Math212a1411 Lebesgue Measure

Outline Lebesgue outer measure. Lebesgue inner measure. Lebesgue’s definition of measurability. Caratheodory’s definition of measurability. Countable additivity. σ-fields, measures, and outer measures. The Borel-Cantelli lemmas

Math212a1411 Lebesgue measure.

Shlomo Sternberg

October 14, 2014

Shlomo Sternberg Math212a1411 Lebesgue measure. Outline Lebesgue outer measure. Lebesgue inner measure. Lebesgue’s definition of measurability. Caratheodory’s definition of measurability. Countable additivity. σ-fields, measures, and outer measures. The Borel-Cantelli lemmas

Reminder

No class this Thursday

Today’s lecture will be devoted to Lebesgue measure, a creation of Henri Lebesgue, in his thesis, one of the most famous theses in the history of mathematics.

Henri Léon Lebesgue

Born: 28 June 1875 in Beauvais, Oise, Picardie, France Died: 26 July 1941 in Paris, France

In today’s lecture we will discuss the concept of measurability of a subset of R. We will begin with Lebesgue’s (1902) definition of measurability, which is easy to understand and intuitive. We will then give Caratheodory’s (1914) definition of measurabiity which is highly non-intuitive but has great technical advantage. For subsets of R these two definitions are equivalent (as we shall prove). But the Caratheodory definition extends to many much more general situations. In particular, the Caratheodory definition will prove useful for us later, when we study Hausdorff measures.

The /2n trick.

An argument which will recur several times is:

∞ X 1 = 1 2n 1 and so ∞ X = . 2n 1 We will call this the “/2n trick” and not spell it out every time we use it.

1 Lebesgue outer measure.

2 Lebesgue inner measure.

3 Lebesgue’s deﬁnition of measurability.

4 Caratheodory’s deﬁnition of measurability.

5 Countable additivity.

6 σ-ﬁelds, measures, and outer measures.

7 The Borel-Cantelli lemmas

The deﬁnition of Lebesgue outer measure.

For any subset A ⊂ R we deﬁne its Lebesgue outer measure by

∗ X [ m (A) := inf `(In): In are intervals with A ⊂ In. (1)

Here the length `(I ) of any interval I = [a, b] is b − a with the same definition for half open intervals (a, b] or [a, b), or open intervals. Of course if a = −∞ and b is finite or +∞, or if a is finite and b = +∞, or if a = −∞ and b = ∞ the length is infinite.

It doesn’t matter if the intervals are open, half open or closed.

∗ X [ m (A) := inf `(In): In are intervals with A ⊂ In. (1)

The inﬁmum in (1) is taken over all covers of A by intervals. By n the /2 trick, i.e. by replacing each Ij = [aj , bj ] by j+1 j+1 (aj − /2 , bj + /2 ) we may assume that the inﬁmum is taken over open intervals. (Equally well, we could use half open intervals of the form [a, b), for example.).

Monotonicity, subadditivity.

If A ⊂ B then m∗(A) ≤ m∗(B) since any cover of B by intervals is a cover of A. For any two sets A and B we clearly have

m∗(A ∪ B) ≤ m∗(A) + m∗(B).

Sets of measure zero don’t matter.

A set Z is said to be of (Lebesgue) measure zero it its Lebesgue outer measure is zero, i.e. if it can be covered by a countable union of (open) intervals whose total length can be made as small as we like. If Z is any set of measure zero, then m∗(A ∪ Z) = m∗(A).

The outer measure of a ﬁnite interval is its length.

If A = [a, b] is an interval, then we can cover it by itself, so

m∗([a, b]) ≤ b − a,

and hence the same is true for (a, b], [a, b), or (a, b). If the interval is infinite, it clearly can not be covered by a set of intervals whose total length is finite, since if we lined them up with end points touching they could not cover an infinite interval. We claim that

m∗(I ) = `(I ) (2) if I is a ﬁnite interval.

Proof.

We may assume that I = [c, d] is a closed interval by what we have already said, and that the minimization in (1) is with respect to a cover by open intervals. So what we must show is that if [ [c, d] ⊂ (ai , bi ) i then X d − c ≤ (bi − ai ). i We ﬁrst apply Heine-Borel to replace the countable cover by a ﬁnite cover. (This only decreases the right hand side of preceding inequality.) So let n be the number of elements in the cover.

