MAT 550 Real Analysis II Spring 2011
Homework 1
Part I. Review of measure theory
1. Let (X, M, µ) be a σ-finite measure space.
(a) Show that X has only countably many atoms (i.e., points x ∈ X such that
µ({x}) > 0).
(b) Suppose X is a metric space and µ is a Borel measure, and let Sr (x) be a sphere
of radius r > 0 centered at x ∈ X. Prove (for given x ∈ X) that µ(Sr (x)) = 0 for
all r > 0 except at most countably many.
Proof. (a) Suppose that on the contrary, we have uncountably many atoms. Since µ
is σ-finite, we can partition X into a countable union of finite-measure sets. One
of these sets, say, X0 , must contain uncountably many atoms x. Partition (0, ∞)
1
into a countable union of sets, say, Sn = ( n1 , n−1
] and S1 = (1, ∞). Then the
uncountable set of atoms in X0 is the union
[
{x ∈ X0 |µ(x) ∈ Sn }.
n
This implies that for at least one n, there are uncountably many atoms x such
that µ(x) ∈ Sn . This means that
infinite number of such
S there are
Pa countablyP
1
atoms {xj }. Then µ(X0 ) ≥ µ( {xj }) =
µ({xj }) >
n = ∞. But X0 was
finite-measure by construction, a contradiction.
(b) Introduce the map p : X → R≥0 by
p(x) = ||x − x0 ||
Partition X into countably many disjoint finite measure sets Xj , and suppose that
there are uncountably many positive measure spheres centeredSat x0 . For each
Xj , consider the sets Xj ∩Sr , which are measurable. Since Sr = j Xj ∩Sr (this is
a countable disjoint union), there are uncountably many positive measure sets of
the form Xj ∩ Sr . This means that there are uncountably many such sets in some
fixed Xj (of finite measure). As in the first part of the problem, for some n, this
means that there are uncountably many (hence countably
infinitely P
many such
P
1
sets in Xj with measure at least n , so that µ(Xj ) ≥ i µ(Xj ∩ Sri ) ≥ i n1 = ∞,
a contradiction.
2. Let A ⊂ Rn with m(A) > 1, where m is the Lebesgue measure. Show that there are
two different points x, y ∈ A such that x − y ∈ Zn .
Proof. Suppose there are no different points x, y ∈ A such that x − y ∈ Zn . Introduce
the map
pzi : Rn −→
Rn
x 7−→ x − zi
where
zi = (zi,1 , . . . , zi,n ) ∈ Zn . Let Ai = A ∩ {x ∈ Rn : zi , j ≤ xj < zi , j + 1}, then
S
i Ai = A, and also by our assumption,
pzi (Ai ) ∩ pzj (Aj ) = ∅,
1
2
for zi 6= zj . Thus
µ(A) =
X
µ(Ai ) =
i
X
µ(pzi (Ai ))
i
≤ µ(I n ) = 1,
which contradicts the fact that µ(A) > 1.
3. Prove a strong version of the Poincaré Recurrence Theorem. Let (X, M, µ) be a finite
measure space (µ(X) < ∞), f : X → X — a measure preserving bijection, and let
Ω ⊂ X be a set of positive measure. Show that for a.e. x ∈ Ω there exists a sequence
nk → ∞ such that f nk (x) ∈ Ω, where f n is the n-th iterate of the mapping f .
Proof. Let
Ek := {x ∈ X : f k (x) ∈ Ω, f k+j (x) ∈
/ Ω},
i.e., Ek is the set of points which last visit Ω in the k-th iteration. Then we have
(1) Ek ∩ El = ∅, whenever k 6= l;
(2) f (Ek+1 ) = Ek , i.e., x last visits Ω at k + 1-st iterate if and only if f (x) last visits
Ω at k-th iterate;
(3) µ(Ek+1 ) = µ(Ek ), this comes from the above and the fact that f preserves measure.
Therefore
X
X
∞ > µ(X) ≥ µ(∪Ek ) =
µ(Ek ) =
µ(E1 ).
k
k
This implies µ(E1 ) as well as all µ(Ek ) = 0. In particular, this implies the set of points
in Ω which leave after some iteration and never come back under further iterations of
f has measure 0.
4. Let (X, M, µ) be a probability space (i.e., µ(X) = 1). Two “events” E1 and E2
(i.e., subsets of X) are called independent if µ(E1 ∩ E2 ) = µ(E1 )µ(E2 ). Two “random
variables” f and g (i.e., measurable real-valued functions on X) are called independent,
if the events f −1 (A) and g −1 (B) are independent for Rany Borel sets A, B ⊂ R. An
expected value (expectation) of a random variable f is X f dµ.
Theorem 1. Let f and g be two independent random variables with finite expectations.
Then their point-wise product f g also has finite expectation and
Z
Z
Z
f gdµ =
f dµ
gdµ.
