By Measure, integration gy miuidaentcra Ac•Ka6pp 2010 72 pagcs MEASURE, INTEGRATION PROBABILITY Ivan F Wilde Mathematics Department Kings College London Chapter 1 0-algebras and Borel functions Preliminary discussion • Suppose that X is a continuous random variable. Then Prob(X – s) = O for any s R. However, the event {X R} has probability one and is the disjoint union of the events {X = s} for s E {X E PACE 1 or72 to View nut*ge Each event on the rig ility zero, so the probabilities of the events on the right hand side do not «add up» to that ofthe left hand side. We wish to understand this. •
Suppose that X is a random variable on a sample space n, and suppose that X takes values XI , x2 , . xi Then EX = Xi prob(X = Xi)Xi Prob(Ai In particular, for 1 Ai (w) = I . Put Chapter 1 • One must (sometimes) ask which subsets of a sample space are deemed to be events. Can one take all subsets of the sample space to be events? The answer is sometimes yes and sometimes no. For example, in the case when the probability of an event within a bounded region of R3 is required to be proportional to the volume associated with the event, then one naturally asks whether every subset of (a bounded region) of R3 actually has a olume.
That this is not so is demonstrated by the Banach-Tarski theorem. (This says that a ball of unit radius in R3 can be cut up into a finite number of pieces which can then be reassembled to form a ball of radius 2. The meaning of ‘Volume» for these pieces is not clear. ) We must be precise about the concept of «event». In the «modern» (Kolmogorov) theory of probability, this is formulated in terms of 0-algebras. Definition 1. 1 . A collection E of subsets of a non-empty set X is called a 0-algebra if (i) X E, (ii) if A EE, (iii) if An E E forn 1, 2, . . , then An
The sets in E are called measurable sets, and (K E) is called a «measurable space». Remarks 1. 2. 1. Since xc , it follows that 2. For anYA1 ,A2, Then . GE, put An+l = – n xc, it follows that e E. 2. For any Al ,A2,. . , An e E, put An+l = An+2 – • = Z. Then we see that Al u • • • U An – Ak E E, by (1) above, and (iii). 1<21 3. LetA1 ,A2, • GE. Then since An n=l An An, we see that Ifwe take An+l = An+2 = • • • = X, then we get Al n • • • n An EE. 4. Let A, B EE. Then and so E. proposition 1. 3. Let { Ea } be an arbitrary collection of 0-algebras ofX.
Then a Ea is a 0-algebra. Proof. We check the requirements. (i) X E Ea for all a and so X aza (ii) Suppose that A E aza . Then A E Ea for all a and so Ac E Ea for all a, that is Ac a Ea . Department of Mathematics 0-algebras and Borel functions (iii) Let Al , belong to a Ea . Then, for each a, An for all n and so n An E Ea . Hence n An E aza . The result follows. Let C be any collection of subsets of X. Then certainly C is contained in the o-algebra consisting of all subsets ofX. Ifwe set MC) = where the intersection is o F of all those o-aleebras the 0-algebra generated by C.
Definition 1. 4. Let C denote the collection of open subsets of R. Then E(C) is called the Sorel 0-algebra of R, usually written B(R). The elements cf B(R) are called Borel sets. Similarly, one defines B(Rn ) as the 0-algebra generated by the open subsets of Rn . Proposition 1. 5. The following subsets of R belong to B(R): (i) (a, b) for any a < b; a) for any a E R; (iii) (a, r") for any a E R; (iv) [a, b] for any a b; (v) a] for any a R; (vi) [a, m) for any a R; (vii) (a, b] for any a < b; (viii) [a, b) for any a b; (ix) any closed subset of R.
Proof. Each of the sets in (i), (ii) or (iii) is open and so belongs to B(R) by construction. (iv) [a, b] = la —n, b+lnln Kings College London 4 (vii) (a, b] (viii) [a, b)- Let E(closed) and E(compact) denote the 0-algebras of subsets of R generated, respectively, by the closed sets and the compact subsets in R. Then E(closed) = E(compact) = B(R). Proof. Every closed subset of R belongs to the 0-algebra B(R). But, by definition, E(closed) is the smallest 0-algebra containing the closed sets, so we must have E(closed) C B(R).
On the other hand, every open set is the complement of a closed set and so belongs to E(closed). gy definition, B(R) is the a-algebra generated by the open sets of R and so we have B(R) E(closed). It follows that B(R) E(closed). Next, we note that since every compact set in R is closed, it follows that E(compact) E(closed). However, any closed set F can be written as the (countable) union Each [-n, n] n F is closed and bounded and so is compact. It follows that F e E(compact) and therefore E(closed) ç E(compact). Hence result. proposition 1. 7.
Let Cl , , C9 denote the collections of subsets of R as given in Proposition 1. 5. Then E(Ci ) = B(R) for each i = 1, 2, . , 9. Proof- Since Ci C B(R), we have E(Ci ) B(R), 1 9. we show that E(compact) ÇE(Ci) WhiCh completes the proof, by Proposition 1. . To show this, we first observe that each E(Ci ) contains al s OF n observe that each E(Ci ) contalns all inten•als (a, b) with a < b. (For example in (v). a] e E(C5 ) implies (by taking complements) that (a, E(C5 ) for any a E R. But then it follows that (a, b] = (a, ) and so (a, b) b- b-a 2n Now let K c R be any given compact set.
