuctio tht the set of ll lower Riem sum is ouded ove d so we hve the existece of the supremum of this set d this is the Lower Riem Itegrl. This is the mechism ehid the meig of the Riem itegrl. The (ouded) fuctio is the sid to e Riem itegrle, if d oly if, the lower d upper Riem itegrls re the sme. The the defiitio of Riem itegrl i terms of the usul Riem sum follows, though ot so esily, s cosequece of the ove. The questio is the sked: Wht out uouded fuctio? Do we hve sesile geerlistio of the re i this cse? We cot use the completeess property of the rel umers. Becuse suset ot ouded ove does ot hve supremum d suset ot ouded elow does ot hve ifimum, for uouded fuctios lower d upper Riem sums cot e defied. Eve if the domi is closed d ouded itervl d the fuctio is defied o this itervl ut ot ouded ove sy, Riem sum for y prtitio c e ritrrily lrge. We c oserve this s follows. Tke y prtitio P: = x 0 < x 1 <... < x = with ritrry orm P. Let A = mi {x i x i-1 : i = 1,..., } Tke Riem sum f ( i )(x i x i1 ) =K, where i c [x i1,x i ]. The let F =mx{f( i ):,...,}. We shll show tht for y Lrge N > K, if f is ot ouded ove, we c fid Riem sum S with respect to the sme prtitio P such tht S > N. Sice f is ot ouded ove, there exists c i [, ] such tht f (c) > N K. A +F The c is i [x j-1, x j ] for some j, 1[j[. The the Riem sum S = i!j f ( i )(x i x i1 ) +f (c)(x j x j1 ) = f ( i )(x i x i1 ) + (f (c) f ( j ))(x j x j1 ) >K + ( N K. A +Ff ( j))(x j x j1 )mk + ( N K A )(x j x j1 )mk +NK =N Similrly, if f is ot ouded elow we c fid ritrrily egtively lrge Riem sum with respect to the sme prtitio P. No mtter how smll the orm of the prtitio, if the fuctio is uouded, the correspodig set of Riem sum is uouded. Therefore, there does ot exists umer L such tht give y ε > 0, we c fid δ > 0 such tht for y prtitio P with orm P < δ, d for y Riem sum S with respect to P, S L < ε. This cosidertio shows ffirmtively tht y fuctio tht is ot ouded CANNOT e Riem itegrle. But some uouded fuctio do hve fiite re. It is difficult to geerlise here. But for fuctios with oe or two discotiuity where the left or right limits is either + or, we c tlk out the existece of improper (Riem) itegrl. For Ng Tze Beg

2 istce, for the fuctio f (x) = 1/ x,x >0, the re uder the grph etwee x = 0 d 0,x =0 1 1 x =1 is give y lim. We c similrly cosider fuctios with td0 + t f (x)dx =td0 lim 2 x + t =2 fiite umer of such discotiuities. Riem Itegrility Theorem 1. For ouded fuctio f:[,]drthe followig sttemets re equivlet. 1. The upper d lower Riem itegrls re the sme. 2. Give ε > 0, there exists prtitio P for the itervl [, ] such tht the differece U(P) L(P) <ε, where U(P) is the upper Riem sum d L(P) is the lower Riem sum with respect to P for f. 3. f is Riem itegrle, i.e., there exists umer L such tht give y ε > 0, we c fid δ > 0 such tht for y prtitio P for [, ] with orm P < δ, d for y Riem sum S with respect to P, S L < ε. Proof. (1) if d oly if (2) is esy. (3) (2). Assume f is Riem itegrle. Give ε > 0, the for y prtitio P for [, ] with orm P < δ, d for y Riem sum S with respect to P, S L < ε/4. Let T ={ Riem sum S: S hs the sme prtitio P}. The for y S i T, L ε/4< S < L + ε/4. Suppose P: = x 0 < x 1 <... < x = with P < δ. Let M i = sup{ f (x): x [x i-1, x