II - REAL ANALYSIS. This property gives us a way to extend the notion of content to finite unions of rectangles: we define

Transcription

1 1 Measures 1.1 Jordan content in R N II - REAL ANALYSIS Let I be an interval in R. Then its 1-content is defined as c 1 (I) := b a if I is bounded with endpoints a, b. If I is unbounded, we define c 1 (I) = +. More generally, Definition Let R = I 1... I N R N be a N-rectangle, where I k R is an interval, k = 1,..., N. Then the N-content of R is defined as c N (R) = c 1 (I 1 )... c 1 (I N ). We always assume that if c 1 (I k ) = 0 for some k, then c N (R) = 0 (so the 2-content of a straightline in R 2 is 0) and if c 1 (I k ) = for some k, with c 1 (I j ) > 0, j k then c N (R) =. One important feature of the N-content (and as we shall see of measures of sets) is its additivity: if R 1, R 2 are disjoint rectangles then c N (R 1 R 2 ) = c N (R 1 ) + c N (R 2 ). This property gives us a way to extend the notion of content to finite unions of rectangles: we define Definition U(R N ) is the class of sets given by finite unions of n-rectangles, E(R N ) is the class of sets given by finite unions of bounded n-rectangles, that is, bounded sets in U(R N ). Elements of E(R N ) are called elementary or simple sets. Note that since R \ S is a rectangle whenever R, S are rectangles, any finite union can assumed to be disjoint: if U = p i=1 R i, then U = p i=1 R i, R 1 = R 1, R k+1 = R k+1 \ k i=1 R i where the rectangles R are mutually disjoint. (This process of turning an arbitrary union k into an union of disjoint sets will be used often.) Definition Let U E(R N ) be such that U = p i=1 R i, where R i are disjoint rectangles. We define the N-content of U by c N (U) := p c N (R i ). i=1 1

2 Of course, one needs to show - see [NotesMR] - that the definition above does not depend on the partition of U into disjoint rectangles. The class of simple sets is quite restrictive: e.g., balls and triangles are not simple sets. The point now is to consider sets that can be approximated, outer and inner, by simple sets. One can see that if J R N is bounded then there exist simple sets U, K such that K J U. Hence the following definition makes sense: Definition Let J R N be bounded. The outer Jordan content of J is defined as The inner Jordan content of J is defined as c N (J) := inf{c N (U) : U J, U E(R N )}. c N (J) := sup{c N (K) : K J, K E(R N )}. The bounded set J is said to be Jordan measurable, J J(R N ), if c N (J) = c(j). In that case we define the Jordan content of J as c N (J) := c N (J) = c(j). Examples Finite sets are simple, hence Jordan measurable, and have content 2. D = Q [0, 1] not Jordan measurable, c(d) = 1 and c(d) = 0. Useful criteria to show Jordan measurability, that relies mainly on the definition of sup and inf: Proposition (i) J J(R N ) if and only if for all ɛ > 0, there exist simple sets U, K E(R N ) such that K J U, c N (U \ K) < ɛ and in this case c N (J) ]c N (U) ɛ, c N (K) + ɛ[. (ii) J J(R N ) if and only if there exist simple sets U n, K n E(R N ) such that K n J U n, c N (U n \ K n ) 0 and in this case c N (J) = lim c N (K n ) = lim c N (U n ). (iii) J J(R N ) with c N (J) = 0 if and only if there is U E(R N ) such that J U, c N (U) < ɛ 2

3 A set with c N (J) = 0 is said to be a null set. Note that since c N (J) c N (J), any set with c N (J) = 0 is Jordan measurable (see also (iii)), in particular, if N J(R N ) is a null-set and J N then J is Jordan measurable and c N (J) = 0. In (i), we can assume that U and K are open or closed without loss of generality (mainly because if U is simple, then also intu and U are simple, and have the same content as U). In particular, taking K and open U closed in (i), we see that U \ K covers the boundary J of J and that J J(R N ) with c N ( J) = 0. The converse is also true: Proposition Let J R N be bounded. Then J J(R N ) J J(R N ) and c N ( J) = 0. In that case, c N (J) = c N (intj) = c N (J) and (i) c N (J) = 0 intj =, (ii) if J E(R N ) then c N (J) = 0 J is finite. Note that a set with non-empty interior cannot be a null set, since in that case it contains a rectangle with positive content. Properties of J(R N ): Proposition The class of J(R N ) is a semi-algebra: 1 (and also A B J(R N )). A, B J(R N ) A B, A \ B J(R N ) 2. Let A, B J(R N ). The Jordan content c N : J(R N ) [0, + ] is: (i) Additive: if A B =, then c N (A B) = c N (A) + c N (B). (ii) Monotonic: if A B, then c N (A) c N (B). (iii) Subadditive: c N (A B) c N (A) + c N (B). (iv) Invariant under translations and reflexions. 3. Products: A J(R N ), B J(R M ) then A B J(R N+M ) and c N+M (A B) = c N (A)c M (B). By induction, any union in the proposition above can be replaced by a finite union: J(R N ) is closed for finite unions and intersections, and is finitely additive and subadditive. We can show that (for any additive, non-negative set function) for A, B J(R N ) (not necessarily disjoint): c N (A) + c N (B) = c N (A B) + c N (B A). (1) Moreover, always have c N (A) = c N (A B) + c N (A \ B). 1 it is not an algebra, since R N J(R N ). 3

