2.5 Outer Measure and Measurable sets. Note The results of this section concern any given outer measure ¸. If an outer measure ¸ on a set X were a measure then it would be additive. In particular, given any two sets A; B ⊆ X we have that A \ B and A \ Bc are disjoint with (A \ B) [ (A \ Bc) = A and so we would have ¸(A) = ¸(A \ B) + ¸(A \ Bc): We will see later that this does not necessarily hold for all A and B but it does lead to the following definition. Definition Let ¸ be an outer measure on a set X. Then E ⊆ X is said to be measurable with respect to ¸ (or ¸-measurable) if ¸(A) = ¸(A \ E) + ¸(A \ Ec) for all A ⊆ X. (7) (This can be read as saying that we take each and every possible “test set”, A, look at the measures of the parts of A that fall within and without E, and check whether these measures add up to that of A.) Since ¸ is subadditive we have ¸(A) · ¸(A\E)+¸(A\Ec) so, in checking measurability, we need only verify that ¸(A) ¸ ¸(A \ E) + ¸(A \ Ec) for all A ⊆ X. (8) Let M = M(¸) denote the collection of ¸-measurable sets. Theorem 2.6 M is a field. Proof Trivially Á and X are in M. Take any E1;E2 2 M and any test set A ⊆ X. Then c ¸(A) = ¸(A \ E1) + ¸(A \ E1): c Now apply the definition of measurability for E2 with the test set A \ E1 to get c c c c ¸(A \ E1) = ¸((A \ E1) \ E2) + ¸((A \ E1) \ E2) c c = ¸(A \ E1 \ E2) + ¸(A \ (E1 [ E2) ): Combining c c ¸(A) = ¸(A \ E1) + ¸(A \ E1 \ E2) + ¸(A \ (E1 [ E2) ): (9) 1 We hope to use the subadditivity of ¸ on the first two term on the right hand side of (9). For the sets there we have c c (A \ E1) [ (A \ E1 \ E2) = A \ (E1 [ (E1 \ E2)) c = A \ ((E1 [ E1) \ (E1 [ E2)) = A \ (X \ (E1 [ E2)) = A \ (E1 [ E2): So c ¸(A \ E1) + ¸(A \ E1 \ E2) ¸ ¸(A \ (E1 [ E2)): Substituting in (9) gives c ¸(A) ¸ ¸(A \ (E1 [ E2)) + ¸(A \ (E1 [ E2) ): So we have verified (8) for E1 [ E2, that is, E1 [ E2 2 M. Observe that the definition of ¸-measurable sets is symmetric in that E 2 M if, and only if, Ec 2 M. Thus c c c E1 n E2 = E1 \ E2 = (E1 [ E2) 2 M. Hence M is a field. ¥ Proposition 2.7 If G; F 2 M(¸) are disjoint then ¸(A \ (G [ F )) = ¸(A \ G) + ¸(A \ F ) for all A ⊆ X: Proof Let A ⊆ X be given. Apply the definition of ¸-measurability to G with the test set A \ (G [ F ). Then ¸(A \ (G [ F )) = ¸((A \ (G [ F )) \ G) +¸((A \ (G [ F )) \ Gc): (10) Yet (G [ F ) \ G = (G \ G) [ (F \ G) = G [ (F \ G) = G 2 since F \ G ⊆ G. Also (G [ F ) \ Gc = (G \ Gc) [ (F \ Gc) = Á [ F = F; because F and G disjoint means F ⊆ Gc and so F \ Gc = F . Thus (9) becomes ¸(A \ (G [ F )) = ¸(A \ G) + ¸(A \ F ): ¥ Using induction it is possible to prove the following. Corollary 2.8 For all n ¸ 1 if fFig1·i·n is a finite collection of disjoint sets from M(¸) then à ! [n Xn ¸ A \ Fi = ¸ (A \ Fi) for all A ⊆ X. i=1 i=1 Proof Left to students. ¥ Corollary 2.9 If fFigi¸1 is a countable collection of disjoint sets from M(¸) then à ! [1 X1 ¸ A \ Fi = ¸ (A \ Fi) for all A ⊆ X. i=1 i=1 Proof S1 Sn For any n ¸ 1 we have A \ i=1 Fi ¶ A \ i=1 Fi and so by monotonicity we have à ! à ! [1 [n ¸ A \ Fi ¸ ¸ A \ Fi i=1 i=1 Xn = ¸ (A \ Fi) i=1 by Corollary 2.8. Let n ! 