X
X
X
(a) Prove the theorem for simple functions f and g (i.e., functions assuming only
finitely many values).
P
εk
(b) Let X = [0, 1] with the Lebesgue measure. For x = ∞
k=1 2k ∈ [0, 1] let fn (x) =
εn . Show that fn and fm are independent random variables for n 6= m.
Proof. (a) From the conditions given in the problem we may assume that f =
P
and g = bj χFj . Then
j
f −1 (ai ) = Ei , g −1 (bj ) = Fj .
P
i
ai χEi
3
Since f and g are independent, we have
Z X
Z
f · gdµ =
ai bj χEi χFj dµ
X i,j
X
=
Z X
ai bj χEi ∩Fj dµ
X i,j
=
X
=
X
ai bj µ(Ei ∩ Fj )
i,j
ai bj µ(Ei ) · µ(Fj )
(note f and g are independent)
i,j
=
X
ai µ(Ei ) ·
i
bj µ(Fj )
j
Z
Z
f dµ ·
=
X
gdµ.
X
X
(b) First one might have observed that the binary expansion of a number in [0, 1] may
not be unique, e.g. 0.1 = 0.01̄. However we may overcome the problem by forcing
the expansion to have the least number of non-zero digits.
By definition we have to show that for any subsets A and B of {0, 1},
−1
−1
µ(fn−1 (A)) · µ(fm
(B)) = µ(fn−1 (A) ∩ fm
(B)).
Now the subsets of {0, 1} are ∅, {0, 1}, {0} and {1}, if one of A and B is either
−1 (B)) = µ(f −1 (A) ∩ f −1 (B)). So we
∅ or {0, 1}, then obviously µ(fn−1 (A)) · µ(fm
m
n
only need to check the cases when A and B are either {0} or {1}. Suppose both
A and B are {0}. Without loss of generality, we may assume n < m. Then all
points in the interval [.x1 · · · xn−1 0, .x1 · · · xn−1 1) are in fn−1 (0), so
µ(fn−1 ({0}) =
1
1
· 2n−1 =
n
2
2
−1 ({0})) = 1 . Let’s look
since we have 2n−1 choices of x1 , · · · , xn−1 . Similarly µ(fm
2
−1 ({0}): their expansions are like
at those points in fn−1 ({0}) ∩ fm
.x1 · · · xn−1 0xn+1 · · · xm−1 0 · · · ,
so by the same reason
−1
({0})) =
µ(fn−1 ({0}) ∩ fm
1
1
· 2m−2 = .
m
2
4
This shows that
−1
−1
µ(fn−1 ({0})) · µ(fm
({0}) = µ(fn−1 ({0}) ∩ fm
({0})).
For other subsets of A and B, the arguments are the same. So finally we get for
any subsets A and B of {0, 1},
−1
−1
µ(fn−1 (A)) · µ(fm
(B)) = µ(fn−1 (A) ∩ fm
(B)),
i.e. fn and fm are independent whenever n 6= m.
5. Let Σ = {x = (x0 , x1 , . . . ) : xi = 0, 1} be the space of sequences of 0 and 1. Define
cylindrical sets by [ε0 , . . . , εn−1 ] = {x ∈ Σ : x0 = ε0 , . . . , xn−1 = εn−1 }.
4
(a) Let d be a metric on Σ defined by d(x, x) = 0 and d(x, y) = 21n , where n is the
smallest index such that xn 6= yn . Check that d is a metric and prove that Σ
is compact and that cylindrical sets are open and closed in the metric topology.
Show that cylindrical sets form a basis of this topology.
(b) Fix numbers p, q ∈ (0, 1) such that p+q = 1 and define µ([ε0 , . . . , εn−1 ]) = pk q n−k ,
where k is the number of zeroes among ε0 , . . . , εn−1 . Show that µ can be extended
to a Borel measure on Σ. (Hint: Use part (a) to prove σ-addititivity.)
(c) Show that (after removing some measure zero sets from Σ and [0, 1]) the measure
space (Σ, M, µ) is isomorphic to [0, 1] with the Lebesgue measure.
Proof. (a) Show d is a metric: From the definition of d it’s easy to see that d is symmetric and non-degenerate. For the triangle inequality, we take distinct x, y, z ∈ Σ
and suppose d(x, y) = 21n and d(y, z) = 21m . Without loss of generality, we may
assume n ≤ m. The lowest index where x differs from y is n while the lowest index
where y differs from z is m. This implies that the lowest index where x and z
differs is n (or possibly greater, in the case n = m). This means that d(x, z)le 21n .
Thus we have
1
1
1
d(x, z)le n < n + m = d(x, y) + d(y, z).