i ]} for i = 1,...,. The for ech i, 1 i, there exists c i i [x i-1, x i ] such tht f (c i ) > M i. The 4( ) f (c i )(x i x i1 ) >(M i, 4( ) )(x i x i1 ) =M i (x i x i1 ) 4 =U(P) 4 where U(P) is the upper Riem sum with respect to the prtitio P. Therefore, U(P) <f (c i )(x i x i1 ) +. Now let m i = if{ f 4 <L =L + 2 (x): x [x i-1, x i ]} for i = 1,...,. The for ech i, 1 i, there exists d i i [x i-1, x i ] such tht f (d i ) < m i +. The 4( ) f (d i )(x i x i1 ), <(m i + 4( ) )(x i x i1 ) =m i (x i x i1 ) + 4 =L(P) + 4 where L(P) is the lower Riem sum with respect to the prtitio P. Thus, L(P) >f (d i )(x i x i1 ). 4 >L 4 4 =L 2 Therefore, U(P) L(P) < ε. Hece (2) follows. (1) (2). Suppose the lower itegrl f is the sme s the upper itegrl _ f. The y the defiitio of the lower itegrl, give ε > 0, there exists prtitio P such tht the lower sum L(P, f ) > f. Similrly y the defiitio of the upper 2 itegrl, there exists prtitio Q such tht the upper Riem sum U(Q, f ) < _ f + 2. Let R e the prtitio P4Q, the Ng Tze Beg

3 f. 2 <L(P, f )[L(P4Q, f )[U(P4Q, f )[U(Q, f ) < _ f + 2 Thus, sice f. So R is the U(P4Q, f ) L(P4Q, f ) < _ f + 2 f + 2 = _ f = required prtitio. Coversely, if for y ε > 0, there exists prtitio R such tht U(R, f ) L(R, f ) <, the _ f [U(R, f ) <L(R, f ) +[ f +. Therefore, _ f < f +for y ε > 0. Thus, _ f[ f. But sice for y prtitio Q, L(P, f )[U(Q, f ), L(P, f )[ _ f _ f ecuse is the ifimum of ll the upper Riem sums. The f f[ _ f _ f = is the supremum of ll the lower Riem sums. Therefore, f. (1)d (2)u(3) sice Give ε > 0, there exists prtitio P : = x 0 < x 1 <... < x L = with L > 1 such tht U(P, f ) L(P, f ) <. Let K = mi {x i x i-1 : i = 1,..., L }. Now sice f is 2 ouded, there exists M > 0 such tht f(x) <M. Let R: = y 0 < y 1 <... < y N = e y prtitio such tht æræ <, where =mi(k,. The ecuse 2(2ML +1) ) æræ <K, N > L d for ech i = 1,2,.., L1 there exist 1[j i [N 1such tht y ji 1[x i <y ji. Let I = {j i :,...,L 1}.The for y Riem sum with respect to the prtitio R, N f ( i )(y i y i1 ), where i c [y i1,y i ], L1 f ( ji )(y ji y ji 1) = i"i f ( i )(y i y i ) + =f ( i )(y i y i ) + L1 f (x i )(y ji y ji 1) + L1 (f ( ji ) f (x i ))(y ji y ji 1) i"i =f ( i )(y i y i ) + L1 f (x i )(y ji x i ) + L1 f (x i )(x i y ji 1) i"i + L1 (f ( ji ) f (x i ))(y ji y ji 1). Note tht the rcketed term is Riem sum S for the prtitio R4P. Thus N f ( i )(y i y i1 ) =S + L1 (f ( ji ) f (x i ))(y ji y ji 1) <U(R4P,f) + L1 2M(y ji y ji 1)[U(P,f) + L1 2M(y ji y ji 1) <L(P,f) L1 MæRæ < f ML æræ < f + sice. 2 +2ML [ f + æræ < [ 2(2ML +1) 2(2ML +1) Also N f ( i )(y i y i1 ) =S + L1 (f ( ji ) f (x i ))(y ji y ji 1) ml(r4p,f) + L1 (f ( ji ) f (x i ))(y ji y ji 1) ml(p,f) + L1 (f ( ji ) f (x i ))(y ji y ji 1) >U(P,f ) 2 L1 2M(y ji y ji 1) >U(P,f ) 2 2 L1 MæRæ > _ f 2 2ML æræ > _ f 2 2ML. 2(2ML +1) = _ f Ng Tze Beg

