MATH 532 Itegrable Fuctios Dr. Neal, WKU We ow shall defie what it meas for a measurable fuctio to be itegrable, show that all itegral properties of simple fuctios still hold, ad the give some coditios for whe a measurable fuctio is itegrable. Defiitio 7.1. Let ( X, F, ) be a measure space ad let f : X (, ) be a measurable fuctio. The f is itegrable with respect to if there exists a sequece of simple fuctios { f } that are mea Cauchy ad such that { f } coverges i measure to f. I this case, we let f d f d for all F. Moreover, the sequece { f } is called a determiig sequece for f. If f is itegrable with respect to, the f d does exist as a fiite real umber for all F by Theorem 6.8. I particular, we write f d for f d. By Theorem X 6.9, f d is well-defied for all F ; that is, ay two determiig sequeces give the same value. Agai by Theorem 6.8, ( ) = f d is a coutably additive set fuctio. Lemma 7.1. Let f be itegrable with respect to ad f 1 d = f d.. For all F, f 1 is itegrable Proof. Let { f } be mea Cauchy itegrable simple fuctios such that f f. The { f 1 } are also mea Cauchy itegrable simple fuctios ad f 1 f 1. Thus, f 1 is itegrable ad f 1 d = f 1 d X X f 1 d = f d = f d. X f d X QD By xercise 5.5, if f f ad g g, the a f a f, f + g f + g, ad f f. Usig properties of limits ad the aalogous results about simple itegrable fuctios from Theorem 6.1, we the ca show that the followig results hold:
Theorem 7.1. Let f ad g be itegrable fuctios. (a) For all a R, the fuctio a f is itegrable ad (a f )d = a f d for all F. (b) The fuctio f + g is itegrable ad ( f + g) d = f d + gd for all F. (c) If f 0 a.e. over, the (d) If f g a.e. over, the f d 0. f d gd. (e) If m f M a.e. over, the m ( ) f d M ( ). (f) The fuctio f is itegrable ad f d f d for all F. (g) For all evets, A with A, f d f d f d. A (h) For all evets, f + g d f d + g d. (i) If () = 0, the f d = 0. Proof. We shall prove (c) ad (f) ad leave the others as exercises. (c) Let f 0 a.e. over, ad let { f } be a detemiig sequece for f so that f f. The f 1 f 1 ad thus f 1 = f 1 f 1 = f 1. But f 1 = f 1 a.e.; thus, f 1 f 1 by Theorem 5.6 (d). Thus, { f 1 } are mea Cauchy itegrable simple fuctios such that f 1 f 1 ad therefore are a determiig sequece for f 1. Now because f d 0 for all, we have f d = f 1 d X X f 1 d f d 0.
(f) Let { f } be mea Cauchy itegrable simple fuctios such that f f. The { f } are also itegrable, Cauchy i mea, ad f f. Thus, f is itegrable (with determiig sequece { f }) ad f d f d for all F. Ad because 0 f d for all, we have 0 f d after takig the limit as. We also have f d f d, or f d f d f d for all ; so, lim f d lim f d lim f d ; thus, f d f d f d or f d f d. Corollary 7.1. Let f ad g be itegrable fuctios. (a) The fuctios max( f, g) ad mi( f, g) are itegrable. (b) The fuctios f + ad f are itegrable. Proof. Because max{ f, g} = 1 2 ( f + g + f g ) ad mi{ f, g} = 1 ( 2 f + g f g ), each are itegrable usig Parts (a), (b), ad (f) of Theorem 7.1. The f + = max{ f, 0} ad f = mi{ f, 0} are also both itegrable. QD Corollary 7.2. A measurable fuctio f is itegrable if ad oly if f is itegrable. Proof. If f is itegrable, the f is itegrable by Theorem 7.1 (f). Now suppose f is itegrable. The f + = f 1 {f 0} ad f = f 1 {f < 0} are both itegrable by Lemma 7.1. So f = f + f is itegrable by Theorem 7.1. QD Corollary 7.3. Let f be itegrable with f 0 a.e. The ( ) = f d is a fiite measure. Proof. We already kow that is coutably additive by Theorem 6.8. But we also have 0 f d f d < for all F ; thus, is a fiite measure. QD X