We need to prove that if

n n [ X [c, d] ⊂ (ai , bi ) then d − c ≤ (bi − ai ). i=1 i=1

We do this this by induction on n. If n = 1 then a1 < c and b1 > d so clearly b1 − a1 > d − c. Suppose that n ≥ 2 and we know the result for all covers (of all intervals [c, d] ) with at most n − 1 intervals in the cover. If some interval (ai , bi ) is disjoint from [c, d] we may eliminate it from the cover, and then we are in the case of n − 1 intervals. So we may assume that every (ai , bi ) has non-empty intersection with [c, d].

Among the the intervals (ai , bi ) there will be one for which ai takes on the minimum possible value. By relabeling, we may assume that this is (a1, b1). Since c is covered, we must have a1 < c. If b1 > d then (a1, b1) covers [c, d] and there is nothing further to do. So assume b1 ≤ d. We must have b1 > c since (a1, b1) ∩ [c, d] 6= ∅. Since b1 ∈ [c, d], at least one of the intervals (ai , bi ), i > 1 contains the point b1. By relabeling, we may assume that it is (a2, b2). But now we have a cover of [c, d] by n − 1 intervals:

n [ [c, d] ⊂ (a1, b2) ∪ (ai , bi ). i=3 Pn So by induction d − c ≤ (b2 − a1) + i=3(bi − ai ). But b2 − a1 ≤ (b2 − a2) + (b1 − a1) since a2 < b1.

We can use small intervals.

We repeat that the intervals used in (1) could be taken as open, closed or half open without changing the definition. If we take them all to be half open, of the form Ii = [ai , bi ), we can write each Ii as a disjoint union of finite or countably many intervals each of length < . So it makes no difference to the definition if we also require the `(Ii ) < (3) in (1). We will see that when we pass to other types of measures this will make a difference.

Summary of where we are so far.

We have veriﬁed, or can easily verify the following properties: 1 m∗(∅) = 0.

2 A ⊂ B ⇒ m∗(A) ≤ m∗(B). 3 By the /2n trick, for any ﬁnite or countable union we have

∗ [ X ∗ m ( Ai ) ≤ m (Ai ). i i 4 If dist (A, B) > 0 then m∗(A ∪ B) = m∗(A) + m∗(B).

5 m∗(A) = inf{m∗(U): U ⊃ A, U open}. 6 For an interval m∗(I ) = `(I ).

The only items that we have not done already are items 4 and 5. But these are immediate: for 4 we may choose the intervals in (1) all to have length < where 2 < dist (A, B) so that there is no overlap. As for item 5, we know from 2 that m∗(A) ≤ m∗(U) for any set U ⊃ A, in particular for any open set U which contains A. We must prove the reverse inequality: if m∗(A) = ∞ this is trivial. Otherwise, we may take the intervals in (1) to be open and then the union on the right is an open set whose Lebesgue outer measure is less than m∗(A) + δ for any δ > 0 if we choose a close enough approximation to the inﬁmum.

Lebesgue outer measure on Rn.

n All the above works for R instead of R if we replace the word “interval” by “rectangle”, meaning a rectangular parallelepiped, i.e a set which is a product of one dimensional intervals. We also replace length by volume (or area in two dimensions). What is needed is the following:

Lemma Let C be a ﬁnite non-overlapping collection of closed rectangles all contained in the closed rectangle J. Then X vol J ≥ vol I . I ∈C If C is any ﬁnite collection of rectangles such that [ J ⊂ I I ∈C then X vol J ≤ vol (I ). I ∈C

This lemma occurs on page 1 of Stroock, A concise introduction to the theory of integration together with its proof. I will take this for granted. In the next few slides I will talk as if we are in R, but n everything goes through unchanged if R is replaced by R . In fact, once we will have developed enough abstract theory, we will not need this lemma. So for those of you who are purists and do not want to go through the lemma in Stroock, just stick to R.

The deﬁnition of Lebesgue inner measure.

Item 5. in our list said that the Lebesgue outer measure of any set is obtained by approximating it from the outside by open sets. The Lebesgue inner measure is deﬁned as

∗ m∗(A) = sup{m (K): K ⊂ A, K compact }. (4)

Clearly ∗ m∗(A) ≤ m (A) since m∗(K) ≤ m∗(A) for any K ⊂ A. We also have

The Lebesgue inner measure of an interval.

Proposition. For any interval I we have

m∗(I ) = `(I ). (5)

Proof. If `(I ) = ∞ the result is obvious. So we may assume that I is a ﬁnite interval which we may assume to be open, I = (a, b). If K ⊂ I is compact, then I is a cover of K and hence from the ∗ deﬁnition of outer measure m (K) ≤ `(I ). So m∗(I ) ≤ `(I ). On 1 the other hand, for any > 0, < 2 (b − a) the interval [a + , b − ] is compact and ∗ m ([a − , a + ]) = b − a − 2 ≤ m∗(I ). Letting → 0 proves the proposition.

