4. Product measure spaces and the Lebesgue integral in R n.

Transcription

1 4 M. M. PELOSO 4. Product measure spaces and the Lebesgue integral in R n. Our current goal is to define the Lebesgue measure on the higher-dimensional eucledean space R n, and to reduce the computations to integrals in lower dimensions. In order to do this, we first present the theor of integration on product measure spaces and the fundamental theorems of Tonelli and Fubini on the equalit of iterated integrals. 4.. Product measure spaces. Definition 4.. Let (, M) and (, N ) be measurable spaces. A subset of of the form A B where A 2Mand B 2Niscalled a measurable rectangle. Wedefinetheproduct -algebra in as the -algebra generated b the collection of measurable rectangles and we denote it b M N. Next, given two measure spaces (, M,µ) and (, N, ) we wish to construct a measure on M N such that the measure of an measurable rectangle A B equals µ(a) (B). Throughout this section, we denote b A the collection of finite unions of disoint measueable rectangles in. Lemma 4.2. The following properties hold: (i) A is an algebra; (ii) if M(A) denotes the -algebra generated b A, then M(A) M N. Proof. In order to prove that A is an algebra we need to show that it is closed under the complement and finite unions of sets in A. Given two measurable rectangles A B and A 2 B 2, we preliminar observe that (A B ) \ (A 2 B 2 )(A \ A 2 ) (B \ B 2 ) is also a measurable rectangle. Now, given a measurable rectangle A B we have that c (A B) ( c B) [ ( c A B), that is, c (A B) is finite union of disoint measurable rectangles, hence in A. Next, let A B and A 2 B 2 be measurable rectangles A not necessaril disoint (otherwise we have nothing to prove). B the first part, A3 c (A 2 B 2 )[ n (E F ), where E F are disoint measurable rectangles. Then, (A B ) \ (A 2 B 2 )(A B ) \ (E F ) [ [(E n F n ) (A \ E ) (B \ F ) [ [ (A \ E n ) (B \ F n ), hence a finite union of disoint measurable rectangles, which belongs to A. Therefore, (A B ) [ (A 2 B 2 ) (A B ) \ c (A 2 B 2 ) [ (A 2 B 2 ) 2A. B induction, we obtain that (A B ) [ [(A n B n ) 2A,

2 MEASURE THEOR AND LEBESGUE INTEGRAL 4 that is, A is closed under finite unions. Finall, let A B 2A, that is, A B [ n (E F ), where E F are disoint measurable rectangles. Then, c (A B) c (E F ) [ [(E n F n ) c (E F ) \ \ c (E n F n ) ( c F ) [ ( c E F ) \ \ ( c F n ) [ ( c E n F n ) n[ n o (A B ) \ (A B): A,A or c E`, B,B F`. Hence, c (A B) is union of intersections of disoint measurable rectangles, hence it is union of disoint measurable rectangles. This shows that A is an algebra, i.e. (i). In order to prove (ii), observe that M N contains the measurable rectangles, hence their finite unions. Thus, M N contains A and therefore the -algebra generated b it, that is, M(A). Conversel, M(A) contains all the measurable rectangles, hence the -algebra generated b them, that is, M N. This completes the proof. Now, let (, M,µ) and (, N, ) be measure spaces and let A be the algebra of finite unions of disoint measurable rectangles. If E and E [ n A B is an element of A (that is, A B are disoint measurable rectangles,,...,n), we set n (E) µ(a ) (B ). (4) Lemma 4.3. With the hpotheses as above, is a premeasure on A. Proof. First of all, we need to show that is well defined, that is, for an E 2A, the value of (E) is independent of the decomposition of E as finite unions of disoint measurable rectangles. Observe that, if A B is a measurable rectangle and A B [ m A B,whereA B are disoint measurable rectangles, we have m A(x) B () A B (x, ) A (x, ) m B A (x) B (). We now integrate w.r.t. x on both sides of the equalities above and see that m m µ(a) B () A(x) B () dµ(x) A (x) B () dµ(x) µ(a ) B (). We proceed b integrating w.r.t. and obtain (A B) µ(a) (B) µ(a) B () d () m µ(a ) B () d () m µ(a ) (B). (5) This shows that (A B) P m µ(a ) (B ) for an decomposition of A B as finite unions of disoint measurable rectangles. This easil implies that is well defined. Now, is finitel additive b construction.

