Math212a1411 Lebesgue measure.people.math.harvard.edu/~shlomo/212a/11.pdf · OutlineLebesgue outer...

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

Math212a1411Lebesgue measure.

Shlomo Sternberg

October 14, 2014

Shlomo Sternberg

Math212a1411 Lebesgue measure.


Reminder

No class this Thursday

Shlomo Sternberg



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

Shlomo Sternberg



Henri Léon Lebesgue

Born: 28 June 1875 in Beauvais, Oise, Picardie, France

Died: 26 July 1941 in Paris, France

Shlomo Sternberg



In today’s lecture we will discuss the concept of measurability of asubset of R. We will begin with Lebesgue’s (1902) definition ofmeasurability, which is easy to understand and intuitive. We willthen give Caratheodory’s (1914) definition of measurabiity which ishighly non-intuitive but has great technical advantage. For subsetsof R these two definitions are equivalent (as we shall prove). Butthe Caratheodory definition extends to many much more generalsituations.In particular, the Caratheodory definition will prove useful for uslater, when we study Hausdorff measures.

Shlomo Sternberg



The ε/2n trick.

An argument which will recur several times is:

∞∑1

1

2n= 1

and so∞∑1

ε

2n= ε.

We will call this the “ε/2n trick” and not spell it out every time weuse it.

Shlomo Sternberg



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

Shlomo Sternberg



The definition of Lebesgue outer measure.

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

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 thesame definition for half open intervals (a, b] or [a, b), or openintervals.

Of course if a = −∞ and b is finite or +∞, or if a is finite andb = +∞, or if a = −∞ and b =∞ the length is infinite.

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

m∗(A) := inf∑

`(In) : In are intervals with A ⊂⋃

In. (1)

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

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Monotonicity, subadditivity.

If A ⊂ B then m∗(A) ≤ m∗(B) since any cover of B by intervals isa cover of A.

For any two sets A and B we clearly have

m∗(A ∪ B) ≤ m∗(A) + m∗(B).

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Sets of measure zero don’t matter.

A set Z is said to be of (Lebesgue) measure zero it its Lebesgueouter measure is zero, i.e. if it can be covered by a countableunion of (open) intervals whose total length can be made as smallas we like.

If Z is any set of measure zero, then m∗(A ∪ Z ) = m∗(A).

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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 intervalis infinite, it clearly can not be covered by a set of intervals whosetotal length is finite, since if we lined them up with end pointstouching they could not cover an infinite interval. We claim that

m∗(I ) = `(I ) (2)

if I is a finite interval.

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Proof.

We may assume that I = [c , d ] is a closed interval by what wehave already said, and that the minimization in (1) is with respectto a cover by open intervals. So what we must show is that if

[c , d ] ⊂⋃i

(ai , bi )

thend − c ≤

∑i

(bi − ai ).

We first apply Heine-Borel to replace the countable cover by afinite cover. (This only decreases the right hand side of precedinginequality.) So let n be the number of elements in the cover.

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We need to prove that if

[c, d ] ⊂n⋃

i=1

(ai , bi ) then d − c ≤n∑

i=1

(bi − ai ).

We do this this by induction on n. If n = 1 then a1 < c andb1 > d so clearly b1 − a1 > d − c .

Suppose that n ≥ 2 and we know the result for all covers (of allintervals [c , d ] ) with at most n − 1 intervals in the cover. If someinterval (ai , bi ) is disjoint from [c , d ] we may eliminate it from thecover, and then we are in the case of n − 1 intervals. So we mayassume that every (ai , bi ) has non-empty intersection with [c , d ].

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Among the the intervals (ai , bi ) there will be one for which ai takeson the minimum possible value. By relabeling, we may assume thatthis is (a1, b1). Since c is covered, we must have a1 < c . If b1 > dthen (a1, b1) covers [c , d ] and there is nothing further to do. Soassume 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 > 1contains 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:

[c , d ] ⊂ (a1, b2) ∪n⋃

i=3

(ai , bi ).

So by induction d − c ≤ (b2 − a1) +∑n

i=3(bi − ai ).