Lebesgue’s deﬁnition of measurability for sets of ﬁnite outer measure.

A set A with m∗(A) < ∞ is said to measurable in the sense of Lebesgue if ∗ m∗(A) = m (A). (6) If A is measurable in the sense of Lebesgue, we write

∗ m(A) = m∗(A) = m (A). (7)

∗ If K is a compact set, then m∗(K) = m (K) since K is a compact set contained in itself. Hence all compact sets are measurable in the sense of Lebesgue. If I is a bounded interval, then I is measurable in the sense of Lebesgue by the preceding Proposition.

Lebesgue’s deﬁnition of measurability for sets of inﬁnite outer measure.

If m∗(A) = ∞, we say that A is measurable in the sense of Lebesgue if all of the sets A ∩ [−n, n] are measurable.

Countable additivity.

Theorem S If A = Ai is a (ﬁnite or) countable disjoint union of sets which are measurable in the sense of Lebesgue, then A is measurable in the sense of Lebesgue and X m(A) = m(Ai ). i

In the proof we may assume that m(A) < ∞ - otherwise apply the result to A ∩ [−n, n] and Ai ∩ [−n, n] for each n.

We have ∗ X ∗ X m (A) ≤ m (An) = m(An). n n

Let > 0, and for each n choose compact Kn ⊂ An with m∗(K ) ≥ m (A ) − = m(A ) − n ∗ n 2n n 2n

which we can do since An is measurable in the sense of Lebesgue. The sets Kn are pairwise disjoint, hence, being compact, at positive distances from one another. Hence

∗ ∗ ∗ m (K1 ∪ · · · ∪ Kn) = m (K1) + ··· + m (Kn)

and K1 ∪ · · · ∪ Kn is compact and contained in A.

m∗(K ) ≥ m (A ) − = m(A ) − n ∗ n 2n n 2n ∗ ∗ ∗ m (K1 ∪ · · · ∪ Kn) = m (K1) + ··· + m (Kn)

and K1 ∪ · · · ∪ Kn is compact and contained in A. Hence

∗ ∗ m∗(A) ≥ m (K1) + ··· + m (Kn), and since this is true for all n we have X m∗(A) ≥ m(An) − . n Since this is true for all > 0 we get X m∗(A) ≥ m(An).

∗ But then m∗(A) ≥ m (A) and so they are equal, so A is P measurable in the sense of Lebesgue, and m(A) = m(Ai ). Shlomo Sternberg Math212a1411 Lebesgue measure. Outline Lebesgue outer measure. Lebesgue inner measure. Lebesgue’s definition of measurability. Caratheodory’s definition of measurability. Countable additivity. σ-fields, measures, and outer measures. The Borel-Cantelli lemmas

Open sets are measurable in the sense of Lebesgue.

Proof. Any open set O can be written as the countable union of open intervals Ii , and n−1 [ Jn := In \ Ii i=1 is a disjoint collection of intervals (some open, some closed, some half open) and O is the disjont union of the Jn. So every open set is a disjoint union of intervals hence measurable in the sense of Lebesgue.

Closed sets are measurable in the sense of Lebesgue.

This requires a bit more work: If F is closed, and m∗(F ) = ∞, then F ∩ [−n, n] is compact, and so F is measurable in the sense of Lebesgue. Suppose that

m∗(F ) < ∞.

For any > 0 consider the sets G := [−1 + , 1 − ] ∩ F 1, 22 22 G := ([−2 + , −1] ∩ F ) ∪ ([1, 2 − ] ∩ F ) 2, 23 23 G := ([−3 + , −2] ∩ F ) ∪ ([2, 3 − ] ∩ F ) 3, 24 24 . .

and set [ G := Gi,. i

The Gi, are all compact, and hence measurable in the sense of Lebesgue, and the union in the deﬁnition of G is disjoint, so G is measurable in the sense of Lebesgue.

The Gi, are all compact, and hence measurable in the sense of Lebesgue, and the union in the deﬁnition of G is disjoint, so is measurable in the sense of Lebesgue. Furthermore, the sum of the lengths of the “gaps” between the intervals that went into the deﬁnition of the Gi, is . So

∗ ∗ ∗ X m(G)+ = m (G)+ ≥ m (F ) ≥ m (G) = m(G) = m(Gi,). i In particular, the sum on the right converges, and hence by considering a ﬁnite number of terms, we will have a ﬁnite sum whose value is at least m(G) − . The corresponding union of sets will be a compact set K contained in F with

∗ m(K) ≥ m (F ) − 2.

