Measures and σ-Algebras

2. Measures and σ-algebras An outer measure such as µ∗ is a rather crude object since, even if the Ai are disjoint, there is generally strict inequality in (1.14). It turns out to be unreasonable to expect equality in (1.14), for disjoint unions, for a function defined on all subsets of X. We therefore restrict attention to smaller collections of subsets. Definition 2.1. A collection of subsets M of a set X is a σ-algebra if (1) φ, X ∈ M (2) E ∈ M =⇒ E C = S X\E ∈ M ∞ (3) {Ei }∞ ⊂ M =⇒ i=1 i=1 Ei ∈ M. For a general outer measure µ∗ we define the notion of µ∗ -measurability of a set. Definition 2.2. A set E ⊂ X is µ∗ -measurable (for an outer measure µ∗ on X) if (2.1) µ∗ (A) = µ∗ (A ∩ E) + µ∗ (A ∩ E { ) ∀ A ⊂ X . Proposition 2.3. The collection of µ∗ -measurable sets for any outer measure is a σ-algebra. Proof. Suppose E is µ∗ -measurable, then E C is µ∗ -measurable by the symmetry of (2.1). Suppose A, E and F are any three sets. Then A ∩ (E ∪ F ) = (A ∩ E ∩ F ) ∪ (A ∩ E ∩ F C ) ∪ (A ∩ E C ∩ F ) A ∩ (E ∪ F )C = A ∩ E C ∩ F C . From the subadditivity of µ∗ µ∗ (A ∩ (E ∪ F )) + µ∗ (A ∩ (E ∪ F )C ) ≤ µ∗ (A ∩ E ∩ F ) + µ∗ (A ∩ E ∪ F C ) + µ∗ (A ∩ E C ∩ F ) + µ∗ (A ∩ E C ∩ F C ). Now, if E and F are µ∗ -measurable then applying the definition twice, for any A, µ∗ (A) = µ∗ (A ∩ E ∩ F ) + µ∗ (A ∩ E ∩ F C ) + µ∗ (A ∩ E C ∩ F ) + µ∗ (A ∩ E C ∩ F C ) ≥ µ∗ (A ∩ (E ∪ F )) + µ∗ (A ∩ (E ∪ F )C ) . The reverse inequality follows from the subadditivity of µ∗ , so E ∪ F is also µ∗ -measurable. ∞ of disjoint µ∗ -measurable sets, set Fn = SnIf {Ei }i=1 is a Ssequence ∞ i=1 Ei . Then for any A, i=1 Ei and F = µ∗ (A ∩ Fn ) = µ∗ (A ∩ Fn ∩ En ) + µ∗ (A ∩ Fn ∩ EnC ) = µ∗ (A ∩ En ) + µ∗ (A ∩ Fn−1 ) . Iterating this shows that ∗ µ (A ∩ Fn ) = n X µ∗ (A ∩ Ej ) . j=1 From the µ∗ -measurability of Fn and the subadditivity of µ∗ , µ∗ (A) = µ∗ (A ∩ Fn ) + µ∗ (A ∩ FnC ) n X ≥ µ∗ (A ∩ Ej ) + µ∗ (A ∩ F C ) . j=1 Taking the limit as n → ∞ and using subadditivity, ∞ X ∗ (2.2) µ (A) ≥ µ∗ (A ∩ Ej ) + µ∗ (A ∩ F C ) j=1 ≥ µ∗ (A ∩ F ) + µ∗ (A ∩ F C ) ≥ µ∗ (A) proves that inequalities are equalities, so F is also µ∗ -measurable. In general, for any countable union of µ∗ -measurable sets, ∞ ∞ [ [ ej , Aj = A j=1 j=1 j−1 j−1 ej = Aj \ A [ Ai = A j ∩ [ i=1 i=1 ej are disjoint. is µ∗ -measurable since the A Ai !C  A measure (sometimes called a positive measure) is an extended function defined on the elements of a σ-algebra M: µ : M → [0, ∞] such that (2.3) (2.4) µ(∅) = 0 and ! ∞ ∞ [ X Ai = µ(Ai ) µ i=1 i=1 if {Ai }∞ i=1 ⊂ M and Ai ∩ Aj = φ i 6= j. The elements of M with measure zero, i.e., E ∈ M, µ(E) = 0, are supposed to be ‘ignorable’. The measure µ is said to be complete if (2.5) E ⊂ X and ∃ F ∈ M , µ(F ) = 0 , E ⊂ F ⇒ E ∈ M . See Problem 4. The first part of the following important result due to Caratheodory