4 Remark (Riemann Integral). Let f : R R be a non-negative, bounded function and let Ω f (E) := {(x, y) : 0 y f (x), x E}. If I = [a, b] is an interval, then c N (Ω f (I)) = b f (x)dx = inf S D ( f ), c N (Ω f (I)) = a b a f (x)dx = sup s D ( f ), where S D ( f ), s D ( f ) are the upper and lowers Darboux sums relative to a decomposition D of I. Hence f is Riemann integrable, f R(I) Ω f (I) J(R N ). Conversely, for E R, let χ E be the characteristic, or indicator, function of E: χ E (x) = 1, x E, and χ E (x) = 0, x E. Asume E is bounded, then E J(R N ) χ E R(I), c N (E) = χ E. R The fact that there are countable sets that are not Jordan measurable yields an example of a sequence of Riemann integrable functions whose pointwise limit is not Riemann integrable: just take E = {p k : k N} J(R N ) and E n = {p k : k n} J(R N ) (finite sets). Then χ E = lim χ En but χ E not Riemann integrable on any interval. We have that finite unions of Jordan measurable sets are always Jordan measurable, but this does not hold even if we take a countable union of points, as Q [0, 1] shows. We would like to extend the definition of Jordan content to countable unions of Jordan measurable sets. This is made possible by the following fundamental result. Theorem Let A n J(R N ), n N, be disjoint. If A = A n J(R N ) then c N (A) = c N (A n ). Proof. Let A n J(R N ), n N, be disjoint and A = A n J(R N ). Given ɛ > 0, let K, K n be closed and U, U n be open such that K A U, c N (U \ K) < ɛ, K n A n U n, c N (U \ K) < ɛ 2 n. We have then that c N (U) ɛ < c N (A) < c N (K) + ɛ and c N (U n ) ɛ 2 c N(A n n ) c N (K n ) + ɛ 2, n so that c N (U n ) ɛ c N (A n ) c N (K n ) + ɛ. 4

5 Now note that by Heine-Borel s theorem, K is compact, as it is closed and bounded. Since {U n } is an open cover, it has a finite subcover: It follows that hence c N (A) c N (A n ). K U n K p U n. c N (A) ɛ < c N (K) c N (A) < p c N (U n ) c N (A n ) + ɛ c N (A n ) + 2ɛ, for all ɛ > 0 Conversely, let again A n J(R N ), n N, be disjoint and A = A n J(R N ). Since, for all n N, we have A n k=1 A k, by monotonicity and finite additivity: ( c N (A) c N n k=1 A ) n k = c N (A k ), n N hence c N (A) k=1 c N (A k ). A function satisfying the condition in the above theorem is said to be a pre-measure, a terminology that will be made clear in the following sections. We now have a way of extending the Jordan content to countable unions of Jordan measurable, need to check that if A = A n = B n for collections of measurable, disjoint, sets (A n ), (B n ) then since A n is Jordan measurable and A n = m=1 A n B m, we get from the previous result that k=1 c N (A n ) = c N (A n B m ) = c N (B m ), m=1 m=1 where we apply to same reasoning to B m to get the last equality. Definition We let J σ (R N ) denote the class of countable unions of Jordan measurable sets. Define the extended Jordan content c N : J σ (R N ) [0, + ] such that if A = A n, with A n J(R N ) disjoint, then c N (A) := c N (A n ). The previous theorem shows that the function defined above is indeed an extension of the Jordan content on J(R N ), and we write in general c N as c N. We let E σ (R N ) denote the class of countable unions of simple sets, that is, the class of countable unions of rectangles. 5