3 42 M. M. PELOSO We need to show is countabl additive, that is, if {A B } is a collection of of disoint measurable rectangles whose union is in A, then + [ + (A B ) (A B ). (6) B the finite additivit of, we first show that it su ces to consider the case in which [ + (A B ) is a single measurable rectangle. The argument is analogous to the one in the proof of Lemma 3.8, but we repeat it for sake of completeness. If A B [ n k (A(k) B (k) ), with A (k) B (k) ) disoint measurable rectangles, it is possible to find subcollections {(A (k) B (k) )}, k, 2,...,n, of {(A B )} such that, for each k, 2,...,n, A (k) B (k) + [ (A (k) If we know that for each k, A (k) B (k) P + from the finite additivit of. A(k) B (k) ). B (k), the conclusion then follows Thus, let us show (6) when [ + (A B )(A B). We proceed as in the argument leading to(5). We have that A(x) B () P + A (x) B (). Integrating first w.r.t. x we have + µ(a) B () A(x) B () dµ(x) µ(a ) B (). Next we integrate w.r.t. and obtain + [ (A B ) (A B) µ(a) (B) + µ(a) B () d () µ(a ) B () d () m µ(a ) (B ). This shows that is a premeasure and we are done. Definition 4.4. Let (, M,µ), (, N, ) be measure spaces, M N the product We define an outer measure (µ ) on P( ) as n + (µ ) (E) inf (A ): E + [ o A, A 2Afor all. -algebra. Lemma 3.5 shows that indeed (µ ) is an outer measure, that A is contained in the -algebra of (µ ) -measurable sets, and (µ ) restricted to A coincides with. As a consequence, we also obtain that M N is contained in the -algebra of (µ ) -measurable sets. Furthermore, if we also assume that (, M,µ) and (, N, ) are -finite, Thm. 3.7 shows that there exists a measure defined on the -algebra of (µ ) -measurable sets.

4 MEASURE THEOR AND LEBESGUE INTEGRAL 43 Thus, we define the product measure µ on the product (µ )(E) (µ ) (E), E 2M N. -algebra M N Finall, µ is the unique measure on M N such that (µ ) A. We observe that the -algebra of (µ ) -measurable sets equals (M N) union the subsets of sets of (µ ) -measure, 4.2. Integration on product measure spaces. Having constructed the product measure space (, M N,µ ) we now have all the results about the integration theor in an abstract measure space at our disposal. However, we wish to relate the calculus of integrals on to the integrals on and on. This is the goal of the current section. Thus, let the measure spaces (, M,µ) and (, N, ) be given. Given a set E 2M N we define its x-section as E x 2:(x, ) 2 E and its -section as E x 2 :(x, ) 2 E. If f is a function on we define the x-section as the function f x () f(x, ) and the -section as f (x) f(x, ). We now have Proposition 4.5. With the notation above, the following hold true: (i) if E 2M N then, for ever x 2, E x 2N and, for ever 2, E 2M; (ii) if f is a function on that is M N-measurable, for ever x 2, f x is N - measurable and, for ever 2, f is M-measurable. Proof. This is eas. Let R E 2M N : E x 2N and 2 for all x 2,2. Then, R contains all measurable rectangles A B, since ( B if x 2 A (A B) x and (A B) ; if x 62 A Next, notice that R is a -algebra, since + [ E x + [ E x and ( A + [ + [ E E, if 2 B ; if 62 B. and also c E x c E x and c E c E. Therefore, R contains the -algebra generated b the measurable rectangles, hence the -algebra generated b A, which equals M N. Hence, R M N, and this shows (i). In order to show (ii), it su ces notice that (f (U)) x (f x ) (U) and (f (U)) (f ) (U). Next, we need the following notion. Definition 4.6. A collection C of subsets of a set is called a monotone class if it is closed with respect to countable increasing unions and countable decreasing intersections. (7)