2
2
2
Show Σ is compact: This can be done by showing that Σ is totally bounded and
d is complete. For any , suppose > 21n , then Σ can be decomposed into a finite
-net: [x0 , · · · , xn ], where x0 , · · · , xn range over 0 and 1. For d being complete, we
have to show that any Cauchy sequence converges. In fact, suppose ȳ0 , · · · , ȳn , · · ·
is such a sequence. For = 21N , there exists an MN such that for n ≥ MN the
first N digits of ȳn should always be the same. Let the limit point z be defined
so that zn = ȳMn ,n . This is well defined, and for n ≥ MN , d(ȳn , z) ≤ 21N . Thus
compactness is proved.
Show cylindrical sets are open and closed: ∀x̄ = (x0 , · · · , xn , · · · ), consider the
open ball
U
1
2n
1
}
2n
{x : the first n + 1-digits of x coincide with those of x̄}
[x0 , · · · , xn ],
(x̄) := {x : d(x, x̄) <
=
=
i.e., the cylindrical sets are open. The closedness is also obvious: The limit point
of any convergent sequence of points in [x0 , · · · , xn ] should lie in [x0 , · · · , xn ],
otherwise the distance between this point and the points in [x0 , · · · , xn ] would be
greater than 1/2n .
Show cylindrical sets form the basis of the topology: Let U be an open set and x̄
in U . Then the 2n1x̄ ball centered at x̄ is contained in U for some nx̄ . This is a
cylindrical set, and U is the union of these sets for all x̄ in U . Therefore every
open set can be written as a union of cylindrical sets.
(b) We will do this in three parts, using Theorem 4.4.7 from Geller (p. 198). We must
show that µ can be extended to a volume function, that this volume function is
inner-regular, and that µ(X) < ∞. Then there is a σ-algebra containing all open
sets on which µ can be extended to a measure. To show that µ can be extended
to a volume function, first we must extend it to a set function, defining it on open
sets. There are countably many cylindrical sets, indexed by finite sequences of 1s
and 0s, so every open set can be written as a countable union of basis sets. These
can be chosen to be disjoint in a canonical manner, as follows. First notice that
5
if two cylindrical sets have nonempty intersection, then one is strictly contained
in the other by definition. Let Cn be the set of cylindrical sets on sequences of
length n, and suppose that subsets Dn ⊂ Cn for n < N have been chosen so that
if C ∈ Dn , then C ⊂ U , and so that no two cylindrical sets in the union of all Dn
intersect. Now let DN be
DN = {C ∈ CN |C ⊂ U, C ∈
/
N[
−1
Dn }
. If x ∈ U then x is contained in some maximal cylindrical subset in U , which
is contained in the set of allPthese DN , so U is the union of all these DN . Now
define µ(U ) to be µ(U ) =
µ(C) for C ranging over this decomposition of U
as a countable disjoint union of cylindrical sets. We must check for convergence.
Consider first a single cylindrical set C of length n. We have µ(C) = pk0 q n−k0 . If
we replace C with the disjoint union of its cylindrical subsets of length n + m, for
each such subset Ci with k new zeros, the volume of Ci is pk0 +k q n+m−k−k0 . There
are 2m nuch choices, m
k for each choice of k. So the total sum of the volumes of
the disjoint pieces is
X m
X m
k0 +k n+m−k−k0
k0 n−k0
p
q
=p q
pk q m−k = µ(C)(p + q)m = µ(C)
k
k
This means, in particular, that any two decompositions of an open set into a
countable union of cylindrical sets sum to the same weight. This is finite additivity
for cylindrical sets. If a decomposition varies from the canonical decomposition,
then it replaces some of the cylindrical sets in the canonical one with a union of
cylindrical sets that equal it. This can be approximated by an iterated sequence
of subdivisions of cylindrical sets into smaller cylindrical sets of a uniform size. At
every step the weight sum is the same so in the limit (which is finite, though that is
unnecessary here) the sum is the same. This also shows that the weight of any open
set is at most one. This is true for the trivial cylindrical set [], and so it is also true
for any different decomposition of the whole space into cylindrical sets. Examine
an open set’s decomposition into cylindrical sets of length n = 1, 2, . . . For each
n, the partial decomposition up through length n is a subset of a decomposition
of [], so the partial sum of the constituent weights is at most 1. Then
µ(U ) =
∞ X
X
n=1
µ(Cn ) = lim
N →∞
N X
X
n=1
µ(Cn ) ≤ lim 1 = 1
N →∞
Essentially the same argument shows monotonicity of µ on cylindrical sets.