4 _ f = Sice f =L, N f ( i )(y i y i1 ) L <. Thus for y prtitio R with orm æræ < d for y Riem sum S with respect to R, S L <. Cosequece of Theorem 1. Theorem 2. Suppose f : [, ] R is rel vlued fuctio. If f is Riem itegrle o [, ], the for y c i [, ], f is Riem itegrle o [, c] d o [c, ]. Proof. Fix ε > 0. The sice f is Riem itegrle o [, ], y Theorem 1, there exists prtitio P : = x 0 < x 1 <... < x = such tht the differece of the upper d lower Riem sum, U(P) L(P) < ε, (1) where U(P) =M i (x i x i1 ), M i = sup{ f (x): x [x i-1, x i ]}, L(P) =m i (x i x i1 ), m i = if{ f (x): x [x i-1, x i ]}. The sice for y refiemet Q of P, tht is y prtitio Q of [, ] such tht P Q, U(Q) U(P) d L(P) L(Q), we hve tht the differece U(Q) L(Q) < U(P) L(P) < ε. Νow if c = or we hve othig to prove. We ow ssume < c <. If c is ot poit of P, the dd c to P d replce P y this ew prtitio d still hve the iequlity (1) stisfied. So we my, without loss of geerlity, ssume tht c is i P tht is c =x k for some 0 < k <. Suppose ow tht P is give y P : = x 0 < x 1 <... < x k = c < x k+1 < < x = The : = x 0 < x 1 <... < x k = c is prtitio for [, c] d ': x k = c < x k+1 < < x = is prtitio for [c, ]. Thus the differece of the upper d lower Riem sum over [, c] is k U( ) L( ) = (M i m i )(x i x i1 )[(M i m i )(x i x i1 ) = U(P) L(P) < ε. Therefore, y Theorem 1, f is Riem itegrle over [, c]. Similrly, U( ') L( ') = (M i m i )(x i x i1 )[(M i m i )(x i x i1 ) = U(P) L(P) < ε. i=k+1 Oce more, y Theorem 1, f is Riem itegrle over [c, ]. This proves the theorem. Corollry 3. Suppose f : [, ] R is rel vlued fuctio. If f is Riem itegrle o [, ], the for y c d d i [, ] such tht < c < d <, f is Riem itegrle o [c, d]. Proof. By Theorem 2, f is Riem itegrle o [, d] d other pplictio of Theorem 2 sys tht f is Riem itegrle o [c, d]. Theorem 4. Suppose f : [, ] R is rel vlued fuctio. If f is Riem itegrle o [, ], the f is lso Riem itegrle o [, ] d f (x)dx [ f (x) dx. Ng Tze Beg

5 Proof. This is just mipultio. We eed to work with the properties of supremum d ifimum. Notice tht for y ouded suset A of R, if A = sup(a). As f is Riem itegrle, there exists prtitio, P : = x 0 < x 1 <... < x = of [, ], such tht the differece of the upper d lower Riem sum with respect to P for f, U(P, f ) L(P, f ) = (M i m i )(x i x i1 ) < ε, where M i = sup{ f (x): x [x i-1, x i ]}, d m i = if{ f (x): x [x i-1, x i ]}= sup{ f (x): x [x i-1, x i ]}. Therefore, M i m i = sup{ f (x): x [x i-1, x i ]} + sup{ f (x): x [x i-1, x i ]} = sup{ f (x) f (y): x, y [x i-1, x i ]}. Also the differece of the upper d lower Riem sum with respect to P for f is U(P, f ) L(P, f ) = (M i m i )(x i x i1 ) < ε, where M i ' = sup{ f (x) : x [x i-1, x i ]}, d m i ' = if{ f (x) : x [x i-1, x i ]}= sup{ f (x) : x [x i-1, x i ]}. Therefore, M i ' m i ' = sup{ f (x) : x [x i-1, x i ]} + sup{- f (x) : x [x i-1, x i ]} = sup{ f (x) f (y) : x, y [x i-1 - x i ]}. I fct, sice { f (x) f (y): x, y [x i-1, x i ]}, if d oly if, { f (x) f (y): x, y [x i-1, x i ]}, M i m i = sup{ f (x) f (y) : x, y [x i-1, x i ]}. Similrly, M i ' m i ' = sup{ f (x) f (y) : x, y [x i-1, x i ]}. Now sice we hve the followig iequlity tht for y, i R, - -, M i ' m i ' M i m i for i = 1,,. It follows tht U(P, f ) L(P, f ) = (M i m i )(x i x i1 ) [ (M i m i )(x i x i1 ) U(P, f ) L(P, f ) < ε. Therefore, the Riem coditio for f is fulfilled. Hece y Theorem 1, f is Riem itegrle. Now sice f f f d f is lso Riem itegrle we hve the tht f (x) dx[ f (x)dx[ f (x) dx which implies tht 4. f (x)dx [ f (x) dx We c use the ove rgumet to prove the followig.. This completes the proof of Theorem Theorem 5. Suppose f : [, ] R is rel vlued fuctio. If f is