Problem 1
First of all, it is immediately clear that each interval (a, ∞) is a Borel set,
because each such interval is open.
By the definition of the Borel σ-algebra on R, it is sufficient to show that
every open set is contained in the σ-algebra A generated by E5 .
First, it is clear that any half-open interval of the form (a, b] is in A. (This
set is given by (a, ∞) ∩ (R \ (b, ∞))).
Let U ⊂ R be an open set. Let {xj }∞
j=1 be an enumeration of the rational
numbers in U . Because U is open, U contains an open interval centered at every
xj , and hence a half-open interval centered at each xj . For each j, let Cj be
the supremum sup{r : (xj − r, xj + r]} ⊂ U . For any ǫ > 0, define the interval
Ij,ǫ := Cj,ǫ = (xj − Cj + ǫ, xj + Cj − ǫ]. I claim that U can be written as:
U=
∞ [
∞
[
Ij, n1 .
j=1 n=1
Once this claim is proven, it is clear that U is in the sigma algebra A.
First, I will show the inclusion
U⊂
∞
∞ [
[
Ij, n1 .
j=1 n=1
Let y ∈ U . Then, for some δ > 0, the open interval of radius δ centered at y is
contained in U . Let B be the open ball of radius 4δ centered at y. B contains
a rational number xj . Furthermore, the half-open interval (xj − 2δ , xj + 2δ ] is
contained in the interval (y − δ, y + δ) by the triangle inequality. Therefore, x is
in the interval Ij,ǫ for any ǫ < 4δ , and therefore x is in the set on the right hand
side.
Now, I will show the inclusion
U⊃
∞ [
∞
[
Ij, n1 .
j=1 n=1
This inclusion is trivial because each Ij n1 is contained in U by definition: Ij n1
1
1
is contained in (xj − Cj + 2n
, xj + Cj − 2n
), which is contained in U by the
definition of the constant Cj .
Problem 2
2
Let {(xj , yj )}∞
j=1 be an enumeration of the rational points in [0, 1] . For every
j, the set Sj := R \ {(xj , yj )} is open. Furthermore, the rectangles Rǫ (−ǫ, 1 +
ǫ) × (−ǫ, 1 + ǫ) are open. Pick some sequence ǫj that goes to zero. Then the set
E can be written
∞
\
Sj ∩ Rǫj ,
E=
j=1
1
which is a countable intersection of open sets.
Problem 3
If A is a σ-algebra, then it is clearly closed under countable increasing unions
because it is closed under countable unions.
Let A be an algebra that is closed under countable increasing unions. The
only condition that needs to be checked to show that A is a σ-algebra is that A
is closed under countable unions.
Let F be a set
F =
∞
[
Fj
j=1
where Fj ∈ A. We’re done if we can show that F can be written as a
countable increasing union of elements of A.
Define Ej in the following manner:
Ej =
j
[
Fj .
k=1
Then
• Ej is an increasing sequence of sets. To see this, notice that if x ∈ Ej ,
then x ∈ Fk S
for some k ≤ j, and therefore x is in Eℓ for all ℓ > j.
S
∞
• ∞
F
⊂
j=1 j
j=1 Ej : If x is in the union on the left hand side, then
x
∈
F
⊂
E
,
and
therefore x is in the union on the right hand side.
jS
S∞ j
∞
• j=1 Fj ⊃ j=1 Ej : If x is in the union on the right hand side, then
x ∈ Ej for some j. So x ∈ Fk for some k ≤ j and therefore x is in the
union on the left-hand side.
Therefore, F is in the algebra A, and A is a σ-algebra.
Problem 4
Part a
The limit inferior of the Ej is defined as
lim inf Ej =
∞ \
∞
[
Ej ,
k=1 j=k
the set of all points x that are “eventually” in all of the Ej .
Note that the sets in the union are increasing in k. Therefore,
by continuity
T∞
of measure, we have that µ(lim inf Ej ) is equal to limk→∞ j=k Ej .
2
T∞
For each k, we have that j=k Ej is contained in Ej for each j > k, so
therefore has measure no more than inf k>j µ(Ek ). This, in turn, is no more
than lim inf µ(Ej ). So for each k, we have


∞
\
µ
Ej  ≤ lim inf µ(Ej ).
j=k
Taking the limit of the left hand side as k → ∞ gives the desired result.
Part b
The limit superior of Ej is defined as
lim sup Ej =
∞ [
∞
\
Ej ,
k=1 j=k
the set of all points x that are in infinitely many of the sets Ej .
We want to use the continuity of measure again. Note that the sets in the
intersection are decreasing in k, so by continuity of measure, and the fact that
the union ofSthe sets Ej has finite measure, we have that µ(lim sup Ej ) is equal
∞
to limk→∞ j=k Ej .
S
For each k, the set ∞
j=k contains the set Ej for every j > k, so therefore
S
∞
is greater than or equal to supj>k µ(Ej ). This, in turn, is larger
µ
j=k Ej
than or equal to lim sup µ(Ej ). Therefore, we have
∞
[
Ej ≥ lim sup µ(Ej )
j=k
for every k. Taking the limit of the left hand side as k → ∞ gives the desired
result.
This isn’t part of the problem statement,
S butit might be worthwhile to look
∞
at an example of a sequence Ej with µ
j=1 Ej = ∞ for which the statement
fails. Consider the family of singleton sets Ej = {j} in the integers. The Ej are
disjoint, so the limit superior of the Ej is the empty set. However, the counting
measure of Ej is equal to 1 for every j.
3