5 44 M. M. PELOSO Clearl, a -algebra is a monotone class, and in particular P( ) is a monotone class. It is eas to see that intersections of monotone classes is still a monotone class. Hence, given an E P( ) there exists the smallest monotone class containing E; namel the intersection of all monotone classes containing E, famil that is not empt since it contains P( ). We denote b C(E) such monotone class and we call it the monotone class generated b E. Theorem 4.7. (Monotone Class Lemma) Let A be an algebra of subsets of. Then, M(A) C(A), that is, the monotone class generated b A coincides with the -algebra generated b A. Proof. ( ) See [Folland], Monotone Class Lemma We use the Monotone Class Lemma to prove the following result, which is fundamental for the proof of the Tonelli Fubini theorem that will follow. Proposition 4.8. Let (, M,µ), (, N, ) be -finite measure spaces. Let E 2M N. Then, for ever x 2, the function 7! µ(e ) is N -measurable and for ever 2, the function x 7! (E x ) is M-measurable. Moreover, (µ )(E) (E x ) dµ(x) µ(e ) d (). Proof. We first assume that µ and are finite measures, that is, µ( ), () < +. Let C be the collection of all sets E 2M N for which the conclusions in the statement are true. We claim that all measurable rectangles A B belong to C. Indeed, b (7) it follows that ( µ (A B) µ(a) if 2 B if 62 B µ(a) B(), so that the function 7! µ (A B) is -measurable. Analogousl, ( (B) if x 2 A (A B) x if x 62 A (B) A(x), so that the function x 7! (A B) x is µ-measurable. Moreover, (µ )(A B) µ(a) (B) µ(a) B () d () µ (A B) d (), and also (µ )(A B) µ(a) (B) (B) A (x) dµ(x) (A B) x dµ(x). This proves the claim. B finite addivit, also finite unions of disoint measurable rectangles also verif the conclusions in the statement, so that C contains the algebra A of finite unions of disoint measurable rectangles. If we show that C is a monotone class, then C would contain the monotone class generated b the algebra A, that, b the Monotone Class Lemma, coincides with M N. This would impl the result, under the assumption that both µ and are finite measures. In order to show that C is a monotone class, we need to show that it closed under countable increasing unions and countable decreasing intersections of sets. Suppose then that {E n } Cand E E 2. Set E [ + n E n. Then {µ E n } is an increasing sequence of non-negative, N -measurable

6 MEASURE THEOR AND LEBESGUE INTEGRAL 45 functions converging to µ(e ). Thus, µ(e ) is also N -measurable, and b the MCT we have that µ(e ) d lim µ E n d lim (µ )(E n)(µ )(E). n!+ n!+ B switching the roles of µ and and arguing in the same wa, we also obtain that (E x ) dµ (µ )(E). Thus, C is closed under countable increasing unions. On the other hand, suppose that {E n } Cand E E 2. Set E \ + n E n. Here we are going to use the assumption that µ and are finite measures. We have that {µ E n } is a decreasing sequence of non-negative, N -measurable functions converging to µ(e ). Thus, µ(e ) is also N -measurable. Moreover, apple µ E n apple µ( ) 2 L ( ), since µ( ), () < +. Thus, we can appl the DCT and obtain µ(e ) d lim n!+ µ E n d lim n!+ (µ )(E n)(µ )(E). Again, b switching the roles of µ and and arguing in the same wa, we have (E x ) dµ (µ )(E). Thus, C is closed under countable decreasing intersections and C is a monotone class. proves the assertion in the case µ, are finite measures. Suppose now that (, M,µ), (, N, ) are -finite measure spaces. Let { } M,{ } N be increasing sequences of sets such that [ +, [ +. Let E be an set in M N. The previous argument applies to the set E E\( ) to give that for ever, the functions 7! µ((e ) ) and x 7! ((E ) x ) are N -measurable and M-measurable, resp., for ever 2, x 2, resp. Therefore, the functions 7! µ((e ) ) () and x 7! ((E ) x ) (x) are N -measurable and M-measurable, resp., for ever 2, x 2, resp. Moreover, (µ )(E ) ((E ) x ) dµ(x) ((E ) x ) (x) dµ(x) µ((e ) ) d () µ((e ) ) () d (). Finall, we appl the MCT to the increasing sequences { ((E ) x ) } and {µ((e ) ) }, that converge to (E x ) and mu(e ), resp., and obtain that This completes the proof. (µ )(E) lim (µ )(E )!+ lim ((E ) x ) (x) dµ (E x ) dµ!+ lim µ((e ) ) () d µ(e ) d.!+ This