The followig result is aalogous to the result of Theorem 6.5 for simple fuctios: Theorem 7.2. Let f be a itegrable fuctio. The the set fuctio ( ) = f d is absolutely cotiuous with respect to. Proof. Let { f } be a detemiig sequece for f. The { f } is a detemiig sequece for f. So by Theorem 6.7, the measures f d are uiformly absolutely cotiuous with respect to. Thus, for every > 0 there is a > 0 such that if () < the f d < / 2 for all. So for () <, we have f d f d f d / 2 <. Thus, lim () 0 f d = 0. QD Other Coditios for Itegrability We ow shall start describig some other coditios that ca be used to determie whe a measurable fuctio is itegrable. Lemma 7.2. Let { f } be a determiig sequece for a itegrable fuctio f. The { f } coverges i mea to f. That is, lim Proof. Let ( f, f ) f f d = 0. > 0 be give. Because { f } is a determiig sequece, { f } is Cauchy i mea. So there exists a iteger N 1 such that f f m d < / 3 whe,m N. Now f f N is also a itegrable fuctio ad { f f N } is a determiig sequece of itegrable simple fuctios for f f N. Thus, f f N d f f N d. So wheever there is a iteger N 1 N such that f f N d / 3 < f f N d N 1. The for N 1 we have f f d f f N d + f N f d < f f N d + / 3 + f N f d < / 3 + / 3 + / 3 =. Hece, { f } coverges i mea to f. QD The ext result shows that a determiig sequece eed ot be a collectio of simple fuctios, but istead may be a sequece of itegrable fuctios.
Theorem 7.3. Let { f } be a mea Cauchy sequece of itegrable fuctios that coverge i measure to the measurable fuctio f. The f is itegrable, { f } coverges i mea to f, ad f d f d for all F. Proof. The results follow by our previous developmet ad Lemma 7.2 if all the { f } are simple fuctios. But if the { f } are ot simple fuctios, the we eed to create a determiig sequece of simple fuctios for f i order to show that f is itegrable. Because each f is itegrable, each f has a determiig sequece of simple fuctios {h k } k=1 that coverges i measure to f (by defiitio of itegrability) ad also coverges i mea to f (by Lemma 7.2). So for each, we ca choose a sigle simple fuctio h such that ( h f 1/ ) <1 / ad f h d < 1. We claim that {h } is a determiig sequece for f. We first show that {h } is mea Cauchy. Let > 0 be give. The choose a iteger N 1 such that f f m d < / 3 whe,m N ad also such that 1 / N < / 3. The for,m N, we have h h m d h f d + f f m d + f m h m d < 1 + 3 + 1 m < 3 + 3 + 3 = ; hece, {h } is mea Cauchy. Next, we show that {h } coverges i measure to f. Let > 0 ad > 0 be give. Choose N 1 such that 1 / N < / 2, 1 / N < / 2, ad ( f f / 2 ) < / 2. The for N we have ( h f ) ( h f / 2 ) + ( f f / 2 ) ( h f 1/ ) + ( f f / 2 ) < 1 + 2 < 2 + 2 = ; thus, h f. So {h } is a determiig sequece for f ad f is itegrable. Al By Lemma 7.2, we kow that {h } coverges i mea to f. Next we show that the origial sequece { f } coverges i mea to f. Let > 0 be give. The choose a iteger N 1 such that 1 / N < / 2 ad such that h f d < / 2 whe N. The for N, we have f f d f h d + h f d < 1 + 2 < 2 + 2 =.