6 A set function λ is said to be σ-additive if λ ( A n ) := λ(a n ), A n disjoint and σ-subadditive if λ ( A n ) λ(a n ). Proposition The classes J σ (R N ) and E σ (R N ) are closed for countable unions and c N is σ-additive and σ-subadditive on J σ (R N ) and E σ (R N ). Examples Any countable set is in J σ (R N ) and if Q = {q n } then ( c N (Q) = c N {q n } ) = c N ({q n }) = Let D = [0, 1] Q. Then D J σ (R N ), since it is countable, and c N (D) = 0. By additivity, if D c := [0, 1] R \ Q J σ (R N ), then c N (D c ) = 1. But int(d c ) =, hence if D c = A n, with A n J(R N ), then int(a n ) =, hence c N (A n ) = 0 and also c N (D c ) = 0. We conclude that D c = [0, 1] \ D = [0, 1] R \ Q J σ (R N ). In particular, J σ (R N ) is not closed for difference of sets, hence not an algebra. NOTE: In general, for A J σ (R N ), c N (A) = 0 int(a) =. In particular, A E σ (R N ), c N (A) = 0 A is countable. 3. Any open set is given by a countable union of rectangles, hence is σ-simple. Therefore J σ (R N ) contains all the open sets. 4. An open set U J(R): let {q n } = [0, 1] Q and ɛ > 0 be given. Define U n :=]q n ɛ 2 n, q n + ɛ 2 n [, U := U n. Then U is open, U E σ (R) J σ (R). By subadditivity, c(u) c(u n ) = ɛ = 2ɛ. 2n 1 On the other hand, since [0, 1] Q U, the upper Jordan content c(u) c([0, 1] Q) = 1. If U J(R), we would have c(u) = c(u) < 2ɛ, c(u) = c(u) 1. For ɛ < 1/2 this is a contradiction. Hence U J(R) for ɛ < 1/2. 6

7 The following two examples will be used and often used as reference during the course. Example (Cantor set). Let I = [a, b]. Define T(I) := I \ ] a+b c(i), [ a+b + c(i) and T( p I k=1 k) := p T(I k=1 k). Let F 0 = [a, b] and F n = T(F n 1 ), n N. Then F n is given by finite unions of closed sets, hence it is a closed, simple set and c(f n ) 0. Define C(I) = F n. Since F n E(R N ) and C(I) F n, with c(f n ) 0, it follows from Proposition (iii) that C(I) J(R) with c(c(i)) = 0. It is an uncountable set with int(c(i)) =. Topological properties: closed, hence compact nowhere dense int(c(i))) = int(c(i))) = perfect set: all points in C(I) are limit points (no isolated points) totally disconnected. Also have that I \ C(I) is a countable union of open intervals, hence is in E σ (R). But C(I) E σ (R) being an uncountable null set. Hence, E σ (R N ) is not closed for difference of sets, hence it is not an algebra. Example (Smith-Volterra-Cantor set). Borel Problem: Find a collection of sets M P(R N ) and m N : M [0, + ] such that (1) M is an algebra closed for countable unions and for A n M, m N A n = m N (A n ). (2) M E(R N ) and m N (E) = c N (E), for E E(R N ). Note that M E σ (R N ), in particular M contains all opens sets, as well as all closed sets in R N. Note also that, since c N is translation invariant, m N is translation invariant on E σ (R N ). If we require that m N is translation invariant on M, we see now that such a collection M is necessarily proper, even though the existence on sets that cannot be measured by m N relies on the axiom of Choice. Example (Vitali s set). Define an equivalence relation on [0, 1[ by x y x y Q. Let V [0, 1[ be such that the intersection of V with [x] contains precisely one element (need Axiom of Choice). We claim that if (M, m N ) is a translation invariant solution for Borel s problem, then V M. It follows from Vitali s example that: If (M, m N ) is a translation invariant solution for Borel s problem then M P(R N ). 7

8 1.2 σ-algebras and measure spaces We now consider an arbitrary base space X. Definition Let M P(X). Then (i) M is an algebra if X M and A, B M A B, A \ B M (ii) M is a σ-algebra if it is an algebra and is closed with respect to countable unions: A n M, n N A n M. We have always, X are in M and that M is closed for countable intersections as well: A n M, n N A n = X \ X \ A n M. Examples P(X) is the largest σ-algebra, {, X} is the smallest. 2. J(R N ), E(R N ) semi-algebra, not closed for countable unions, hence not σ-algebras. 3. J σ (R N ), E σ (R N ): closed for countable unions but not algebras: not closed for difference of set. Hence, are not σ-algebras. 4. Let (X, τ) be a topological space. Then the collection τ of open sets in X is closed for countable unions, but fails to be an algebra, as it is not closed for complements. Even if a given class A is not s σ-algebra, we can always consider the smallest σ- algebra that contains it. It is easy to see that the intersection of σ-algebras is still a σ-algebra, and P(X) is a σ-algebra containing any collection of sets A. We define the σ-algebra generated by A by M(A) := M, M σ-algebra, A M. Definition Let X be a topological space. The Borel σ-algebra B(X) is the σ-algebra generated by the class of opens sets. Sets in B(X) are called Borel sets, include all open sets, all closed sets, all countable unions of closed sets - such a set is called a Fσ-set - and all countable intersections of open sets - such a set is called a G δ -set. When X = R N, it is easily seen that B(R N ) is generated by open/closed/half- open rectangles. Definition Let M be a σ-algebra. A set function µ : M [0, + ] is a measure on M if 8