1 to get à ! [1 X1 ¸ A \ Fi ¸ ¸ (A \ Fi) : i=1 i=1 The reverse inequality follows from subadditivity. ¥ 3 Theorem 2.10 M(¸) is a σ-field and ¸ restricted to M(¸) is a measure. Proof Let fEigi¸1 be a countable collection from M. They might not be disjoint so define F1 = E1 and i[¡1 Fi = Ei n Ej j=1 for all i > 1. The Fi are disjoint and Fi 2 M since M is a field. Let Sm S1 S1 Gm = j=1 Fj and G = j=1 Fj = i=1 Ei. Then for any A ⊆ X and for any n ¸ 1 we have c ¸(A) = ¸(A \ Gn) + ¸(A \ Gn) since Gn 2 M Xn c = ¸(A \ Fi) + ¸(A \ Gn) by Corollary 2.8 i=1 Xn c c c ¸ ¸(A \ Fi) + ¸(A \ G ) since G ⊆ Gn. i=1 True for all n means that X1 c ¸(A) ¸ ¸(A \ Fi) + ¸(A \ G ) i=1à ! [1 c = ¸ A \ Fi + ¸(A \ G ) i=1 by Corollary 2.8, since the Fi are disjoint, = ¸ (A \ G) + ¸(A \ Gc): True for all A ⊆ X means that G 2 M(¸). Choosing A = X in Corollary 2.9 shows that ¸ is σ-additive on M(¸). Hence ¸ is a measure on M(¸). ¥ ¤ Example y With the Lebesgue-Stieltjes outer measure ¹F of example 8 we ¤ can now form the σ-field M(¹F ). This is known as the collection of Lebesgue- Stieltjes measurable sets and is denoted by LF . If F (x) = x, it is simply known as the collection of Lebesgue measurable sets and is denoted by L. We now specialise to those outer measures constructed, as in (4), from measures defined in a ring. 4 S1 Theorem 2.11 Let R be a ring of sets in X such that X = i=1 Ei for ¤ some Ei 2 R. Let ¹ be a measure on R and let ¹ be the outer measure on X constructed from ¹ as in (4). Then (i) the elements of R are ¹¤-measurable sets, (ii) ¹¤ = ¹ on R. Proof (i) Let E 2 R and a test set A ⊆ X be given. If ¹¤(A) = +1 then (7) is trivially satisfied so assume that ¹¤(A) < +1. Let " > 0 be given. By theS definition (4) there exists a countable collection fEigi¸1 ⊆ R such that A ⊆ i¸1 Ei and X1 ¤ ¤ ¹ (A) · ¹(Ei) < ¹ (A) + ": i=1 Yet ¹ is a measure on R and Ei;E 2 R so c ¹(Ei) = ¹(Ei \ E) + ¹(Ei \ E ): Combining we see X1 ¤ c ¹ (A) + " > (¹(Ei \ E) + ¹(Ei \ E )) i=1 ¸ ¹¤(A \ E) + ¹¤(A \ Ec); c c since fEi \ Egi¸1 and fEi \ E gi¸1 are covers for A \ E and A \ E respec- tively. True for all " > 0 means that (7) is satisfied and so E 2 M and thus R ⊆ M. (ii) Let E 2 R be given. Then since E is a cover from R for E we have that X ¤ ¹ (E) = inf ¹(Ei) · ¹(E). all covers Take any other cover fEigi¸1 of E. As in Theorem 2.10 replace the Ei S1 S1 by disjoint sets Fi ⊆ Ei;Fi 2 R and where i=1 Fi = i=1 Ei. Then 5 à ! [1 [1 ¹(E) = ¹ E \ Fi since E ⊆ Fi, à i=1 ! i=1 [1 = ¹ (E \ Fi) i=1 X1 = ¹ (E \ Fi) since ¹ is additive on R, i=1 X1 X1 · ¹ (Fi) · ¹ (Ei) i=1 i=1 since E \ Fi ⊆ Fi ⊆ Ei and ¹ is monotonic on R. So ¹(E) is a lower bound for the sums for which ¹¤(E) is the greatest lower bound and thus ¹(E) · ¹¤(E). Hence ¹(E) = ¹¤(E). ¥ We now see that it is reasonable to say that ¹¤ extends ¹. Further, let A be a collection of subsets of X: Definition We say that the extended real-valued function Á : A! R¤ is σ-finite ifS for all A 2 A there exists a countable collection fAngn¸1 ⊆ A such that A ⊆ n¸1 An and jÁ(An)j < 1 for all n ¸ 1. Then it can be shown that Theorem 2.12 If, in addition to the conditions of Theorem 2.11, ¹ is σ-finite on R then the extension to M is unique and is also σ-finite. Proof not given here. ¥ Example 8y Recall that ¹F was first defined on P and then extended to E ¤ in Corollary 2.3. In the last example the σ-field M (¹F ), known as LF ; was ¤ constructed. Now Theorem 2.11 implies that E ⊆ LF and ¹F = ¹F on E. So ¤ it is reasonable to write ¹F simply as ¹F on LF , called the Lebesgue-Stieltjes measure. If F (x) = x; we write ¹F simply as ¹, called the Lebesgue measure on R. Notey LF ¶E¶P so LF is a σ-field containing P. But B, the Borel sets of R, is the smallest σ-field containing P. Hence LF ¶B, true for all distribution functions F . 6.