Hence all closed sets are measurable in the sense of Lebesgue.

The inner-outer characterization of measurability.

Theorem A is measurable in the sense of Lebesgue if and only if for every > 0 there is an open set U ⊃ A and a closed set F ⊂ A such that

m(U \ F ) < .

Proof in one direction. Suppose that A is measurable in the sense of Lebesgue with m(A) < ∞. Then there is an open set U ⊃ A with m(U) < m∗(A) + /2 = m(A) + /2, and there is a compact set F ⊂ A with m(F ) ≥ m∗(A) − /2 = m(A) − /2. Since U \ F is open, it is measurable in the sense of Lebesgue, and so is F as it is compact. Also F and U \ F are disjoint. Hence m(U \ F ) = m(U) − m(F ) < m(A) + − m(A) − = . 2 2 Shlomo Sternberg Math212a1411 Lebesgue measure. Outline Lebesgue outer measure. Lebesgue inner measure. Lebesgue’s definition of measurability. Caratheodory’s definition of measurability. Countable additivity. σ-fields, measures, and outer measures. The Borel-Cantelli lemmas

If A is measurable in the sense of Lebesgue, and m(A) = ∞, we can apply the above to A ∩ I where I is any compact interval, in particular to the interval [−n, n]. So there exist open sets Un ⊃ A ∩ [−n, n] and closed sets Fn ⊂ A ∩ [−n, n] with n S m(Un/Fn) < /2 . Let U := Un and [ F := (Fn ∩ ([−n, −n + 1] ∪ [n − 1, n])) . A convergent sequence in F must eventually lie in an interval of length one and hence to at most the union of two entries in the union deﬁning F , so F is closed. We have U ⊃ A ⊃ F and [ U/F ⊂ (Un/Fn). Indeed, any p ∈ U must belong to some ([−n, −n + 1] ∪ [n − 1, n]), and so if it does not belong to F then it does not belong to Fn. So X m(U/F ) ≤ m(Un/Fn) < .

Proof in the other direction.

Suppose that for each , there exist U ⊃ A ⊃ F with m(U \ F ) < . Suppose that m∗(A) < ∞. Then m(F ) < ∞ and m(U) ≤ m(U \ F ) + m(F ) < + m(F ) < ∞. Then

∗ m (A) ≤ m(U) < m(F ) + = m∗(F ) + ≤ m∗(A) + .

∗ Since this is true for every > 0 we conclude that m∗(A) ≥ m (A) so they are equal and A is measurable in the sense of Lebesgue. If m∗(A) = ∞, we have U ∩ (−n − , n + ) ⊃ A ∩ [−n, n] ⊃ F ∩ [−n, n] and

m((U ∩ (−n − , n + ) \ (F ∩ [−n, n]) < 2 + = 3

so we can proceed as before to conclude that ∗ m∗(A ∩ [−n, n]) = m (A ∩ [−n, n]). Shlomo Sternberg Math212a1411 Lebesgue measure. Outline Lebesgue outer measure. Lebesgue inner measure. Lebesgue’s definition of measurability. Caratheodory’s definition of measurability. Countable additivity. σ-fields, measures, and outer measures. The Borel-Cantelli lemmas

Consequences of the theorem.

Proposition. If A is measurable in the sense of Lebesgue, so is its complement c A = R \ A.

Proof. Indeed, if F ⊂ A ⊂ U with F closed and U open, then F c ⊃ Ac ⊃ Uc with F c open and Uc closed. Furthermore, F c \ Uc = U \ F so if A satisﬁes the condition of the theorem so does Ac .

More consequences.

Proposiition. If A and B are measurable in the sense of Lebesgue so is A ∩ B

Proof.

For > 0 choose UA ⊃ A ⊃ FA and UB ⊃ B ⊃ FB with m(UA \ FA) < /2 and m(UB \ FB ) < /2. Then

(UA ∩ UB ) ⊃ (A ∩ B) ⊃ (FA ∩ FB )

and (UA ∩ UB ) \ (FA ∩ FB ) ⊂ (UA \ FA) ∪ (UB \ FB ).

Still more consequences of the theorem.

Proposition. If A and B are measurable in the sense of Lebesgue then so is A ∪ B.

Proof. Indeed, A ∪ B = (Ac ∩ Bc )c .

Since A \ B = A ∩ Bc we also get Proposition. If A and B are measurable in the sense of Lebesgue then so is A \ B.

Caratheodory’s deﬁnition of measurability.