But b2 − a1 ≤ (b2 − a2) + (b1 − a1) since a2 < b1. �

Shlomo Sternberg



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 takethem all to be half open, of the form Ii = [ai , bi ), we can writeeach Ii as a disjoint union of finite or countably many intervalseach of length < ε. So it makes no difference to the definition ifwe also require the

`(Ii ) < ε (3)

in (1). We will see that when we pass to other types of measuresthis will make a difference.

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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 ε/2n trick, for any finite or countable union we have

m∗(⋃i

Ai ) ≤∑i

m∗(Ai ).

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 ).

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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 < ε where 2ε < dist (A,B) so that there is nooverlap. As for item 5, we know from 2 that m∗(A) ≤ m∗(U) forany 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 thenthe union on the right is an open set whose Lebesgue outermeasure is less than m∗(A) + δ for any δ > 0 if we choose a closeenough approximation to the infimum.

Shlomo Sternberg



Lebesgue outer measure on Rn.

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

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Lemma

Let C be a finite non-overlapping collection of closed rectangles allcontained in the closed rectangle J. Then

vol J ≥∑I∈C

vol I .

If C is any finite collection of rectangles such that

J ⊂⋃I∈C

I

thenvol J ≤

∑I∈C

vol (I ).

Shlomo Sternberg



This lemma occurs on page 1 of Stroock, A concise introduction tothe theory of integration together with its proof. I will take this forgranted. In the next few slides I will talk as if we are in R, buteverything goes through unchanged if R is replaced by Rn.

In fact, once we will have developed enough abstract theory, wewill not need this lemma. So for those of you who are purists anddo not want to go through the lemma in Stroock, just stick to R.

Shlomo Sternberg



The definition of Lebesgue inner measure.

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

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

Clearlym∗(A) ≤ m∗(A)

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

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The Lebesgue inner measure of an interval.

Proposition.

For any interval I we have

m∗(I ) = `(I ). (5)

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Proof.

If `(I ) =∞ the result is obvious. So we may assume that I is afinite interval which we may assume to be open, I = (a, b). IfK ⊂ I is compact, then I is a cover of K and hence from thedefinition of outer measure m∗(K ) ≤ `(I ). So m∗(I ) ≤ `(I ). Onthe other hand, for any ε > 0, ε < 1

2 (b − a) the interval[a + ε, b − ε] is compact andm∗([a− ε, a + ε]) = b − a− 2ε ≤ m∗(I ). Letting ε→ 0 proves theproposition.

Shlomo Sternberg



Lebesgue’s definition of measurability for sets of finiteouter measure.

A set A with m∗(A) <∞ is said to measurable in the sense ofLebesgue 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 compactset contained in itself. Hence all compact sets are measurable inthe sense of Lebesgue. If I is a bounded interval, then I ismeasurable in the sense of Lebesgue by the preceding Proposition.

Shlomo Sternberg



Lebesgue’s definition of measurability for sets of infiniteouter measure.

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

Shlomo Sternberg



Countable additivity.

Theorem

If A =⋃

Ai is a (finite or) countable disjoint union of sets whichare measurable in the sense of Lebesgue, then A is measurable inthe sense of Lebesgue and

m(A) =∑i

m(Ai ).

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

Shlomo Sternberg



We havem∗(A) ≤

∑n

m∗(An) =∑n

m(An).

Let ε > 0, and for each n choose compact Kn ⊂ An with

m∗(Kn) ≥ m∗(An)− ε

2n= m(An)− ε

2n

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

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

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

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m∗(Kn) ≥ m∗(An)− ε

2n= m(An)− ε

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

m∗(A) ≥∑n

m(An)− ε.

Since this is true for all ε > 0 we get

m∗(A) ≥∑

m(An).

But then m∗(A) ≥ m∗(A) and so they are equal, so A ismeasurable in the sense of Lebesgue, and m(A) =

∑m(Ai ). �

Shlomo Sternberg



Open sets are measurable in the sense of Lebesgue.

Proof.