was shown above. Theorem 2.4. If µ∗ is an outer measure on X then the collection of µ∗ -measurable subsets of X is a σ-algebra and µ∗ restricted to M is a complete measure. Proof. We have already shown that the collection of µ∗ -measurable subsets of X is a σ-algebra. To see the second part, observe that taking A = F in (2.2) gives µ∗ (F ) = X j µ∗ (Ej ) if F = ∞ [ Ej j=1 and the Ej are disjoint elements of M. This is (2.4). Similarly if µ∗ (E) = 0 and F ⊂ E then µ∗ (F ) = 0. Thus it is enough to show that for any subset E ⊂ X, µ∗ (E) = 0 implies E ∈ M. For any A ⊂ X, using the fact that µ∗ (A ∩ E) = 0, and the ‘increasing’ property of µ∗ µ∗ (A) ≤ µ∗ (A ∩ E) + µ∗ (A ∩ E C ) = µ∗ (A ∩ E C ) ≤ µ∗ (A) shows that these must always be equalities, so E ∈ M (i.e., is µ∗ measurable).  Going back to our primary concern, recall that we constructed the outer measure µ∗ from 0 ≤ u ∈ (C0 (X))0 using (1.11) and (1.12). For the measure whose existence follows from Caratheodory’s theorem to be much use we need Proposition 2.5. If 0 ≤ u ∈ (C0 (X))0 , for X a locally compact metric space, then each open subset of X is µ∗ -measurable for the outer measure defined by (1.11) and (1.12) and µ in (1.11) is its measure. Proof. Let U ⊂ X be open. We only need to prove (2.1) for all A ⊂ X with µ∗ (A) < ∞.2 2Why? Suppose first that A ⊂ X is open and µ∗ (A) < ∞. Then A ∩ U is open, so given  > 0 there exists f ∈ C(X) supp(f ) b A ∩ U with 0 ≤ f ≤ 1 and µ∗ (A ∩ U ) = µ(A ∩ U ) ≤ u(f ) +  . Now, A\ supp(f ) is also open, so we can find g ∈ C(X) , 0 ≤ g ≤ 1 , supp(g) b A\ supp(f ) with µ∗ (A\ supp(f )) = µ(A\ supp(f )) ≤ u(g) +  . Since A\ supp(f ) ⊃ A ∩ U C , 0 ≤ f + g ≤ 1 , supp(f + g) b A , µ(A) ≥ u(f + g) = u(f ) + u(g) > µ∗ (A ∩ U ) + µ∗ (A ∩ U C ) − 2 ≥ µ∗ (A) − 2 using subadditivity of µ∗ . Letting  ↓ 0 we conclude that µ∗ (A) ≤ µ∗ (A ∩ U ) + µ∗ (A ∩ U C ) ≤ µ∗ (A) = µ(A) . This gives (2.1) when A is open. In general, if E ⊂ X and µ∗ (E) < ∞ then given  > 0 there exists A ⊂ X open with µ∗ (E) > µ∗ (A) − . Thus, µ∗ (E) ≥ µ∗ (A ∩ U ) + µ∗ (A ∩ U C ) −  ≥ µ∗ (E ∩ U ) + µ∗ (E ∩ U C ) −  ≥ µ∗ (E) −  . This shows that (2.1) always holds, so U is µ∗ -measurable if it is open. We have already observed that µ(U ) = µ∗ (U ) if U is open.  Thus we have shown that the σ-algebra given by Caratheodory’s theorem contains all open sets. You showed in Problem 3 that the intersection of any collection of σ-algebras on a given set is a σ-algebra. Since P(X) is always a σ-algebra it follows that for any collection E ⊂ P(X) there is always a smallest σ-algebra containing E, namely \ ME = {M ⊃ E ; M is a σ-algebra , M ⊂ P(X)} . The elements of the smallest σ-algebra containing the open sets are called ‘Borel sets’. A measure defined on the σ-algebra of all Borel sets is called a Borel measure. This we have shown: Proposition 2.6. The measure defined by (1.11), (1.12) from 0 ≤ u ∈ (C0 (X))0 by Caratheodory’s theorem is a Borel measure. Proof. This is what Proposition 2.5 says! See how easy proofs are.  