7 46 M. M. PELOSO Theorem 4.9. (The Tonelli Fubini Theorem.) Let (, M,µ), (, N, ) be -finite measure spaces. () (Tonelli) Let f be a non-negative M N-measurable function on. Then, the functions g() R f (x) dµ(x) and h(x) R f x() d () are N -measurable and M-measurable, resp., and f(x, ) d(µ )(x, ) g() d () h(x) dµ(x). (8) (2) (Fubini) Let f 2 L, (µ )(x, ). Then, the function g() R f (x) dµ(x) belgongs to L (, ) for a.e. x and the function h(x) R f x() d () belgongs to L (,µ) for a.e. and equalities (8) hold. Notice that (8) can also be written in terms of iterated integrals f(x, ) d(µ )(x, ) f(x, ) dµ(x) d () It is custumar to use the notations f(x, ) d(µ )(x, ) f(x, ) d () dµ(x) f(x, ) d () dµ(x). fdµd. Proof. () Observe that Tonelli s theorem reduces to Prop. 4.8 in the case of characteristic functions. B finite addivit of the integrals and measures, the result follows for non-negative simple functions. Let f be(m N)-measurable. B Thm. 2.9 () there exists an increasing sequence {s n } of non-negative simple functions, such that s n! f pointwise. Then, we can appl the MCT to obtain f(x, ) d(µ )(x, ) lim s n (x, ) d(µ )(x, ) n!+ lim s n (x, ) dµ(x) d () f(x, ) dµ(x) d (), (9) n!+ and also This proves (). lim n!+ f(x, ) d(µ )(x, ) lim s n (x, ) d(µ )(x, ) n!+ s n (x, ) d () dµ(x) f(x, ) d () dµ(x). (2) In order to prove Fubini s theorem, we proceed in an analogous fashion. For f 2 L (µ ), we Thm. 2.9 (2) to find a sequnce {' n } of simple functions, such that apple ' apple ' n apple f, ' n! f pointwise. Then, we argue as in (9) and (2), using the DCT, instead of the MCT. This completes the proof of Tonelli Fubini s theorem. Example 4.. The hpotheses in the Tonelli Fubini theorem cannot be relaxed, as the following examples show. () Let f :(, ) (, )! R be given b f(x, ) /x. Then, clearl x d dx dx,