Any open set O can be written as the countable union of openintervals Ii , and

Jn := In \n−1⋃i=1

Ii

is a disjoint collection of intervals (some open, some closed, somehalf open) and O is the disjont union of the Jn. So every open setis a disjoint union of intervals hence measurable in the sense ofLebesgue.

Shlomo Sternberg



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, andso F is measurable in the sense of Lebesgue. Suppose that

m∗(F ) <∞.

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For any ε > 0 consider the sets

G1,ε := [−1 +ε

22, 1− ε

22] ∩ F

G2,ε := ([−2 +ε

23,−1] ∩ F ) ∪ ([1, 2− ε

23] ∩ F )

G3,ε := ([−3 +ε

24,−2] ∩ F ) ∪ ([2, 3− ε

24] ∩ F )

...

and setGε :=

⋃i

Gi ,ε.

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

Shlomo Sternberg



The Gi ,ε are all compact, and hence measurable in the sense ofLebesgue, and the union in the definition of Gε is disjoint, so ismeasurable in the sense of Lebesgue. Furthermore, the sum of thelengths of the “gaps” between the intervals that went into thedefinition of the Gi ,ε is ε. So

m(Gε)+ε = m∗(Gε)+ε ≥ m∗(F ) ≥ m∗(Gε) = m(Gε) =∑i

m(Gi ,ε).

In particular, the sum on the right converges, and hence byconsidering a finite number of terms, we will have a finite sumwhose value is at least m(Gε)− ε. The corresponding union of setswill be a compact set Kε contained in F with

m(Kε) ≥ m∗(F )− 2ε.

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

Shlomo Sternberg



The inner-outer characterization of measurability.

Theorem

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

m(U \ F ) < ε.

Proof in one direction. Suppose that A is measurable in thesense of Lebesgue with m(A) <∞. Then there is an open setU ⊃ A with m(U) < m∗(A) + ε/2 = m(A) + ε/2, and there is acompact set F ⊂ A with m(F ) ≥ m∗(A)− ε/2 = m(A)− ε/2.Since U \ F is open, it is measurable in the sense of Lebesgue, andso is F as it is compact. Also F and U \ F are disjoint. Hence

m(U \ F ) = m(U)−m(F ) < m(A) +ε

2−(

m(A)− ε

2

)= ε.

Shlomo Sternberg



If A is measurable in the sense of Lebesgue, and m(A) =∞, wecan apply the above to A ∩ I where I is any compact interval, inparticular to the interval [−n, n]. So there exist open setsUn ⊃ A ∩ [−n, n] and closed sets Fn ⊂ A ∩ [−n, n] withm(Un/Fn) < ε/2n. Let U :=

⋃Un and

F :=⋃

(Fn ∩ ([−n,−n + 1] ∪ [n − 1, n])) .

A convergent sequence in F must eventually lie in an interval oflength one and hence to at most the union of two entries in theunion 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

m(U/F ) ≤∑

m(Un/Fn) < ε. �

Shlomo Sternberg



Proof in the other direction.

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

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

Since this is true for every ε > 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 haveU ∩ (−n − ε, n + ε) ⊃ A ∩ [−n, n] ⊃ F ∩ [−n, n] and

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

so we can proceed as before to conclude thatm∗(A ∩ [−n, n]) = m∗(A ∩ [−n, n]). �

Shlomo Sternberg



Consequences of the theorem.

Proposition.

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

Proof.

Indeed, if F ⊂ A ⊂ U with F closed and U open, thenF 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 sodoes Ac .

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More consequences.

Proposiition.

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

Proof.

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

(UA ∩ UB) ⊃ (A ∩ B) ⊃ (FA ∩ FB)

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

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Still more consequences of the theorem.

Proposition.

If A and B are measurable in the sense of Lebesgue then so isA ∪ 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 isA \ B.

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Caratheodory’s definition of measurability.

A set E ⊂ R is said to be measurable according toCaratheodory 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 otherwords, A ∩ E c = A \ E . This definition has many advantages, aswe shall see. Our first task will be to show that it is equivalent toLebesgue’s.

Notice that we always have m∗(A) ≤ m∗(A ∩ E ) + m∗(A \ E ) socondition (8) is equivalent to

m∗(A ∩ E ) + m∗(A \ E ) ≤ m∗(A) (9)

for all A.Shlomo Sternberg



Caratheodory = Lebesgue.