A set E ⊂ R is said to be measurable according to Caratheodory if for any set A ⊂ R we have

m∗(A) = m∗(A ∩ E) + m∗(A ∩ E c ) (8)

where (recall that) E c denotes the complement of E. In other words, A ∩ E c = A \ E. This deﬁnition has many advantages, as we shall see. Our ﬁrst task will be to show that it is equivalent to Lebesgue’s. Notice that we always have m∗(A) ≤ m∗(A ∩ E) + m∗(A \ E) so condition (8) is equivalent to

m∗(A ∩ E) + m∗(A \ E) ≤ m∗(A) (9)

for all A. Shlomo Sternberg Math212a1411 Lebesgue measure. Outline Lebesgue outer measure. Lebesgue inner measure. Lebesgue’s definition of measurability. Caratheodory’s definition of measurability. Countable additivity. σ-fields, measures, and outer measures. The Borel-Cantelli lemmas

Caratheodory = Lebesgue.

Theorem A set E is measurable in the sense of Caratheodory if and only if it is measurable in the sense of Lebesgue.

Proof that Lebesgue ⇒ Caratheodory. Suppose E is measurable in the sense of Lebesgue. Let > 0. Choose U ⊃ E ⊃ F with U open, F closed and m(U/F ) < which we can do by the “inner-outer” Theorem. Let V be an open set containing A. Then A \ E ⊂ V \ F and A ∩ E ⊂ (V ∩ U) so

A \ E ⊂ V \ F and A ∩ E ⊂ (V ∩ U) so

m∗(A \ E) + m∗(A ∩ E) ≤ m(V \ F ) + m(V ∩ U) ≤ m(V \ U) + m(U \ F ) + m(V ∩ U) ≤ m(V ) + .

(We can pass from the second line to the third since both V \ U and V ∩ U are measurable in the sense of Lebesgue and we can apply the proposition about disjoint unions.) Taking the inﬁmum over all open V containing A, the last term becomes m∗(A) + , and as is arbitrary, we have established (9) showing that E is measurable in the sense of Caratheodory.

Caratheodory ⇒ Lebesgue, case where m∗(E) < ∞.

Then for any > 0 there exists an open set U ⊃ E with m(U) < m∗(E) + . We may apply condition (8) to A = U to get

m(U) = m∗(U ∩ E) + m∗(U \ E) = m∗(E) + m∗(U \ E)

so m∗(U \ E) < . This means that there is an open set V ⊃ (U \ E) with m(V ) < . But we know that U \ V is measurable in the sense of Lebesgue, since U and V are, and

m(U) ≤ m(V ) + m(U \ V )

so m(U \ V ) > m(U) − .

m(U \ V ) > m(U) − . So there is a closed set F ⊂ U \ V with m(F ) > m(U) − . But since V ⊃ U \ E = U ∩ E c , we have

U \ V = U ∩ V c ⊂ U ∩ (Uc ∪ E) = E.

So F ⊂ E. So F ⊂ E ⊂ U and

m(U \ F ) = m(U) − m(F ) < .

Hence E is measurable in the sense of Lebesgue.

Caratheodory ⇒ Lebesgue, case where m∗(E) = ∞.

If m(E) = ∞, we must show that E ∩ [−n, n] is measurable in the sense of Caratheodory, for then it is measurable in the sense of Lebesgue from what we already know. We know that the interval [−n, n] itself, being measurable in the sense of Lebesgue, is measurable in the sense of Caratheodory. So we will have completed the proof of the theorem if we show that the intersection of E with [−n, n] is measurable in the sense of Caratheodory.

Unions and intersections.

More generally, we will show that the union or intersection of two sets which are measurable in the sense of Caratheodory is again measurable in the sense of Caratheodory. Notice that the deﬁnition (8) is symmetric in E and E c so if E is measurable in the sense of Caratheodory so is E c . So it suﬃces to prove the next lemma to complete the proof. Lemma

If E1 and E2 are measurable in the sense of Caratheodory so is E1 ∪ E2.

Proof of the lemma, 1.

For any set A we have

∗ ∗ ∗ c m (A) = m (A ∩ E1) + m (A ∩ E1 )

c by (8) applied to E1. Applying (8) to A ∩ E1 and E2 gives

∗ c ∗ c ∗ c c m (A ∩ E1 ) = m (A ∩ E1 ∩ E2) + m (A ∩ E1 ∩ E2 ).