8 while, MEASURE THEOR AND LEBESGUE INTEGRAL 47 x dx d x dx d g() d where g() + if << and g() if <<. This last integral does not exist and the iterated integrals are not equal. (2) Let T (x, ) : <x<, <<x, T 2 (x, ) : <<, <x<, and define f :(, ) (, )! R be given f(x, ) x 2 T (x, ) 2 T 2 (x, ). Then, f(x, ) d dx On the other hand, f(x, ) dx d.. x x 2 d x 2 d dx x + d x x + d x + Thus, again the iterated integrals are not equal. 2 dx + x 2 dx d x d + Remark 4.. The measure µ is not complete, even if (µ, M) and(, N ) are complete measures. As an example consider (, µ,m) (c,, N ) (R, m, L), the Lebesgue measure on R. Let A be an set of measure and B the example of a non-measurable set constructed in Example.. Then (A B) 62 L L, but it is contained in a set of (m m)-measure. This is not a significant drawback, since we can alwas complete the measure (µ ). However, the Tonelli Fubini theorem takes a slightl more complicated form in the case of complete measures. Theorem 4.2. (The Tonelli Fubini Theorem for complete measures) Let (, M, µ), (, N, ) be complete -finite measure spaces. Let (, T, ) be the completion of the product measure space (, M N,µ ) and f be T -measurable. (-Tonelli) If f then f x is N -measurable for µ-a.e. x 2 and f is M-measurable for -a.e. 2 and f(x, ) d (x, ) f(x, ) dµ(x) d () f(x, ) d () dµ(x). (2) (2-Fubini) If f 2 L ( ) then f x is N -measurable for µ-a.e. x 2 and f is M-measurable for -a.e. 2. Moreover f x 2 L ( ) for µ-a.e. x 2, f 2 L (µ) for -a.e. 2 and (2) holds. d

9 48 M. M. PELOSO Proof. ( ) We sketch the proof, leaving the details to the reader. Let f be T -measurable. Since (T, ) is the completion of (M N,µ ), b Prop. 2.2 it follows that there exists an (M N)-measurable function g such that g f -a.e. Let h f g. We can appl Tonelli Fubini theorem to g in either case f 2 L + ( ) or f 2 L ( ) and obtain that g(x, ) d (x, ) g(x, ) d(µ )(x, ) g(x, ) dµ(x) d () g(x, ) d () dµ(x). Hence, it su ces to show that if h is T -measurable and h -a.e., then h x, h are in L ( ) and in L (µ), hence measurable, resp., and h x d h dµ (22) for µ-a.e. x, and -a.e., resp. Indeed, using the observation at the end of Section 2, (22) would impl that f(x, ) g(x, ) d (x, ) h(x, ) d (x, ) h(x, ) d(µ )(x, ) h(x, ) dµ(x) d () h(x, ) d () dµ(x). Thus, we prove (22). Since h -a.e., there exists E 2M N such that (µ )(E) and h on c E.Wethenhave h(x, ) d h(x, ) d(µ ) h(x, ) d(µ ). Since (µ )(E), we have which impl that µ(e ) d E (E x ) dµ µ(e ) a.e. and (E x ) µ a.e.. (23) Then, there exists a sequence of simple functions -measurable {' n } such that apple ' apple ' 2 apple apple h and ' n! h. Notice that ' nx! h x and ' n! h, for all x,. Observe also that, clearl, ' n on c E, so that ' nx on c (E x ) and ' n on c (E ), for all x,. Thus, b (23), ' nx Ex µ-a.e. and ' n E -a.e. Since E x and E are N and M-measurable, resp., and µ and are complete, b Prop.?? it follows that also ' nx and ' n are N and M-measurarble, resp. This easil implies that h x 2 L ( ) and h 2 L (µ) for µ-a.e. x and -a.e. and thus (22), and hence the theorem, follow.