Theorem

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

Proof that Lebesgue ⇒ Caratheodory. Suppose E ismeasurable in the sense of Lebesgue. Let ε > 0. ChooseU ⊃ E ⊃ F with U open, F closed and m(U/F ) < ε which we cando by the “inner-outer” Theorem. Let V be an open setcontaining A. Then A \ E ⊂ V \ F and A ∩ E ⊂ (V ∩ U) so

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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 ) + ε.

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

Shlomo Sternberg



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

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

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

som∗(U \ E ) < ε.

This means that there is an open set V ⊃ (U \ E ) with m(V ) < ε.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 )

som(U \ V ) > m(U)− ε.

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m(U \ V ) > m(U)− ε.

So there is a closed set F ⊂ U \ V with m(F ) > m(U)− ε. Butsince 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 ) < ε.

Hence E is measurable in the sense of Lebesgue.

Shlomo Sternberg



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

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

Shlomo Sternberg



Unions and intersections.

More generally, we will show that the union or intersection of twosets which are measurable in the sense of Caratheodory is againmeasurable 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 ofCaratheodory so is E c . So it suffices to prove the next lemma tocomplete the proof.

Lemma

If E1 and E2 are measurable in the sense of Caratheodory so isE1 ∪ E2.

Shlomo Sternberg



Proof of the lemma, 1.

For any set A we have

m∗(A) = m∗(A ∩ E1) + m∗(A ∩ E c1 )

by (8) applied to E1. Applying (8) to A ∩ E c1 and E2 gives

m∗(A ∩ E c1 ) = m∗(A ∩ E c

1 ∩ E2) + m∗(A ∩ E c1 ∩ E c

2 ).

Substituting this back into the preceding equation gives

m∗(A) = m∗(A∩E1) + m∗(A∩E c1 ∩E2) + m∗(A∩E c

1 ∩E c2 ). (10)

Since E c1 ∩ E c

2 = (E1 ∪ E2)c we can write this as

m∗(A) = m∗(A ∩ E1) + m∗(A ∩ E c1 ∩ E2) + m∗(A ∩ (E1 ∪ E2)c).

Shlomo Sternberg



Proof of the lemma, 2.

m∗(A) = m∗(A ∩ E1) + m∗(A ∩ E c1 ∩ E2) + m∗(A ∩ (E1 ∪ E2)c).

Now A ∩ (E1 ∪ E2) = (A ∩ E1) ∪ (A ∩ (E c1 ∩ E2) so

m∗(A ∩ E1) + m∗(A ∩ E c1 ∩ E2) ≥ m∗(A ∩ (E1 ∪ E2)).

Substituting this for the two terms on the right of the previousdisplayed equation gives

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

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

Shlomo Sternberg



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 thesebeing equivalent. Notice by induction starting with two terms as inthe lemma, that any finite union of sets in M is again in M

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The first main theorem in the subject is the following descriptionof M and the function m on it:

Theorem

M and the function m :M→ [0,∞] have the followingproperties:

R ∈M.

E ∈M⇒ E c ∈M.

If En ∈M for n = 1, 2, 3, . . . then⋃

n En ∈M.

If Fn ∈M and the Fn are pairwise disjoint, thenF :=

⋃n Fn ∈M and

m(F ) =∞∑n=1

m(Fn).

Shlomo Sternberg



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

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Recall

m∗(A) = m∗(A∩E1)+m∗(A∩E c1 ∩E2)+m∗(A∩E c

1 ∩E c2 ). (10)

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

A = F1 ∪ F2, E1 = F1, E2 = F2

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

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

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

Shlomo Sternberg



m∗(A) = m∗(A∩E1)+m∗(A∩E c1 ∩E2)+m∗(A∩E c

1 ∩E c2 ). (10)

More generally, if we let A be arbitrary and take E1 = F1, E2 = F2

in (10) we get

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

If F3 ∈M is disjoint from F1 and F2 we may apply (8) with Areplaced by A ∩ (F1 ∪ F2)c and E by F3 to get

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

since

(F1 ∪ F2)c ∩ F c3 = F c

1 ∩ F c2 ∩ F c

3 = (F1 ∪ F2 ∪ F3)c .

