Stat 310A/Math 230A Theory of Probability
Practice Midterm Solutions
Andrea Montanari
October 28, 2010
Problem 1
Consider the measurable space (Ω
,
F
), with:
Ω =
{
0
,
1
}
N
the set of (infinite) binary sequences
ω
=
(
ω
1
, ω
2
, ω
3
, . . .
);
F
the
σ
algebra generated by cylindrical sets (equivalently the
σ
algebra generated by
sets of the type
A
i,x
=
{
ω
:
ω
i
=
x
}
for
i
∈
N
and
x
∈ {
0
,
1
}
).
(a)
Let
f
: Ω
→
[0
,
1] be defined by
f
(
ω
)
≡
∞
X
n
=1
ω
n
2
n
.
(1)
for
ω
= (
ω
1
, ω
2
, . . .
). Is
f
measurable? Prove your answer. (As usual, assume [0
,
1] to be endowed with the
Borel
σ
algebra.)
Solution :
Yes,
f
is measurable. Indeed consider the following collection of intervals
P
=
{
B
k,n
= [
k/
2
n
,
(
k
+ 1)
/
2
n
) :
, n, k
∈
N
, n
≥
0
,
0
≤
k
≤
2
n

1
} ∪
n
{
1
}
o
.
(2)
This is a
π
system (indeed if
n
≥
m
,
B
k,n
∩
B
l,m
is either empty or equal to
B
k,n
). Further
σ
(
P
) =
B
[0
,
1)
.
Indeed any interval of the form [
k/
2
n
, l/
2
n
) is disjoint union of a finite colection of intervals in
P
.
As a
consequence any interval (
a, b
), 0
≤
a < b
≤
1 is in
σ
(
P
) since we can find monotone sequences
a
n
=
k
n
/
2
n
> a
,
b
n
=
l
n
/
2
n
< b
with
a
n
↓
a
,
b
n
↑
b
, whence (
a, b
) =
∪
n
≥
n
0
[
a
n
, b
n
). By taking union intevals in
P
, we conclude that all intervals of the form [0
, b
), (
a,
1] are in
σ
(
P
) as well. This completes the proof that
σ
(
P
) =
B
[0
,
1)
.
It is therefore sufficient to show that
f

1
(
B
k,n
)
∈ F
for
k, n
as above. If the binary expansion of
k
is
k
=
k
1
2
n

1
+
k
2
2
n

2
+
· · ·
+
k
n
, it is elementary to see that
f

1
(
B
k,n
) =
n
ω
such that
ω
1
=
k
1
, . . . , ω
n
=
k
n
o
\
n
ω
= (
k
1
, k
2
, . . . , k
n
,
1
,
1
,
1
,
1
, . . .
)
o
,
(3)
which is the intersection of two measuraable sets.
(b)
Let
g
: [0
,
1]
→
Ω be defined by
g
(
x
)
≡
(
g
1
(
x
)
, g
2
(
x
)
, g
3
(
x
)
, . . .
)
,
(4)
g
i
(
x
)
≡
bb
2
i
x
cc
mod 2
,
(5)
where
bb
a
cc ≡
max
{
n
∈
N
:
n < a
}
. Is
g
a measurable mapping? Prove your answer. (Assume [0
,
1] and Ω
to be endowed the same
σ
algebras as above.)
Solution :
Consider the collection of sets
P
=
{
C
n
(
u
n
1
) :
n
∈
N
, u
n
1
∈ {
0
,
1
}
n
} ⊆ F
defined by
C
n
(
u
n
1
) =
{
ω
:
ω
1
=
u
1
, ω
2
=
u
2
, . . . , ω
n
=
u
n
}
.
(6)
1
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It is clear that
P
is a
π
system (indeed, for
n
≥
m C
n
(
u
n
1
)
∩
C
m
(
v
m
1
) is either empty or coincides with
C
n
(
u
n
1
))
with
σ
(
P
) =
F
(indeed the events
A
i,x
can be constructed as union of finitely many events
C
n
(
u
n
1
)).
We have
g

1
(
C
n
(
u
n
1
)) =
x
∈
[0
,
1] :
bb
2
x
cc
=
u
1
mod 2
\
· · ·
\
x
∈
[0
,
1] :
bb
2
n
x
cc
=
u
n
mod 2
.
(7)
It is therefore sufficient to show that any set of the form
{
x
∈
[0
,
1] :
,
bb
2
k
x
cc
=
u
k
mod 2
is Borel. Such
a set is the union of a finite number of sets of the form
J
k,j
=
{
x
∈
[0
,
1] :
,
bb
2
k
x
cc
=
j
, with
j
∈
N
. But
bb
2
k
x
cc
=
j
if and only if 2
k
x
∈
(
j, j
+ 1], and therefore
J
k,j
are just intervals.
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 '11
 MONTANARI,A.
 Probability, Continuous function, measure, Types of functions, FN, Lebesgue integration

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