Substituting this back into the preceding equation gives

∗ ∗ ∗ c ∗ c c m (A) = m (A ∩ E1) + m (A ∩ E1 ∩ E2) + m (A ∩ E1 ∩ E2 ). (10)

c c c Since E1 ∩ E2 = (E1 ∪ E2) we can write this as

∗ ∗ ∗ c ∗ c m (A) = m (A ∩ E1) + m (A ∩ E1 ∩ E2) + m (A ∩ (E1 ∪ E2) ).

Proof of the lemma, 2.

∗ ∗ ∗ c ∗ c m (A) = m (A ∩ E1) + m (A ∩ E1 ∩ E2) + m (A ∩ (E1 ∪ E2) ).

c Now A ∩ (E1 ∪ E2) = (A ∩ E1) ∪ (A ∩ (E1 ∩ E2) so

∗ ∗ c ∗ m (A ∩ E1) + m (A ∩ E1 ∩ E2) ≥ m (A ∩ (E1 ∪ E2)).

Substituting this for the two terms on the right of the previous displayed equation gives

∗ ∗ ∗ c m (A) ≥ m (A ∩ (E1 ∪ E2)) + m (A ∩ (E1 ∪ E2) )

which is just (9) for the set E1 ∪ E2. This proves the lemma and the theorem.

The class M of measurable sets.

We let M denote the class of measurable subsets of R - “measurability” in the sense of Lebesgue or Caratheodory these being equivalent. Notice by induction starting with two terms as in the lemma, that any ﬁnite union of sets in M is again in M

The ﬁrst main theorem in the subject is the following description of M and the function m on it: Theorem M and the function m : M → [0, ∞] have the following properties: R ∈ M. E ∈ M ⇒ E c ∈ M. S If En ∈ M for n = 1, 2, 3,... then n En ∈ M. If Fn ∈ M and the Fn are pairwise disjoint, then S F := n Fn ∈ M and

∞ X m(F ) = m(Fn). n=1

We already know the first two items on the list, and we know that a finite union of sets in M is again in M. We also know the last assertion. But it will be instructive and useful for us to have a proof starting directly from Caratheodory’s definition of measurablity:

Recall

∗ ∗ ∗ c ∗ c c m (A) = m (A∩E1)+m (A∩E1 ∩E2)+m (A∩E1 ∩E2 ). (10)

If F1 ∈ M, F2 ∈ M and F1 ∩ F2 = ∅ then taking

A = F1 ∪ F2, E1 = F1, E2 = F2

in (10) gives m(F1 ∪ F2) = m(F1) + m(F2).

Induction then shows that if F1,..., Fn are pairwise disjoint elements of M then their union belongs to M and

m(F1 ∪ F2 ∪ · · · ∪ Fn) = m(F1) + m(F2) + ··· + m(Fn).

∗ ∗ ∗ c ∗ c c m (A) = m (A∩E1)+m (A∩E1 ∩E2)+m (A∩E1 ∩E2 ). (10)

More generally, if we let A be arbitrary and take E1 = F1, E2 = F2 in (10) we get ∗ ∗ ∗ ∗ c m (A) = m (A ∩ F1) + m (A ∩ F2) + m (A ∩ (F1 ∪ F2) ).

If F3 ∈ M is disjoint from F1 and F2 we may apply (8) with A c replaced by A ∩ (F1 ∪ F2) and E by F3 to get ∗ c ∗ ∗ c m (A ∩ (F1 ∪ F2) )) = m (A ∩ F3) + m (A ∩ (F1 ∪ F2 ∪ F3) ), since c c c c c c (F1 ∪ F2) ∩ F3 = F1 ∩ F2 ∩ F3 = (F1 ∪ F2 ∪ F3) . Substituting this back into the preceding equation gives ∗ ∗ ∗ ∗ ∗ c m (A) = m (A∩F1)+m (A∩F2)+m (A∩F3)+m (A∩(F1∪F2∪F3) ).

∗ ∗ ∗ ∗ ∗ c m (A) = m (A∩F1)+m (A∩F2)+m (A∩F3)+m (A∩(F1∪F2∪F3) ).

Proceeding inductively, we conclude that if F1,..., Fn are pairwise disjoint elements of M then

n ∗ X ∗ ∗ c m (A) = m (A ∩ Fi ) + m (A ∩ (F1 ∪ · · · ∪ Fn) ). (11) 1

Now suppose that we have a countable family {Fi } of pairwise disjoint sets belonging to M. Since

n !c ∞ !c [ [ Fi ⊃ Fi i=1 i=1 we conclude from (11) that

n ∞ !c ! ∗ X ∗ ∗ [ m (A) ≥ m (A ∩ Fi ) + m A ∩ Fi 1 i=1 and hence passing to the limit

∞ ∞ !c ! ∗ X ∗ ∗ [ m (A) ≥ m (A ∩ Fi ) + m A ∩ Fi . 1 i=1

Now given any collection of sets Bk we can ﬁnd intervals {Ik,j } with [ Bk ⊂ Ik,j j and X m∗(B ) ≤ `(I ) + . k k,j 2k j So [ [ Bk ⊂ Ik,j k k,j and hence ∗ [ X ∗ m Bk ≤ m (Bk ), the inequality being trivially true if the sum on the right is inﬁnite.