Substituting this back into the preceding equation gives

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

Shlomo Sternberg



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

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

m∗(A) =n∑1

m∗(A ∩ Fi ) + m∗(A ∩ (F1 ∪ · · · ∪ Fn)c). (11)

Shlomo Sternberg



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

n⋃i=1

Fi

)c

⊃

( ∞⋃i=1

Fi

)c

we conclude from (11) that

m∗(A) ≥n∑1

m∗(A ∩ Fi ) + m∗

(A ∩

( ∞⋃i=1

Fi

)c)

and hence passing to the limit

m∗(A) ≥∞∑1

m∗(A ∩ Fi ) + m∗

(A ∩

( ∞⋃i=1

Fi

)c).

Shlomo Sternberg



Now given any collection of sets Bk we can find intervals {Ik,j}with

Bk ⊂⋃j

Ik,j

andm∗(Bk) ≤

∑j

`(Ik,j) +ε

2k.

So ⋃k

Bk ⊂⋃k,j

Ik,j

and hencem∗(⋃

Bk

)≤∑

m∗(Bk),

the inequality being trivially true if the sum on the right is infinite.

Shlomo Sternberg



So∑∞

i=1 m∗(A ∩ Fk) ≥ m∗ (A ∩ (⋃∞

i=1 Fi )) . Thus

m∗(A) ≥∞∑1

m∗(A ∩ Fi ) + m∗

(A ∩

( ∞⋃i=1

Fi

)c)≥

≥ m∗

(A ∩

( ∞⋃i=1

Fi

))+ m∗

(A ∩

( ∞⋃i=1

Fi

)c).

The extreme right of this inequality is the left hand side of (9)applied to

E =⋃i

Fi ,

and so E ∈M and the preceding string of inequalities must beequalities since the middle is trapped between both sides whichmust be equal.

Shlomo Sternberg



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

m∗(A) =∑i

m∗(A ∩ Fi ) + m∗

(A ∩

( ∞⋃i=1

Fi

)c). (12)

If we take A =⋃

Fi we conclude that

m(F ) =∞∑n=1

m(Fn) (13)

if the Fj are disjoint and

F =⋃

Fj .

So we have reproved the last assertion of the theorem usingCaratheodory’s definition.

Shlomo Sternberg



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 acountable union of sets in M can be always written as a countabledisjoint union of sets in M. Indeed, set

F1 := E1, F2 := E2 \ E1 = E1 ∩ E c2

F3 := E3 \ (E1 ∪ E2)

etc. The right hand sides all belong to M since M is closed undertaking complements and finite unions and hence intersections, and⋃

j

Fj =⋃

Ej .

We have completed the proof of the theorem.

Shlomo Sternberg



Some consequences - symmetric differences.

The symmetric difference between two sets is the set of pointsbelonging 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).

Shlomo Sternberg



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 havemeasure zero. Similarly for B \ A. Also (A ∩ B) ∈M since

A ∩ B = A \ (A \ B).

ThusB = (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).

Shlomo Sternberg



More consequences: increasing limits.

Proposition.

Suppose that An ∈M and An ⊂ An+1 for n = 1, 2, . . . . Then

m(⋃

An

)= lim

n→∞m(An).

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).

Shlomo Sternberg



Proof.

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

m(⋃

An

)=∞∑i=1

m(Bi ) = limn→∞

n∑i=1

m(Bn)

= limn→∞

m

(n⋃

i=1

Bi

)= lim

n→∞m(An).

Shlomo Sternberg



More consequences: decreasing limits.

Proposition.

If Cn ⊃ Cn+1 is a decreasing family of sets in M and m(C1) <∞then

m(⋂

Cn

)= lim

n→∞m(Cn).

Proof.

Let C =⋂

Cn and An := C1/Cn. Then An is a monotoneincreasing family of sets as in the preceding proposition som(C1)−m(Cn) = m(An)→ m(C1/C ) = m(C1)−m(C ). Nowsubtract m(C1) from both sides.