For each x R and n N, 1 1 let In (x) be the interval (x -n , x + n For each fixed n, the collection { In (x) : x K) is an open cover of K and so has a finite subcover, Kc In (XI ) u u In (xm(n) ) E Jn for suitable points XI , xm(n) in K. Evidently, Jn E(Ci We claim that K – m Jn . Clearly, KC coJn . For the converse, (n) let x m Jn . Then x e Jn for all n. Hence x belongs to some In n=l that is, there is yn =xjEK such that x-yn I < 1/ and, since K is closed, it follows n. We see that yn x as n that x E K whlch proves the clam.
Therefore K – and so n=l B(R) = E(compact) E(Ci ) and the proof is complete. Definition 1. 8. Let (X, E) be a measurable space and f: X Ra given function. f is Said to be Borel measurable if f —1 6 OF n measurable space and f: X Ra given function. f is said to be Borel measurable iff-l (G) E E for each G open in R Proposition . 9. The function f: X R is Borel measurable if and only if f —1 (A) EX for B(R). Proof. Iff-l (A) EX for each A E B(R), then certainly f —1 (G) E E for each open set G in R (because such G belongs to B(R)). Conversely, suppose that f —1 (G) for any open set G in R.
Let S denote the collection of subsets of R given by S – R : f —1 (É) Then S is a 0-algebra. To see this, we note the following. (i) f -1 (R) = X E E, so R ES. (i)) f -1 (R E) -Xf-l so ifEêSthen so is E. (iii) If El , a, . belongtos, then f-l (El (El f-l (E2)U which belongs to E and so El U E2 U • • • e S. This shows that S is, indeed, a 0-algebra, as claimed. But S contains all open sets, by hypothesis, and therefore B(R) S. Hence, for any A G B(R), f —1 (A) This completes the proof. Remark 1 . 10. Note that S need not be equal to 3(R).
For example, ifX is the 0-algebra of all subsets of X then every function f: X R is gorel measurable and f -1 (E) EE for any subset E in R whatsoever. As another example, suppose that f is constant. Then f —1 (E) is either equal in R whatsoever. As another example, suppose that f is constant. Then f -1 (E) is either equal to X or else is empty depending on hether E contains the value assumed by f or not. In any event, f —1 (E) E E, whatever E is. King’s College London 6 We can improve somewhat on the previous proposition, Still using the same idea. roposition 1 Let C be a collection of subsets of R such that E(C) = B(R) and let f: X — R. Then fis Borel measurable if and only iff —1 (A) e E for all A C. Proof. Suppose that f is Borel measurable. Since C B(R), it follows by Proposition 1 that f —1 (A) E for any A C. Now suppose that f —1 (A) for allAE C. AS before, let us set (E) EE}. Then S is a 0-algebra which contains C and so we have B(R) E(C) ç S. It follows that f -1 (A) EE for any A E B(R). Remark 1. 12. This means that we can take C to be any of the collections of sets indicated in proposition 1. 5.
For example, we can say that f is Bore’ measurable if and only iff -1 a]) E E for each a G R. We can choose any convenient collection to work with. Proposition 1 . 13. Let f: X — R be Borel measurable and g: R — R continuous. hen g • f: X R is gore’ measurable. Proof. Let G be any open set in R. heng R R continuous. Then g 0 f: X R is Borel measurable. Proof. Let G be any open set In R. Then g -1 (G) is open in R and so (g )—l (G) f —1 (g —1 (G)) belongs to E. Proposition 1. 14. Let f: X R and g: X R be Borel functions. Then the set E = {x f (x) < g(x)} is measurable. Proof.
For each rational number r G Q, let Er = {x f (x) < r < g(x) Then Er = {x: f (x) < r} n {x: r < g(x) ) which is the intersection of two measurable sets in X and so is itself measurable. Finally, we note that reQ Er which is a countable union of measurable sets and so is measurable. Proposition 1. 15. Let(X, E) be a measurable space and letf. X R and g: X R be Borel functions. Then (i) af+ b is a Borel function for any a, b R; (ii) f + g is a Borel function; (iii) If la is a Borel function for any a O; (iv) iff never vanishes, then 1/f is a Borel function; (v) f g is a Borel function; (vi) If l, max{ f, g} and min{ f, g} are Borel functions.
Proof. We shall consider each statement one by one. (i) For any c É R, {x E X: (af (c b)/a a DtlCl x: f (x) (c — b)/a > (c- b)/a ha < O- ü X, a = O and c > b a = O and c < b. In any event, the left hand side belongs to E, which proves (i). (ii) LetcE R. Then {x: f (x) + g(x) > c} = {x: f (x) > -g(x) + c}. But -g c is a Bore’ function by (i) and so the right hand side belongs to E by Proposition 1. 14. (iii) The function t la is a continuous function and so f la is a Borel function by Proposition 1. 13. Alternatively, we note that for c > 0 {x: If {x : —cl/ Forc < O, the left hand side Z. (iv) If c > 0, then {x: — {x • f (x) > c f (x) ) E E, using (i). If c < O, then {x: 1/f (x) c) = {x: 1 àcf e E, again by (i). 8 (v) This follows from the identity g)24 together with (i), (ji) and (iii). (vi) f I is a Borel function by (iii) with a = 1. Now 1 max{ f, g = 1 (f (x) + g(x)) +2 If(x) - 2 and 1 (f (X) + g(X)) -2 If(X) - . 2 The result now follows from (i), (ii) and (iii). Theorem 1. E) be a measurable space and 16. Let (X,