10 MEASURE THEOR AND LEBESGUE INTEGRAL The Lebesgue integral in R n. Clearl, the results of the previous section on product measure spaces can be extended to the case of cartesian products of an finite number of measure spaces. We assume the validit of all these results, without formall state an of them. We define the Lebesgue measure in m n on R n as the completion of m m (n copies of m) on B R B R, or, equivalentl, on L L. We denote b L n the -algebra domain of m n, and call these sets Lebesgue measurable sets in R n. In this section we prove some properties of m n that extend similar properties valid in the case n. If R R R n R n is a rectangle, we call R its sides,,...,n. Theorem 4.3. Let E 2L n. Then, the following properties hold true. (i) m n (E) inf m n (U) : E U, U open sup m n (K) : K E, K compact ; (ii) E A \ N, where A is a G -set and m n (N ),andalsoe A 2 [ N 2, where A 2 is an F -set and m n (N 2 ); (iii) if m n (E) < +, then for ever "> there exists a finite collection of rectangles {R },,...,N, whose sides are intervals and such that m n E 4[ N R <". Proof. (i) B construction of the product measure, given E 2L n and ">, there exists a collection {R } of rectangles such that E [ + R and P + m n(r ) apple m n (E)+". In order to prove the first equalit, notice that we ma assume that m n (E) < +, since otherwise we have nothing to prove. Then, P + m n(r ) is a convergent series, so that there exists M> such that m n (R ) apple M for all. For each fixed, R R () R (n),withr (k) 2L, k,...,n. B Prop. 3., for each fixed, we can find open sets U (k) such that R (n) U (k) 2 "/(NM n ), where N is a su cientl large fixed positive integer. Then R U : + [ U () U (n). Notice that the U are open rectangles. Hence U is open, and m n (U) apple apple + + m n (U ) + n k m n (R )+"2 apple m n (E)+2". m U (k) apple + n k and m U (k) m R (k) + " <mr (k) + Since ">was arbitrar, the first equalit in (i) is proved. In order to prove the second equalit, we proceed exactl as in the proof of (ii) in Prop. 3.. The proofs of (ii), which exactl the same as the ones in Prop. 3.4, is again left as an exercise. (iii) Let m n (E) < +, and let "> be given. We can find open rectangles {U } such that P + m n(u ) apple m n (E)+"/2. Hence, m n (U ) < + for all. For each, the sides of U are countable unions of disoint open intervals, sa U (`) [ + I(`) k`,k`, `,...,n. Thus, for each

11 5 M. M. PELOSO fixed, + [ U k I () + [,k k n I (n),k n + [ k,...,k n I (),k I (n),k n. Hence, for each, we can find a finite subunion V of the I (),k I (n),k n s such that m n (U ) apple m n (V )+2 "/2. Notice that the sides of the V s are finite unions of intervals. Then, if N is large enough so that P + N+ m n(u ) <"/2wehave On the other hand, m n N [ m n E \ N[ [ N V apple m n apple m n N [ <". + [ V \ E apple m n This proves (iii), and we are done. U \ V [ + [ N+ U U \ V + m n + [ + U \ E apple Next, we prove that m n is also translation invariant. N+ U m n (U ) m n (E) <". Proposition 4.4. Let E 2L n,andf 2 L + (m n ) or 2 L (m n ). For a 2 R define E + a x + ax2 E 2L n and f a f( + a). Then, m n (E + a) m n (E) and R f a dm n R fdm n. Proof. B composition of translations along each each axis, it su ces to consider the case of translations along one axis, that is, a (,...,a,,...,). Then, if E E E n is a rectangle, then m n (E + a) m(e ) m(e + a ) m(e n )m n (E), b the -dimensional result Prop Then, the result is true for the elements of algebra A of finite unions of disoints measurable rectangles. B the construction of the product measure, the invariance now follows. Notice also that in particular the class of sets of measure zero is invariant under translations. Next, let f be Lebesgue measurable and f a f( + a). Let B be a Borel set in R (or in C, if f is complex-valued). Then, f (B) is a Lebesgue measurable set, hence f (B) E [ N, where E is a Borel set in R n and N is contained in a Borel set of measure. Then, denoting b a the traslation b a 2 R n, so that f( + a) f a, (f a ) (B) a f (B) a E [ N (E a)+(n a),, that is, (f a ) (B) is union of a Borel set and of a set of measure. Hence, it is Lebesgue measurable, and f a is measurable. Finall, the identit R f a dm n R fdm n holds true when f E, and E is a measurable set. Hence, it holds true for simple functions, and b the MCT and the DCT it holds true for f 2 L + (m n ) and f 2 L (m n ), resp.