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

Shlomo Sternberg



Constantin Carathéodory

Born: 13 Sept 1873 in Berlin, Germany

Died: 2 Feb 1950 in Munich, Germany

Shlomo Sternberg



Axiomatic approach.

We will now take the items in Theorem 6 as axioms: Let X be aset. (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

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, andhence given any collection C of subsets of X , there is a smallestσ-field F which contains it. Then F is called the σ-fieldgenerated by C.

Shlomo Sternberg



Borel sets.

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

Shlomo Sternberg



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 inF then

m

(⋃n

Fn

)=∑n

m(Fn).

In the countable additivity condition it is understood that bothsides might be infinite.

Shlomo Sternberg



Outer measures.

An outer measure on a set X is a map m∗ to [0,∞] defined onthe collection of all subsets of X which satisfies

m∗(∅) = 0,

Monotonicity: If A ⊂ B then m∗(A) ≤ m∗(B), and

Countable subadditivity: m∗ (⋃

n An) ≤∑

n m∗(An).

Shlomo Sternberg



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 justnow asserts that the collection of measurable sets is a σ-field, andthe restriction of m∗ to the collection of measurable sets is ameasure which we shall usually denote by m.

Shlomo Sternberg



Terminological conflict.

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

Shlomo Sternberg



The next task.

An obvious task, given Caratheodory’s theorem, is to look for waysof constructing outer measures.

This will be the subject of the next lecture.

Shlomo Sternberg



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

An ↑ A⇒ m(An)→ m(A)

and (with the obvious notation)

Cn ↓ C and m(C1) <∞⇒ m(Cn)→ m(C ).

Shlomo Sternberg



Some notation

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

E := {En i.o.} = {En infinitely often}:= lim sup En

:=⋂k

⋃n≥k

En

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

Shlomo Sternberg



The first Borel-Cantelli lemma (BC1)

Lemma

Suppose that∑

m(En) <∞ then m(lim sup En) = 0.

Proof.

Let Gk =⋃

n≥k En. So⋃

k≤n≤p En ↑ Gk . But

m

⋃k≤n≤p

En

≤ ∑k≤n≤p

m(En)

and so m(Gk) ≤∑∞

k m(En). Since lim sup En ⊂ Gk for all k and∑m(En) <∞ we conclude that m(lim sup En) is less than any

positive number by choosing k sufficiently large.

Shlomo Sternberg



Some probabilistic language

We now restrict to the case where m(X ) = 1 and use P instead ofm. A measurable set is now called an event. A sequence of eventsEn is called independent if, for every k ∈ N, if all the i1, . . . ik aredistinct then

P(Ei1 ∩ · · · ∩ Eik ) =∏

P(Eij ).

Shlomo Sternberg



The second Borel-Cantelli lemma (BC2)

Lemma

Let En be a sequence of independent events. Then∑P(En) =∞⇒ P(En i.o.) = 1.

Notice that En i.o. =⋂

k

⋃n≥k En so its complement is⋃

k

⋂n≥k

E cn .

We must show that this has probability zero, and for this it isenough to show that

⋂n≥k E c

n has probability zero. Now⋂k≤n≤`

E cn ↓

⋂n≥k

E cn

so it is enough to show that P(⋂

k≤n≤` E cn

)→ 0 as `→∞.

Shlomo Sternberg



Proof.

Let pn := P(En) so that P(E cn ) = 1− pn. Then, by independence,

P

⋂k≤n≤`

E cn

=∏

k≤n≤`(1− pn).

Now for x ≥ 0, 1− x ≤ e−x so∏k≤n≤`

(1− pn) ≤ e−∑

k≤n≤` pn

and since∑

pn =∞ this last expression → 0.

Shlomo Sternberg



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

Shlomo Sternberg


Math212a1411 Lebesgue measure.people.math.harvard.edu/~shlomo/212a/11.pdf · OutlineLebesgue outer...

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Transcript of Math212a1411 Lebesgue measure.people.math.harvard.edu/~shlomo/212a/11.pdf · OutlineLebesgue outer...