9 (i) µ( ) = 0, (ii) µ is σ-additive: for A n M, n N, disjoint, µ A n = µ(a n ). A measure space is a triple (X, M, µ) where µ is a measure on M. Sets in M are said to be µ-measurable. The measure is finite if µ(x) < and σ-finite if there exist X n with µ(x n ) < and X = X n. Note that µ( ) = 0 µ(a) <, for some A M. Moreover, for A, B M, not necessarily disjoint, we always have µ(a) + µ(b) = µ(a B) + µ(b A), µ(a) = µ(a B) + µ(a \ B). (2) Examples Counting measure: # : P(X) [0, + ] #(E) = the number of elements of E, if E is finite, and #(E) = +, if E is infinite. # is finite measure X is finite, and a σ-finite measure X is countable. In particular, it is an example of non σ-finite measure on R. 2. Dirac measure given x 0 X, define δ x0 : P(X) [0, ] by 1, if x 0 E, δ x0 (E) = 0, if x 0 E 3. Dirac comb: µ : P(X) [0, + ] such that is a σ-finite measure. µ(e) = #(E Z) = δ n (E). (NOTE: the sum of measures is always a measure -Exercise.) 4. Probability measure: p : M [0, 1] such that p(x) = 1, M is the space of events and (X, M, p) is called a probability space. For instance, take p(e) = #(E) #(X), X finite. 9 n Z

10 5. Borel measures: if X is a topological space, a Borel measure is a measure defined on σ-algebra of the Borel sets B(X), generated by the opens sets. In R N, the most important Borel measure is the Lebesgue measure, which is translation invariant and yields a solution to Borel s problem. 6. Let f : R R be increasing, define µ f (]a, b[) := f (b) f (a). Then µ is σ-additive on the σ-algebra generated by the open intervals, that is in B(R), so (R, B(R), µ f ) is a measure space. The Lebesgue measure corresponds to the case f (x) = x. (The Dirac measure is a particular case, with H(x) = 1, x x 0, H(x) = 0, x < x 0.) 7. Haar measures: invariant measures on locally compact topological groups, defined on Borel sets. An additive, non-negative, function is always monotonic, as µ(b) = µ(b A)+µ(B\A). Using monotonicity, we can see that σ-additivity implies σ-subadditivity. Moreover, a measure is always continuous with respect to monotonic sequences, in a sense made clear by the next result. Proposition Let (X, M, µ) be a measure space. 1. µ is σ-subadditive; 2. Let E n M, n N such that µ(e 1 ) <, 2 E n+1 E n and E = E n (write E n E). Then µ(e) = lim µ(e n ). 3. Let E n M, n N such that E n E n+1 and E = E n (write E n E). Then µ(e) = lim µ(e n ). Properties 2. and 3. are sometimes called continuity from above and continuity from below, respectively. (In [NotesR], Property 3. is called Lebesgue s monotone convergence theorem.) Proof. To see σ-subadditivity, let A = A n and write A = A n with A n = A n \ n 1 k=1 A k, then by σ-additivity and monotonicity µ(a) = µ(a n) µ(a n ). To prove 2., let E n M, n N such that µ(e 1 ) <, E n+1 E n and E = E n then E 1 can be written as a disjoint union E 1 = E ( k=1 E k \ E k+1 ) 2 (N, P(N), #), with E n = {k n} then E n =, µ(e n ) =. 10

11 hence Since µ(e 1 ) = µ(e) + µ(e k \ E k+1 ) = lim k=1 µ(e k \ E k+1 ) <. k=1 n µ(e k ) µ(e k+1 ) = µ(e 1 ) lim µ(e n+1 ), k=1 and µ(e 1 ) <, it follows that µ(e) = lim µ(e n ). To prove 3., let now E n M, n N such that E n E n+1 and E = E n. Then E can be written as a disjoint union and in the same way E = k=0 E k+1 \ E k, E 0 = µ(e) = n µ(e k+1 \ E k ) = lim µ(e k+1 \ E k ) = lim µ(e n+1 ). k=0 k=0 Let (X, M, µ) be a measure space. Sets in M with µ(e) = 0 are usually called null sets and play an important role in measure theory, as null sets are used as a means of approximation: a property that holds except on a null-set is said to hold µ-almost everywhere, µ-a.e. (and we often look for characterizations of measurable sets minus a null-set). Note that a countable union of null-sets is also a null-set, by σ-subadditivity. Let N be a null set and E N. If E M, then µ(e) µ(n) = 0, hence E is also a null set. There is however no reason in general for E M. Definition A measure space (X, M, µ) is said to be complete if E N M, µ(n) = 0 E M, µ(e) = 0. Even if a given measure space is not complete, we can always form its completion, extending µ to a larger σ-algebra. Let N M be the collection of all null sets. Define M := {E F : E M, F N, N N}, µ : M [0, + ], µ(e F) := µ(e). First check that µ is well-defined: if E 1 F 1 = E 2 F 2 then E 1 \ E 2 F 2 N 2 for some null set N 2, hence is also a null set (since E 1 \ E 2 M) and µ(e 1 ) = µ(e 1 \ E 2 ) + µ(e 1 E 2 ) = µ(e 1 E 2 ) = µ(e 2 ), by the same reasoning with E 2. Moreover, if A E F M with µ(e F) = 0, then E is a µ-null set, so A is a subset of a µ-null set and hence A M. An alternative definition of M is the following: 11