Now to show that this is a volume function, we must show it to be countably
subadditive and countably superadditive, not just on cylindrical sets but also on
open sets. To see subadditivity, suppose S ⊂ ∪Sj . Then examine the cylindrical
sets in the decompositions on both sides. There are two possibilities for each
CS in the decomposition of S. CS could be contained in a single Cj from the
decomposition of some Sj . Suppose it is not. Then it is contained in a union of
Cj , none of which contain it. In this case, because intersection of cylindrical sets
implies inclusion, one direction or the other, it is necessary that every Cj that
intersects CS is contained in it. Then the second possibility is that CS is equal to
a union of some collection of Cj . This can be made disjoint by iterating along the
length of the cylindrical sets involved. Now we will build a disjoint collection of
cylindrical sets, each of which is contained in the canonical decomposition of some
Sj and whose union contains S. Let {CS } be divided into D and E, corresponding
6
to the first and second type of CS in the paragraph above. For d ∈ D1 , let Cd be
the unique shortest Cj containing d. For e ∈ E, let Ee be a disjoint collection of Cj
equal to e. Then because each d or e in D ∪ E is disjoint from all the others (they
all come from a disjoint decomposition of S), none of the Cd or elements of Ee can
intersect one another, except if two Cd coincide. This can be seen in cases. If two
distinct Cd intersected, then one would strictly contain the other and correspond
to a shorter cylindrical set, violating the choice of Cd . If a Cd intersected an
element of Ee , it would intersect the e that generated it. Cd cannot contain e
because then e would belong in D, not E; Cd cannot be contained in e because
then e and d would intersect. Finally, since all the e are disjoint, all the EeS
, which
union to them, are disjoint from one another. Then Z = {Cd : d ∈ D} ∪ Ee is
a disjoint union containing every CS , every set in which is some Cj . This means
that
X
XX
X
µ(S) ≤
µ(C) ≤
µ(Cj ) =
µ(Sj )
C∈Z
j
Cj
S
For countable superadditivity, consider again S which contains Sj , all disjoint.
Then taking the decomposition of both sides into cylindrical sets, we get that
all the Cj on the right are distinct by disjointness of Sj , and that each one is
contained in some CS (if instead Cj was contained in a union of multiple CS , then
this would violate the construction of the canonical decomposition; Cj would
rightfully belong rather than all the CS ). This CS is unique because all the CS
are disjoint. Now for each CS in the decomposition, we can consider the set DCS
of Cj contained in it. The union over all CS of DCS exhausts the Cj since every
one is subservient to some CS . By finite additivity and monotonicity of cylindrical
sets, we have superadditivity of cylindrical sets, so that
X
µ(CS ) ≥
µ(d)
d∈DCS
because all of the sums involved converge nicely, we can then write
X
XX
X
µ(S) =
µ(CS ) ≥
µ(Cj ) =
µ(Sj )
j
j
which is superadditivity.
Now we have shown that µ is a volume function, and that µ(X) = 1 < ∞. If
we can show that µ is inner-regular, then by Theorem 4.4.7 in Geller (p. 198), µ
extends to a measure. Inner regularity of the volume function would mean that
µ(U ) = sup µe (K)
K⊂U
for open sets U , where µe is the
S outer measure associated to µSWrite U in its
canonical decomposition, U = Ci . Then the partial unions N Ci are finite
unions of closed sets, hence
PNa closed subset of a compact space, hence compact.
Therefore, sup µe (K) ≥
µ(Ci ) for all N , so that the supremum is at least
the limit of this sum, namely µ(U ) = µe (U ). Note that µe is monotone, since
the infimum defining µe for T ⊂ U is taken over a larger set of open sets. Then
sup µe (K) ≤ µe (U ) = µ(U ) and the supremum is exactly µ(U ), proving that µe is
inner-regular. This concludes the proof.
(c) We construct the isomorphism by comparing the Lebesgue measure and µ. Consider the usual binary expansion of the elements in the unit interval I: Divide I
into two equal intervals, then for each subinterval, divide it again into two equal
subintervals, and keep going on. In each step, denote the left subinterval by 0,
7
and the right one by 1. Except those points in the boundary of subintervals, each
point x ∈ I has a unique binary expansion according to the division. For example,
x = .010101 · · ·
means x lies on the left in the first subdivision, and then on the right in the second
subdivision, etc. Now it is easy to see that the measure induced from Euclidean
distance is the Lebesgue measure.
Now let’s give another division of I, which gives a (p, q) binary expansion of
elements in I. Divide I into two subintervals with the left subinterval having
length p, and the right subinterval having length q. And for these two subintervals
divide them in the same way as the first division, i.e. the quotient of the length
of the left subinterval and that of the right subinterval is pq , and keep going on.
We can give a binary expansion ((p, q) expansion) of elements in I the same way
as the usual binary expansion. This gives the model of Σ. It is not difficult to
check that the measure, still the one induced from Euclidean distance (not d), is
exactly µ.
Note the boundary points in both cases have measure 0, by identifying the same
points in both divisions, we get an isomorphism from (Σ, M, µ) to I with the
Lebesgue measure.
Part II. Iterated integrals
All of these problems are standard.