Riem itegrle o [, ], the f 2 = f f is lso Riem itegrle o [, ]. Proof. Let K = sup{ f (x) : x [, ]}. If K = 0, we hve othig to prove sice f would e the zero costt fuctio. Assume K > 0. As i the proof of the lst theorem, there exists prtitio P : = x 0 < x 1 <... < x = of [, ] such tht the differece of the upper d lower Riem sum with respect to P for f, U(P, f ) L(P, f ) = (M i m i )(x i x i1 ) < ε/(2κ ), where M i = sup{ f (x) : x [x i-1, x i ]}, d m i = if{ f (x) : x [x i-1, x i ]}. For i = 1,,, let M i ' = sup{ f 2 (x) : x [x i-1, x i ]}, d m i ' = if{ f 2 (x) : x [x i-1, x i ]}= sup{ f 2 (x) : x [x i-1, x i ]}. The for i = 1,,, Ng Tze Beg

6 M i ' m i ' = sup{ f 2 (x): x [x i-1, x i ]} + sup{ f 2 (x): x [x i-1, x i ]} = sup{f 2 (x) f 2 (y): x, y [x i-1, x i ]} = sup{( f (x) f (y))(f(x) + f (y)): x, y [x i-1, x i ]} 2K sup{( f (x) f (y)): x, y [x i-1, x i ]} = 2K (M i m i ). Therefore, the differece of the upper Riem sum d the lower Riem sum with respect to P for f 2, is U(P, f 2 ) L(P, f 2 ) = (M i m i )(x i x i1 )[2K(M i m i )(x i x i1 ) 2K ε/(2κ ) = ε. Τherefore, y Theorem 1, f 2 is Riem itegrle o [, ]. A esy cosequece of the ove theorem is the followig. Corollry 6. Suppose f d g re Riem itegrle o [, ], the f g = f g is lso Riem itegrle o [, ], Proof. Note tht f g = 1/2 [( f + g ) 2 f 2 g 2 ]. The result the follows from Theorem 5. Theorem 7. Suppose f is Riem itegrle o [, ]. If there exists K > 0 such tht for ll x i [, ], f (x) K, the 1/ f is lso Riem itegrle o [, ]. Proof. Fix ε > 0. The sice f is Riem itegrle o [, ], y Theorem 1, there exists prtitio P : = x 0 < x 1 <... < x = such tht the differece of the upper d lower Riem sum, U(P, f ) L(P, f ) < εκ 2, (2) where U(P, f) =M i (x i x i1 ), M i = sup{ f (x): x [x i-1, x i ]}, L(P, f) =m i (x i x i1 ), m i = if{ f (x) : x [x i-1, x i ]}. The U(P, f ) L(P, f ) = (M i m i )(x i x i1 ) (3) Let M i ' = sup{1/ f (x): x [x i-1, x i ]}, d m i ' = if{1/ f (x): x [x i-1, x i ]}= sup{1/ f (x): x [x i-1, x i ]}. The for i = 1,,, M i ' m i ' = sup{1/ f (x): x [x i-1, x i ]} + sup{1/ f (x): x [x i-1, x i ]} = sup{1/ f (x) 1/ f (y): x, y [x i-1, x i ]} = sup{ 1/ f (x) 1/ f (y) : x, y [x i-1, x i ]} = sup{ f (x) f (y) / f (x) f (y) : x, y [x i-1, x i ]} sup{ f (x) f (y) / K 2 : x, y [x i-1, x i ]}, ecuse for y x i [, ], 1/ f (x) 1/ K, = sup { f (x) f (y) : x, y [x i-1, x i ]}/ K 2 = (M i m i )/ K (4) Now U(P,1/f) =M i (x i x i1 ) d L(P,1/f), the differece of the =m i (x i x i1 ) upper d lower Riem sum with respect to P for 1 / f, U(P,1/ f ) L(P,1/ f ) = (M i m i )(x i x i1 )[1/K 2 (M i m i )(x i x i1 ), y (4), = (1/K 2 )(U(P, f ) L(P, f )), y (3), Ng Tze Beg

7 < (1/K 2 )εκ 2 = ε. Hece, Riem's coditio for 1/ f is stisfied o [, ] d so y Theorem 1, 1/ f is Riem itegrle o [, ]. This completes the proof. The followig results cocerig supremum d ifimum of ouded susets re very useful d we hve used it ofte here. Let A d B e ouded susets of R. 1. If for every i A, there exists poit i B such tht, the sup A sup B. 2. sup(a + B) = sup A + sup B. 3. if A = sup ( A). 4. If k 0, the sup (k A) = k supa. Coclusio. Together with the dditivity of Riem itegrle fuctios, Corollry 6 sys tht the set of ll Riem itegrle fuctios o [, ] form commuttive rig. Ng Tze Beg