Fially, for all F, we have f d f d = ( f f ) d f f d f f d 0 as. QD Lemma 7.3. Let g be a itegrable fuctio. (a) For all a > 0, ( g > a ) <. (b) { g > 0} is a deuerable uio of ested icreasig evets of fiite measure. Proof. By Corollary 7.2, g is itegrable. Because g d { g = 0} = 0 ( g = 0) = 0, we have g d = g d + g d { g > 0} { g = 0} = g d + g d {0< g a} { g > a} g d + a ( g > a). {0< g a} So, ( g >a) 1 g d g d a {0< g a} <. The { g > 0} = U { g > 1/ k}. QD k =1 The ext result is a famous result that gives aother coditio of itegrability: Theorem 7.4. (Lebesgue s Domiated Covergece Theorem). Let { f } be a sequece of itegrable fuctios that coverge i measure to the measurable fuctio f. Let g be a o-zero itegrable fuctio such that f g a.e. for every. The f is itegrable, { f } coverges i mea to f, ad f d f d for all F. Proof. By Theorem 7.3, it suffices to show that { f } is mea Cauchy. Accordig to Lemma 7.3, let = { g > 0} = U { g > 1/ k} = U k so that { k } is ested icreasig k=1 k =1 ad ( k ) < for all k. Now let > 0 be give. For all,m,k we have f f m d = f f m d + f f m d k k 2 g d + f f m d. k k
Because the defiite itegral of 2 g is a fiite measure ad { k } k is ested decreasig with ull itersectio, we have 2 g d 0 as k icreases. Thus we ca k fix a k large eough so that 2 g d < / 3. We also ca assume that ( k ) > 0 k because () > 0 due to g beig o-zero. Now let G m = { f f m / (3 ( k ))}. We ow have f f m d = f f m d + f f m d k G m k G m k 2 g d + / (3 ( k )) d k G m 2 g d + 3 ( k ) G m ( k ) = 2 g d +. G 3 m Fially, because { f } coverges i measure to f, { f } is Cauchy i measure. So (G m ) 0 as,m icrease. The because the defiite itegral 2 g d is absolutely ( ) cotiuous with respect to, we have 2 g d 0 as,m icrease. So we ca G m make this itegral less tha / 3 by choosig,m large eough. So we the have f f m d < for,m large eough. Hece, { f } is mea Cauchy ad the Theorem follows by By Theorem 7.3. QD The ext two results give coditios that allow us to replace covergece i measure with covergece almost everywhere whe defiig a determiig sequece. Lemma 7.4. Let ( X, F, ) be a fiite measure space. Suppose { f } is a mea Cauchy sequece of itegrable fuctios that coverge almost everywhere to a measurable fuctio f. The f is itegrable, { f } coverges i mea to f, ad f d f d for all F. Proof. By gorov s Theorem (Th. 5.9), f f. The results follow from Th. 7.3. QD The ext result shows that we ca replace the coditio of a fiite measure space with { f } beig bouded by a itegrable fuctio.
Theorem 7.5. (Lebesgue s Domiated Covergece Theorem). Let { f } be a sequece of itegrable fuctios that coverge almost everywhere to the measurable fuctio f. Let g be a o-zero itegrable fuctio such that f g a.e. for every. The f is itegrable, { f } coverges i mea to f, ad f d f d for all F. Proof. We shall show that the assumptios imply that f f. The the results will follow from results follow from Theorem. 7.4. Let N 0 = {x : f (x ) / f (x )} ad N = {x : f (x ) > g(x ) }. The each N i is a ull set; thus, N = U N i is a ull set. The for all x outside of N, f ( x) g(x ) for all, i = 0 ad f (x ) f (x ). But the f (x ) (x) g(x) also. Now let > 0 be give ad let U {x : f i (x ) f ( x) }. The { } is a ested i = decreasig sequece of sets. If x I, the for all, there is a i such that f i ( x) f (x ) ; thus, f (x ) / f (x ). Thus, I N 0, a ull set. Moroever, ( 1 ) = ( 1 N ) because N is a ull set. But if x ( 1 N ), the f i ( x) f (x ) for some i. But f i ( x) f (x ) 2 g(x ), so if g( x) < / 2 the we would have f i ( x) f (x ) < which is ot the case. So if x ( 1 N ), the we must have g( x) / 2 > / 3. Thus, ( 1 N ) { g > / 3}, which has fiite measure by Lemma 7.3. ( 1 ) = ( 1 N ) ( g > / 3) <. Now we have a ested decreasig sequece of sets { } with ( 1 ) <. So by Propositio 1.3, lim ( ) = ( I ) ( N 0 ) = 0. But {x : f (x ) f ( x) } ; so ({x : f ( x) f (x ) }) ( ) 0 ad f f. QD Lemma 7.5. Let h be a simple fuctio ad let g be a itegrable fuctio such that h g a.e. The h is itegrable. Proof. Let h = a i 1 Ai ad h = a i 1 Ai where we may assume all a i 0. Let A be a i= 1 i= 1 ull set such that h g outside of A. The for each i, a i 1 Ai A c g 1 Ai A c. By Theorem 7.1 (e), we have a i ( A i ) = a i ( A i A c ) g d <. Thus, (A i ) < A i A c for all i which makes h itegrable. QD