12 A M there exist U, K M with K A U and µ(u \ K) = 0, which illustrates that sets in M are precisely the ones that can be approximated by sets in M, modulo a µ-null set. (Exercise.) Theorem (X, M, µ) is a complete measure space and is the smallest complete extension of (X, M, µ). 1.3 Outer measures Now we turn to the issue of defining measure spaces. One way of achieving this is to consider first outer approximations, which should be defined for any set, and then define measurability from there. Definition An outer measure is a set function µ : P(X) [0, + ] such that (i) µ ( ) = 0, (ii) µ is monotonic, (iii) µ is σ-subadditive: for any A n X, n N, µ A n µ (A n ). So, we drop σ-additivity, requiring only the weaker σ-subadditivity ( approximation from the outside ) but require on the other hand that µ is defined on the whole of P(X). Any measure defined on P(X) is also an outer measure, so the counting measure and the Dirac measures are outer measures. Example Outer Jordan content is subadditive but not σ-subadditive: take D = Q [0, 1], then c N (D) = 1, but D = {q n }, and c N ({q n }) = 0. Hence it is not an outer measure. The following proposition gives a common way of obtaining outer measures: Proposition Let E P(X) be such that, X E and λ : E [0, ] such that λ( ) = 0. For A X, let λ : P(X) [0, ] given by Then λ is an outer measure. λ (E) = inf λ(e n ) : E j E, A E n. 12

13 Now we want to associate a measure space to a given outer measure µ, that is, a σ- algebra M µ of measurable sets and a measure µ = µ on M µ, so µ is σ-additive on M µ. Noting that a finitely additive function is σ-additive if and only if it is σ-subadditive, we want to find a σ-algebra M µ where µ is additive. Definition Let µ be an outer measure on X. A set A X is said to be µ -measurable if µ (E) = µ (E A) + µ (E A c ), for all E P(X). We denote by M µ the class of all µ -measurable sets. Note that we have always, by subadditivity, µ (E) µ (E A) + µ (E A c ). The next lemma shows that we get finite additivity of µ restricted to M µ (note that B can be any set). Lemma Let A M µ, B P(X). Then µ (A B) = µ (A) + µ (B \ A). If A B = then µ (A B) = µ (A) + µ (B). Proof. Taking E = A B in the definition of µ -measurability, we get µ (A B) = µ ((A B) A) + µ ((A B) A c ) = µ (A) + µ (B A c ). In fact, more is true, as we can prove in the same way that if A M µ, B, C X, with A, B disjoint, then µ (C (A B)) = µ (C A) + µ (C B). We have then, by induction, that µ is (finitely) additive on any algebra contained in M µ, hence σ-additive on any σ-algebra contained in M µ. Lemma M µ is a σ-algebra. Proof. It is closed for complements, by definition, and, X M µ. Let now A, B M µ. We show that A B M µ, want to see that µ (E) µ (E (A B)) + µ (E (A B) c ) for any E X. Now E (A B) c = (E A c ) (E A B c ) (disjoint union), hence µ (E (A B) c ) + µ (E (A B)) µ (E A c ) + µ (E A B c ) + µ (E A B) = µ (E A c ) + µ (E A) = µ (E) where we used µ -mensurability of B and A in the last equalities. Hence M µ is an algebra, in particular, closed for finite unions. 13

14 We show it is closed for countable unions. Let A = A n, with A n M µ, n N, assume disjoint. Let E X. For finite unions, we know that for any n N, µ (E) = µ (E ( n k=1 A k)) + µ (E ( n k=1 A k) c ) µ (E ( n k=1 A k)) + µ (E A c ) (since ( n A k=1 k) c A c ). Moreover, by measurability of each A n, using induction, one can see (Exercise) that n µ (E ( n k=1 A n)) = µ (E A k ). Letting n, it follows µ (E) k=1 µ (E A k ) + µ (E A c ) µ (E A) + µ (E A c ), k=1 by σ-subadditivity 3. Since we always have, µ (E) µ (E A) + µ (E A c ), the measurability of A follows, and that finishes the proof. Given an outer measure µ, we have constructed a σ-algebra M µ where µ is σ- additive (in fact, we only needed additive), that is, so that the restriction of µ to M µ is a true measure. Moreover, this measure is always complete. In fact any set with µ (A) = 0 is always measurable : if E X, then µ (E A) µ (A) = 0, so that µ (E) µ (A E) + µ (A c E) = µ (A c E) µ (E). Hence µ (E) = µ (A E) + µ (A c E) and A M µ. In particular, if A N where N M µ such that µ(n) = µ (N) = 0, then µ (A) µ (N) = 0, so A is measurable and the space is complete. We have proved: Theorem (Caratheodory). Let µ be an outer measure on X. Then there exists a complete measure space (X, M µ, µ) such that where M µ is as in µ(e) = µ (E), for E M µ. The first application of Caratheodory s construction is to extend additive functions on algebras to measures on σ-algebras. Definition Let A P(X) be an algebra, or a semi-algebra such that X is a countable union of elements of A. A premeasure on A is a function µ 0 : A [0, ] such that µ ( ) = 0, and for A n A disjoint, n N, then A = A n A µ 0 (A) = µ 0 (A n ). 3 In fact, in this case we have σ-additivity: prove that µ (E A) = k=1 µ (E A k ) 14

