Homework 2, MAT 215
Problem 1. Prove an analogue of Theorem 1.8 for n functions..
Solution. We need to prove the following: if u1 , ..., un are real measurable functions on a
measurable space X, and Φ is a continuous map of Rn into a topological space Y , then
h(x) = Φ(u1 (x), ..., un (x)) is a measurable function from X to Y . Define f : X → Rn
by x → (u1 (x), ..., un (x)). By Theorem 1.7(b), to prove that h is measurable, it is enough
to prove that f is measurable. If R is any open rectangle in Rn which is the Cartesian
−1
product of n segments I1 , . . . , In , then f −1 (R) = u−1
1 (I1 ) ∩ · · · ∩ un (In ), which is measurable
since u1 , ..., un are measurable. Finally, every open set of Rn is the countable union of such
rectangles, so we are done.
✷
Problem 2. Prove that if f is a real-valued function on a measurable space X, such that
{x|f (x) ≥ r} is measurable for every rational r, then f is measurable.
Solution. Let U ⊆ R1 be an open set. First, U can be written as a union of countably
many open balls with rational radii that are centered at rational points. So to prove that
f −1 (U) is measurable, it is enough to prove this when U is an open ball of this form, say
with radius r and center c. Since the set of measurable sets is closed under complements
and finite intersections, every set of the form {x|r1 > f (x) ≥ r2 } is measurable for rational
r1 , r2 . Now note that {x|c + r > f (x) > c − r} can be written as the countable union
S
−1
(U) is measurable.
n1 {x|c + r > f (x) ≥ c − r + 1/n}, so f
✷
Problem 3. Let {an } and {bn } be sequences in [−∞, ∞], and prove the following assertions:
(a) lim sup(−an ) = − lim inf an .
n→∞
n→∞
(b) lim sup(an + bn ) ≤ lim sup an + lim sup bn , provided none of the sums is of the form
n→∞
n→∞
n→∞
∞ − ∞.
(c) If an ≤ bn for all n, then lim inf an ≤ lim inf bn .
n→∞
n→∞
Show by an example that strict inequality can hold in (b).
Solution. The supremum Ak of the set {−ak , −ak+1 , ...} is the negative of the infimum A′k
of the set {ak , ak+1, ...}. Hence inf k {Ak } = − supk {A′k }, which implies (a). The relation
sup{ak + bk , ak+1 + bk+1 , ...} ≤ sup{ak , ak+1 , ...} + sup{bk , bk+1 , ...}
is clear, so this implies (b). To see that the inequality in (b) can be strict, consider a1 = 1,
ai = 0 for i > 1, and b1 = −1, bi = 0 for i > 1. Then lim sup(an + bn ) = 0, but
lim sup an + lim sup bn = 1. Now suppose that an ≤ bn for all n. Then inf{ak , ak+1 , ..} ≤
inf{bk , bk+1 , ...} for all k, so (c) follows.
✷
1
2
Problem 4.
(a) Suppose f : X → [−∞, ∞] and g : X → [−∞, ∞] are measurable. Prove that the
sets {x|f (x) < g(x)}, {x|f (x) = g(x)} are measurable.
(b) Prove the set of points at which a sequence of measurable real-valued functions converges (to a finite limit) is measurable.
Solution. Let Y+ and Y− be the sets where g(x) = ∞ and −∞, respectively, and define Z+
and Z− analogously for f . Then these subsets are measurable: for example, Y+ is a countable
intersection of the sets {x ∈ X|g(x) ≥ n} as n ranges over the positive integers. Let X ′ be
the complement of these sets, i.e., the subset where both f and g take finite values.
So we can define the function h = f − g on X ′ , and it is a measurable function. The first
set of (a) is
h−1 ([−∞, 0)) ∪ (Y+ \ Z+ ) ∪ (Z− \ Y− ),
so is measurable. Also, the set where f and g agree is
(X ′ \ h−1 ([−∞, 0) ∪ (0, ∞])) ∪ (Y+ ∩ Z+ ) ∪ (Y− ∩ Z− ),
which is also measurable. As for (b), let fn be a sequence of measurable real functions, and
let E be the set of x such that fn (x) converges as n → ∞. Define f = lim sup fn . Then f is
measurable (Theorem 1.14), and f agrees with lim fn on E. For each n, the function f − fn
is measurable (1.22), so the set En,r which is defined to be the preimage of f − fn of (−r, r)
T S∞
is measurable. Then E = ∞
n=1 En,r , so is measurable.
r=1
✷
Problem 5. Let X be an uncountable set, let M be the collection of all sets E ⊂ X such
that either E or E c is at most countable, and define µ(E) = 0 in the first case, µ(E) = 1 in
the second. Prove that M is a σ-algebra in X and that µ is a measure on M. Describe the
corresponding measurable functions and their integrals.
Solution. Since X c =Ø is at most countable, X ∈ M. Also, if E ∈ M, then either E or E c is
at most countable, so the same is true for E c since (E c )c = E, and so E c ∈ M. Now suppose
S
En ∈ M for all n, and put E = n≥1 En . Let I be the set of n for which En is at most
countable, and let J be the set of n for which En is uncountable, but Enc is at most countable,
S
S
so that E = n∈I En ∪ n∈J En . If J =Ø, then E is a countable union of countable sets,
T
T
T
and hence is countable. Otherwise, E c = n∈I Enc ∩ n∈J Enc , so E c ⊆ n∈J Enc , which is
countable since J 6=Ø, so E ∈ M. Thus, M is a σ-algebra.
Now write a measurable set A as a disjoint union of measurable sets An . If A is at most
P
countable, then so is each An , so µ(A) =
µ(An ) = 0. In case Ac is at most countable,
then A is uncountable, so at least one Ai is uncountable. Suppose that Ai and Aj are both
uncountable for i 6= j. Then Aci ∪Acj is countable and equal to X since Ai and Aj are disjoint.
3
But this contradicts that X is uncountable, so exactly one Ai is uncountable, which means
P
that µ(A) =
µ(An ) = 1. Hence µ is a measure on M.
The measurable functions on M consist of those functions f : X → R1 , such that for each
r ∈ R1 , f −1 (r) is either at most countable, or f −1 (R1 \ {r}) is at most countable. If we let
A ⊂ R1 denote the set of points such that f −1 (r) is not countable, then the integral of f is
P
r∈A r.
✷