Corollary 7.4. Let f be a measurable fuctio ad let g be a itegrable fuctio such that f g a.e. The f is itegrable. Proof. By Theorem 5.11, there exists a sequece of o-egative simple fuctios { f } that coverge poitwise to f ad such that 0 f f g for all. By Lemma Lemma 7.5, each f is itegrable. Thus by Domiated Covergece (Theorem 7.5), f is itegrable ad so f is also itegrable by Corollary 7.2. QD We ow show that the Lebesque itegral gives the stadard calculus defiite itegral for cotiuous real-valued fuctios. Theorem 7.6. Let f : [a, b] R be a cotiuous fuctio. Let F be the Lebesgue measurable subsets of [a, b]. The f is itegrable with respect to Lebesgue measure b ad f d = f (x ) dx, the stadard Riema itegral. [a,b] a Proof. Recall that every cotiuous fuctio o the Real lie is actually Borel measurable. To show that f is itegrable, we shall create a determiig sequece of itegrable fuctios { f } for f by applyig Theorem 7.5. For each 1, we first partitio [a, b] ito subitervals of equal legth 1 / : a = a 0 < a 1 <... < a 1 < a = b, where a i = a + (b a)i / for 0 i. The because f is cotious o each closed sub-iterval [a i 1, a i ], there exists poits c i [a i 1, a i ] such that f (c i ) = mi{ f (x ): a i 1 x a i }. The for each, we defie f by 1 f (x ) = f (c i )1 [ai 1, a i ) + f (c )1 [a 1, a ]. i=1 For all, f is a simple fuctio that is itegrable with respect to Lebesgue measure, f is 0 outside of [a, b], ad 1 f d = f 1 [a, b] d = f (c i ) ([a i 1, a i )) + f (c ) ([a 1, a ]) = 1 [a, b] i =1 Moreover, by the choice of f (c i ), the value b b itegral f ( x) dx, ad f ( x) dx a a 1 f (c i ). i=1 f d is a lower Reima sum for the [a, b] f (c i ). i=1
Now for all, f ( x) g( x) = max{ f ( x) : a x b}1 [a, b] for all x. Also, the simple fuctio g is itegrable ad gd = max{ f ( x) : a x b} (b a). To apply the domiated covergece theorem, we shall show that f f o [a, b]. Let x [a, b]. For each, x is i exactly oe of the subitervals used to defie f. We the let c x be the poit chose for this subiterval i the defiitio of f. The f (x ) = f (c x ). Moreover, x is always withi 1 / of c x. Hece, the sequece { c x } =1 coverges to x o the real lie. Because f is cotiuous, f (c x ) f(x). That is, f (x ) f (x ). Thus, by Theorem 7.5, f d [a,b] f d [a, b] 1 b f(c i ) = f (x)dx. i=1 a We coclude with aother importat itegral covergece theorem. Theorem 7.7. (Mootoe Covergece Theorem) Let ( X, F, ) be a fiite measure space. Let { f } be a mootoe icreasig sequece of o-egative itegrable fuctios that coverge almost everywhere to the o-egative measurable fuctio f. The lim f d = f d for all F. Proof. If f is itegrable, the the result follows by Domiated Covergece (Theorem 7.5). But suppose we do ot kow i advace if f is itegrable. We do kow that the values f d form a mootoe icreasig sequece of fiite o-egative umbers. X Suppose the that lim f d exists. The the sequece { f d } is Cauchy. So if X > 0, the for large,m with m, we have f f m d = ( f f m ) d = f d f m d <. Thus, { f } is a mea Cauchy sequece of itegrable fuctios that coverges almost everywhere to f. Thus, by Lemma 7.4, f is itegrable ad lim f d = f d for all F. If lim f d = +, for ay evet, the f will ot be itegrable over, though we ca defie f d = + to obtai the result.
xercises xercise 7.1. Prove the remaiig parts (other tha (c) ad (f)) of Theorem 7.1. xercise 7.2. Let f be a itegrable fuctio such that f > 0 a.e. over a measurable set. Suppose f d = 0. Prove that () = 0. xercise 7.3. Let f be a itegrable fuctio such that that f = 0 a.e. f d = 0 for all F. Prove