P∞ ∗ ∗ S∞ So i=1 m (A ∩ Fk ) ≥ m (A ∩ ( i=1 Fi )) . Thus

∞ ∞ !c ! ∗ X ∗ ∗ [ m (A) ≥ m (A ∩ Fi ) + m A ∩ Fi ≥ 1 i=1

∞ !! ∞ !c ! ∗ [ ∗ [ ≥ m A ∩ Fi + m A ∩ Fi . i=1 i=1 The extreme right of this inequality is the left hand side of (9) applied to [ E = Fi , i and so E ∈ M and the preceding string of inequalities must be equalities since the middle is trapped between both sides which must be equal.

Hence we have proved that if Fn is a disjoint countable family of sets belonging to M then their union belongs to M and

∞ !c ! ∗ X ∗ ∗ [ m (A) = m (A ∩ Fi ) + m A ∩ Fi . (12) i i=1 S If we take A = Fi we conclude that

∞ X m(F ) = m(Fn) (13) n=1

if the Fj are disjoint and [ F = Fj .

So we have reproved the last assertion of the theorem using Caratheodory’s definition. Shlomo Sternberg Math212a1411 Lebesgue measure. Outline Lebesgue outer measure. Lebesgue inner measure. Lebesgue’s definition of measurability. Caratheodory’s definition of measurability. Countable additivity. σ-fields, measures, and outer measures. The Borel-Cantelli lemmas

S Proof of 3:If En ∈ M for n = 1, 2, 3,... then n En ∈ M.

For the proof of this third assertion, we need only observe that a countable union of sets in M can be always written as a countable disjoint union of sets in M. Indeed, set

c F1 := E1, F2 := E2 \ E1 = E1 ∩ E2

F3 := E3 \ (E1 ∪ E2) etc. The right hand sides all belong to M since M is closed under taking complements and ﬁnite unions and hence intersections, and [ [ Fj = Ej . j

We have completed the proof of the theorem.

Some consequences - symmetric diﬀerences.

The symmetric diﬀerence between two sets is the set of points belonging to one or the other but not both:

A∆B := (A \ B) ∪ (B \ A).

Proposition. If A ∈ M and m(A∆B) = 0 then B ∈ M and m(A) = m(B).

Proof. By assumption A \ B has measure zero (and hence is measurable) since it is contained in the set A∆B which is assumed to have measure zero. Similarly for B \ A. Also (A ∩ B) ∈ M since

A ∩ B = A \ (A \ B).

Thus B = (A ∩ B) ∪ (B \ A) ∈ M. Since B \ A and A ∩ B are disjoint, we have

m(B) = m(A∩B)+m(B\A) = m(A∩B) = m(A∩B)+m(A\B) = m(A).

More consequences: increasing limits.

Proposition.

Suppose that An ∈ M and An ⊂ An+1 for n = 1, 2,... . Then [ m An = lim m(An). n→∞

S If A := An we write the hypotheses of the proposition as An ↑ A. In this language the proposition asserts that

An ↑ A ⇒ m(An) → m(A).

Proof.

Setting Bn := An \ An−1 (with B1 = A1) the Bi are pairwise disjoint and have the same union as the Ai so

∞ n [ X X m An = m(Bi ) = lim m(Bn) n→∞ i=1 i=1

n ! [ = lim m Bi = lim m(An). n→∞ n→∞ i=1

More consequences: decreasing limits.

Proposition.

If Cn ⊃ Cn+1 is a decreasing family of sets in M and m(C1) < ∞ then \ m Cn = lim m(Cn). n→∞

Proof. T Let C = Cn and An := C1/Cn. Then An is a monotone increasing family of sets as in the preceding proposition so m(C1) − m(Cn) = m(An) → m(C1/C) = m(C1) − m(C). Now subtract m(C1) from both sides.

Taking Cn = [n, ∞) shows that we need m(Ck ) < ∞ for some k.

Constantin Carathéodory

Born: 13 Sept 1873 in Berlin, Germany Died: 2 Feb 1950 in Munich, Germany

Axiomatic approach.