15 In particular, µ 0 is additive on A. Now to each premeasure, we can associate an outer measure, according to µ (E) = inf µ 0 (A n ) : A j A, E A n. (3) Proposition Let µ 0 be a premeasure on an algebra / semi-algebra 4 A and µ be defined as above, M µ the σ-algebra of µ -measurable sets. Then (i) µ A = µ 0 (ii) A M µ if and only if µ 0 (B) = µ (B A) + µ (B A c ), for all B A, (iii) A M µ. Proof. (i). Let A A. It is clear from the definition that µ (A) µ 0 (A). Let A A n, A n A. Then A = A A n A hence, by the premeasure property, µ 0 (A) = µ 0 (A A n ) µ 0 (A n ). Hence µ 0 (A) µ (A), so equality follows. (ii) Let A X such that µ 0 (B) = µ (B A) + µ (B A c ), for all B A,. For E X such that E B n, B n A we have E A (B n A), E A c (B n A c ) hence µ (E A) + µ (E A c ) µ (B n A) + µ (B n A c ) = µ 0 (B n ) It follows µ (E) µ (E A) + µ (E A c ) µ (E), so A is measurable. (iii) is an easy consequence of (i) and (ii). Theorem (Hahn s extension theorem). Let µ 0 be a premeasure on an algebra / semialgebra A and M be the σ-algebra generated by A. Then (i) µ 0 extends to a σ-additive function µ on M and (X, M, µ) is a measure space. (ii) If ν is also an extension of µ 0 to M, then µ(e) = ν(e), if µ(e) <, E M. In particular, the extension is unique if µ 0 is σ-finite. 4 In this case, we assume that X = A n for some collection A n A. 15

16 Proof. (i) is Caratheodory s extension restricted to M, noting that if A M µ, then also M M µ (as M µ is a σ-algebra). As for (ii), if ν is another extension, then ν(e) µ (E) = µ(e) as if E A i, with A i disjoint, then ν(e) ν( A i ) = ν(a i ) = µ 0 (A i ). If µ(e) <, then for any ɛ > 0, can take E A = A i, A i A disjoint such that µ(a \ E) < ɛ. Have µ(a) = ν(a) by σ-additivity of µ, ν, as the measures coincide on A. Hence µ(e) µ(a) = ν(a) = ν(e) + ν(a \ E) ν(e) + µ(a \ E) ν(e) + ɛ, ɛ > 0 so µ(e) ν(e). If the space is σ-finite, any set can be written as a countable union of sets with finite measure, hence the two measures coincide. Note that in fact we always have a complete extension of µ 0 and A considering µ on the class M µ of µ -measurable sets. Remark (Inner measures). If µ is an outer measure obtained by extending a premeasure on an algebra, then we can define an inner measure by µ (E) = µ 0 (X) µ (E c ). Then E is µ -measurable iff µ (E) = µ (E). (Exercise [Fol] uses regularity) 1.4 Lebesgue measure We are now back in R N and will use the results in the previous section to provide a solution (in fact, two) to Borel s problem: we want to define a measure space (R N, M N, m N ) such that M N E(R N ), that is, simple sets, and m N extends the Jordan content: m N (E) = c N (E), if E E(R N ). The outer Jordan content is subadditive but not σ-subadditive (just take D = Q [0, 1] = {q n }), hence Caratheodory s construction cannot be applied directly. Nevertheless, it follows from Theorem that c N is a premeasure on the semi-algebra E(R N ): we saw in partcular that, if A = A n E(R N ), with A n disjoint, then c N (A) = c N (A n ). From the results in the previous section, it induces an outer measure m N and a σ-algebra where it becomes a measure. Definition The Lebesgue outer measure m N : P(RN ) [0, + ] is defined as m N (A) = inf{ c N (R n ) : R n bounded rectangle, A R n}. 16