We can even continue in the same vein. A Borel measure is said to be outer regular on E ⊂ X if µ(E) = inf {µ(U ) ; U ⊃ E , U open} . (2.6) Thus the measure constructed in Proposition 2.5 is outer regular on all Borel sets! A Borel measure is inner regular on E if µ(E) = sup {µ(K) ; K ⊂ E , K compact} . (2.7) Here we need to know that compact sets are Borel measurable. This is Problem 5. Definition 2.7. A Radon measure (on a metric space) is a Borel measure which is outer regular on all Borel sets, inner regular on open sets and finite on compact sets. Proposition 2.8. The measure defined by (1.11), (1.12) from 0 ≤ u ∈ (C0 (X))0 using Caratheodory’s theorem is a Radon measure. Proof. Suppose K ⊂ X is compact. Let χK be the characteristic function of K , χK = 1 on K , χK = 0 on K C . Suppose f ∈ C0 (X) , supp(f ) b X and f ≥ χK . Set U = {x ∈ X ; f (x) > 1 − } where  > 0 is small. Thus U is open, by the continuity of f and contains K. Moreover, we can choose g ∈ C(X) , supp(g) b U , 0 ≤ g ≤ 1 with g = 1 near3 K. Thus, g ≤ (1 − )−1 f and hence µ∗ (K) ≤ u(g) = (1 − )−1 u(f ) . Letting  ↓ 0, and using the measurability of K, µ(K) ≤ u(f ) ⇒ µ(K) = inf {u(f ) ; f ∈ C(X) , supp(f ) b X , f ≥ χK } . In particular this implies that µ(K) < ∞ if K b X, but is also proves (2.7).  Let me now review a little of what we have done. We used the positive functional u to define an outer measure µ∗ , hence a measure µ and then checked the properties of the latter. This is a pretty nice scheme; getting ahead of myself a little, let me suggest that we try it on something else. 3Meaning in a neighborhood of K. Let us say that Q ⊂ Rn is ‘rectangular’ if it is a product of finite intervals (open, closed or half-open) n Y (2.8) Q= (or[ai , bi ]or) ai ≤ bi i=1 we all agree on its standard volume: n Y (2.9) v(Q) = (bi − ai ) ∈ [0, ∞) . i=1 Clearly if we have two such sets, Q1 ⊂ Q2 , then v(Q1 ) ≤ v(Q2 ). Let us try to define an outer measure on subsets of Rn by (∞ ) ∞ X [ ∗ (2.10) v (A) = inf v(Qi ) ; A ⊂ Qi , Qi rectangular . i=1 i=1 We want to show that (2.10) does define an outer measure. This is pretty easy; certainly v(∅) = 0. Similarly if {Ai }∞ i=1 are (disjoint) sets and {Qij }∞ is a covering of A by open rectangles then all the Qij i i=1 S together cover A = i Ai and XX v ∗ (A) ≤ v(Qij ) i ⇒ v ∗ (A) ≤ j X v ∗ (Ai ) . i So we have an outer measure. We also want Lemma 2.9. If Q is rectangular then v ∗ (Q) = v(Q). Assuming this, the measure defined from v ∗ using Caratheodory’s theorem is called Lebesgue measure. Proposition 2.10. Lebesgue measure is a Borel measure. To prove this we just need to show that (open) rectangular sets are v -measurable. ∗