We will now take the items in Theorem 6 as axioms: Let X be a set. (Usually X will be a topological space or even a metric space). A collection F of subsets of X is called a σ field if: X ∈ F, If E ∈ F then E c = X \ E ∈ F, and S If {En} is a sequence of elements in F then n En ∈ F, The intersection of any family of σ-fields is again a σ-field, and hence given any collection C of subsets of X , there is a smallest σ-field F which contains it. Then F is called the σ-field generated by C.

Borel sets.

If X is a metric space, the σ-ﬁeld generated by the collection of open sets is called the Borel σ-ﬁeld, usually denoted by B or B(X ) and a set belonging to B is called a Borel set.

Measures on a σ-ﬁeld.

Given a σ-ﬁeld F a (non-negative) measure is a function

m : F → [0, ∞]

such that m(∅) = 0 and

Countable additivity: If Fn is a disjoint collection of sets in F then ! [ X m Fn = m(Fn). n n In the countable additivity condition it is understood that both sides might be inﬁnite.

Outer measures.

An outer measure on a set X is a map m∗ to [0, ∞] deﬁned on the collection of all subsets of X which satisﬁes m∗(∅) = 0, Monotonicity: If A ⊂ B then m∗(A) ≤ m∗(B), and ∗ S P ∗ Countable subadditivity: m ( n An) ≤ n m (An).

Measures from outer measures via Caratheordory.

Given an outer measure, m∗, we deﬁned a set E to be measurable (relative to m∗) if

m∗(A) = m∗(A ∩ E) + m∗(A ∩ E c )

for all sets A. Then Caratheodory’s theorem that we proved just now asserts that the collection of measurable sets is a σ-ﬁeld, and the restriction of m∗ to the collection of measurable sets is a measure which we shall usually denote by m.

Terminological conﬂict.

There is an unfortunate disagreement in terminology, in that many of the professionals, especially in geometric measure theory, use the term “measure” for what we have been calling “outer measure”. However we will follow the above conventions which used to be the old fashioned standard.

The next task.

An obvious task, given Caratheodory’s theorem, is to look for ways of constructing outer measures. This will be the subject of the next lecture.

Over the next few slides I want to give remarkable applications of our two “monotone convergence” propositions:

An ↑ A ⇒ m(An) → m(A)

and (with the obvious notation)

Cn ↓ C and m(C1) < ∞ ⇒ m(Cn) → m(C).

Some notation

Let En be a sequence of measurable sets. The following deﬁnitions of the set E are synonymous:

E := {En i.o.} = {En inﬁnitely often}

:= lim sup En \ [ := En k n≥k

:= {x|∀k∃n(k) ≥ k such that x ∈ En(k)}

:= {x|x ∈ En for inﬁnitely many En}.

The ﬁrst Borel-Cantelli lemma (BC1)

Lemma P Suppose that m(En) < ∞ then m(lim sup En) = 0.

Proof. S S Let Gk = n≥k En. So k≤n≤p En ↑ Gk . But   [ X m  En ≤ m(En) k≤n≤p k≤n≤p

P∞ and so m(Gk ) ≤ k m(En). Since lim sup En ⊂ Gk for all k and P m(En) < ∞ we conclude that m(lim sup En) is less than any positive number by choosing k suﬃciently large.

Some probabilistic language

We now restrict to the case where m(X ) = 1 and use P instead of m. A measurable set is now called an event. A sequence of events En is called independent if, for every k ∈ N, if all the i1,... ik are distinct then Y P(Ei1 ∩ · · · ∩ Eik ) = P(Eij ).

The second Borel-Cantelli lemma (BC2)

Lemma

Let En be a sequence of independent events. Then P P(En) = ∞ ⇒ P(En i.o.) = 1. T S Notice that En i.o. = k n≥k En so its complement is [ \ c En . k n≥k We must show that this has probability zero, and for this it is T c enough to show that n≥k En has probability zero. Now \ c \ c En ↓ En k≤n≤` n≥k

T c so it is enough to show that P k≤n≤` En → 0 as ` → ∞.

Proof. c Let pn := P(En) so that P(En ) = 1 − pn. Then, by independence,   \ c Y P  En  = (1 − pn). k≤n≤` k≤n≤`

Now for x ≥ 0, 1 − x ≤ e−x so

P Y − pn (1 − pn) ≤ e k≤n≤` k≤n≤` P and since pn = ∞ this last expression → 0.

Notice we get a “0 or 1 law” in that for a sequence En of independent events we conclude that P(En inﬁnitely often) is either zero or one.

Shlomo Sternberg Math212a1411 Lebesgue measure.