17 The Lebesgue measurable sets L(R N ) are those sets A such that for any E R N, m N (E) = m N (E A) + m N (E Ac ). The Lebesgue measure m N : L(R N ) [0, ] as m N (A) = m N (A), if A L(RN ). It is easy to see that m N is indeed the outer measure induced by c N as in (3). Note that the σ-algebra generated by E(R N ) coincides with the σ-algebra generated by the collection of bounded rectangles, which coincides with the σ-algebra generated by the open sets, that is, with the Borel algebra B(R N ). It follows Proposition that A L(R N ) iff c N (R) = m N (R A) + m N (R Ac ), R bounded rectangle. We summarize the results from the previous section: (i) L(R N ), B(R N ) are σ-algebras and m N is a measure such that for any E E(R N ), m N (E) = c N (E). (ii) m N is the unique extension of c N to B(R N ) (as R N is σ-finite). (iii) L(R N ) is a complete extension of B(R N ) (will see that it is the completion). Recalling Borel s problem of extending c N we now have: Theorem (R N, B(R N ), m N ) and (R N, L(R N ), m N ) are solutions to Borel s problem. (R N, B(R N ), m N ) is the smallest solution, and (R N, L(R N ), m N ) is a complete solution. 5 Note that completeness of L(R N ) follows from the stronger crucial property, that we use often: m N (A) = 0 A L(RN ). Examples Countable sets 2. All open sets and all the closed sets are in B(R N ) L(R N ), as are all sets of the form G δ = U n, U n open, F σ = K n, K n closed. Sets that are countable unions of closed sets are called F-sigma sets, and countable intersections of open sets are called G-delta sets. Also, the closure and the interior of any set are always measurable. 5 it will follow from B(R N ) = L(R N ) that it is also the smallest complete solution. 17

18 3. The set U = ]q n ɛ, q 2 n n + ɛ [, where q 2 n n are the rationals in [0, 1]: open (not in J(R N )) 4. Cantor and Volterra ( fat Cantor): closed. Volterra s set shows that in L(R N ), a set with positive measure can have empty interior, in fact be nowhere dense. 5. Cardinality: the Cantor set C(I) and all subsets of Cantor are Lebesgue measurable (in fact, also Jordan measurable), as they are null sets. Since C(I) is uncountable, it follows that #J(R N ) = #L(R N ) = #P(R). On the other hand, the Borel sets are generated by a countable basis of open sets, and it can be proved that #B(R N ) = #P(N) = #(R) So there are many more Lebesgue (and Jordan) measurable sets than Borel sets. It is not easy however to give an explicit description of a set in L(R N ) \ B(R N ). We have, as we see below, J(R N ) L(R N ). We give now two equivalent definitions of m N (we take the extension of c N to E σ (R N ) given by σ-additivity). Proposition Let E σ (R N ) be the collection of countable unions of simple sets m N (A) = inf{c N(E) : E E σ (R N ), A E} = inf{c N (U) : U is open, A U}. Proof. We have m N (A) = inf{ c N (E n ) : E n E(R N ), A E n}. The first equality follows noting that we can assume without loss of generality that E n are disjoint and in this case if E = E n E σ (R N ), then c N (E) = c N (E n ). As for the second, any open set U E σ (R N ), as it can be written as a countable union of (disjoint) rectangles, so inf{c N (E) : E E σ (R N ), A E} inf{c N (U) : U is open, A U}. Moreover, given E E σ (R N ) and ɛ > 0, can always find open U E such that c N (U) c N (E) + ɛ: if c N (E) < and E = R n, take open rectangles R n R n such that c N (R n \ R n ) < ɛ/2 n and U = R n. If c N (E) = then any open U E also has infinite content. Hence, the reverse inequality holds, and equality follows. The first definition given above is quite similar in form to that of Jordan outer content: just replace E(R N ) by E σ (R N ). In particular, we can see that c N (A) m N (A) c N(A) for any A hence if A J(R N ) then m N (A) = c N(A). Moreover, 18

19 Proposition J σ (R N ) L(R N ) and m N = c N on J σ (R N ) Proof. We show that J(R N ) L(R N ), and so m N = c N on J ( R N ). Given A J(R N ) then for any bounded rectangle R, R A, R A C J(R N ), by additivity c N (R) = c N (R A) + c N (R A c ) = m N (R A) + m N (R Ac ). Then J σ (R N ) L(R N ), as L(R N ) is a σ-algebra. Exercise: show m N extends c N. Not all Borel sets are Jordan measurable: saw in Example an open non Jordan measurable set and we have seen that not all Jordan measurable are Borel measurable. It follows from the second definition that if A L(R N ) then m N (A) = inf{c N (U) : U is open, A U} = inf{m N (U) : U is open, A U}. A Borel measure with the above property is said to be (outer) regular. Regularity means that the measure is completely determined by its values on open sets, by approximation. This property, together with the fact that sets with 0 outer measure are measurable (completeness), give a number of useful characterizations of Lebesgue measurable sets. Proposition E L(R N ) (i) given ɛ > 0, there is U open, with E U and m N (U \ E) < ɛ (ii) given ɛ > 0, there is K closed, with E K and m N (E \ K) < ɛ (iii) given ɛ > 0, there are U open, K closed, with K E U and m N (U \ K) < ɛ. Proof. (iii) is equivalent to (i) + (ii), while (ii) is equivalent to (i) taking complements. So we prove equivalence (i). Let E L(R N ) and ɛ > 0 be given. If m N (E) <, then since m N (E) = inf{m N(U) : U is open, E U}, we can take open U E such that m N (E) m N (U) < m N (E) + ɛ. Since m N (U) = m N (E) + m N (E \ U), by measurability of E, and m N (E) <, m N (U) <, it follows that m N (U \ E) < ɛ. If m N (E) =, write E = E n, where E n = E R n, with R n bounded rectangles, disjoint such that X = R n. Then m N (E n) <, and by the above there is open U n E n such that m N (U n \ E n ) < ɛ. Let U = U 2 n n, open. Then m N (U \ E) < ɛ. Conversely, if there is U n E such that m N (U n \ E) < 1 n, then letting U = U n, have m N (U \ E) m N (U n \ E) < 1 n m N (U \ E) = 0 U \ E L(RN ). Hence E = U \ (U \ E) L(R N ). 19

20 The following consequence is very useful: Corollary E L(R N ) if there exist U n open, with U n E such that m N (U n \ E) 0 and in that case m N (U n ) m N (E). Recall that a F σ set is a countable union of closed sets and a G δ set is a countable intersection of open sets; F σ and G δ sets are in the Borel algebra. Proposition E L(R N ) there are a F σ set B and a G δ set A such that A E B and m N (B \ A) = 0. In this case, m N (A) = m N (B) = m N (E). Proof. For each n N, let K n be closed, U n be open, such that K n E U n and m N (U n \ K n ) < 1. Let B = K n n and A = U n, then B \ A U n \ K n for each n, and the result follows. Conversely, if there are such A, B, then E \ A B \ A B(R N ) L(R N ), hence by completeness, E \ A L(R N ) and also E = A E \ A L(R N ). It follows straightaway from the definition of completion that: Corollary L(R N ) is the completion of the Borel algebra B(R N ). Proposition (R N, L(R N ), m N ) is the unique complete and regular (it is the largest regular and the smallest complete) solution to Borel s problem. Proof. Any complete solution contains B(R N ), hence contains B(R N ) = L(R N ). Let now (X, M, µ) be a regular solution to Borel s problem and A M. By regularity, µ(a) = m N (A) as µ = m N = c N on open sets. Let R be a bounded rectangle, then c N (R) = µ(r) = µ(a R) + µ(a c R) = m N (A R) + m N (Ac R) hence A L(R N ) and in this case µ(a) = µ (A) = m N (A). To finish this section, we now investigate some properties, of geometrical nature, of the Lebesgue measure that were known for the Jordan content: invariance, and behavior with respect to products. Proposition Let A L(R N ) and x R N. Then A + x = {a + x : a A} L(R N ) and m N (A) = m N (A + x). 2. Let A B(R N ) and x R N. Then A + x = {a + x : a A} B(R N ) and m N (A) = m N (A + x). 20

21 Proof. It follows from regularity and the fact that c N is translation invariant that m N (A + x) = m N (A). Now for E X, x R N, E (A + x) = ((E + ( x)) A) + x hence m N (E (A + x)) + m N (E (Ac + x)) = m N ((E + ( x)) A) + m N ((E + ( x)) Ac ) = m N ((E + ( x)) = m N (E), by measurability of A. As for the Borel case: let A = {A R N : A + x B(R N ), x R N }. Then A contains the open sets and A is a σ-algebra, hence A B(R N ). Moreover, the Lebesgue measure is also invariant with respect to unitary transformations [Foll Thm 2.44] We have seen, when introducing Borel s problem, that no tranlsation invariant solution could be defined on the whole P(R N ): the classical example is Vitali s set, which provides an example of a non- Lebesgue measurable set. Examples Vitali s construction can be adapted so as to show that any A with m N (A) > 0 contains a non-measurable subset. 2. Lebesgue measurable, not Borel measurable. follows form existence of nonmeasurable. Let f : [0, 1] [0, 1] be Cantor function, that is f ( x i 3 i ) := x i 2 i if x i 1, and f constant on subintervals with x j = 1 for some (the smallest) j, f (x) = j xi 2 i. Then f increasing, continuous, and f (C) = [0, 1], where C is the Cantor set in [0, 1]. (Also known as devil s staircase ). Such a function maps Borel sets to Borel sets. Let V [0, 1] be Vitail s set and take E C such that E = f 1 (V). Then E is Lebesgue measurable (even Jordan measurable), as a subset of the Cantor set, but not Borel measurable, as in this case f (E) = V would be as well. Behavior with respect to products: see first Lemma Let A R N and B R M, then m N+M (A B) m N (A)m M (B). In particular, it follows that if m N (A) = 0 (or m M (B) = 0) then m N+M (A B) = 0, hence A B L(R N+M ). Proposition Let A L(R N ) and B L(R M ). Then A B L(R N+M ) and m N+M (A B) = m N (A)m M (B). 2. Let A B(R N ) and B B(R M ). Then A B B(R N+M ) and m N+M (A B) = m N (A)m M (B). 21