2Measurable functions and random variables

II Probability and Measure

2.1 Measurable functions
The definition of a measurable function is somewhat like the definition of a
continuous function, except that we replace “open” with “in the σ-algebra”.
Definition
(Measurable functions)
.
Let (
E, E
) and (
G, G
) be measure spaces.
A map f : E G is measurable if for every A G, we have
f
1
(A) = {x E : f(x) E} E.
If (G, G) = (R, B), then we will just say that f is measurable on E.
If (
G, G
) = ([0
,
]
, B
), then we will just say that
f
is non-negative measurable.
If E is a topological space and E = B(E), then we call f a Borel function.
How do we actually check in practice that a function is measurable? It turns
out we are lucky. We can simply check that
f
1
(
A
)
E
for
A
in any generating
set Q of G.
Lemma.
Let (
E, E
) and (
G, G
) be measurable spaces, and
G
=
σ
(
Q
) for some
Q. If f
1
(A) E for all A Q, then f is measurable.
Proof. We claim that
{A G : f
1
(A) E}
is a
σ
-algebra on
G
. Then the result follows immediately by definition of
σ
(
Q
).
Indeed, this follows from the fact that
f
1
preserves everything. More
precisely, we have
f
1
[
n
A
n
!
=
[
n
f
1
(A
n
), f
1
(A
C
) = (f
1
(A))
C
, f
1
() = .
So if, say, all A
n
A, then so is
S
n
A
n
.
Example.
In the particular case where we have a function
f
:
E R
, we know
that
B
=
B
(
R
) is generated by (
−∞, y
] for
y R
. So we just have to check that
{x E : f(x) y} = f
1
((−∞, y])) E.
Example.
Let
E, F
be topological spaces, and
f
:
E F
be continuous. We
will see that
f
is a measurable function (under the Borel
σ
-algebras). Indeed,
by definition, whenever
U F
is open, we have
f
1
(
U
) open as well. So
f
1
(
U
)
B
(
E
) for all
U F
open. But since
B
(
F
) is the
σ
-algebra generated
by the open sets, this implies that f is measurable.
This is one very important example. We can do another very important
example.
Example.
Suppose that
A E
. The indicator function of
A
is
1
A
(
x
) :
E
{0, 1} given by
1
A
(x) =
(
1 x A
0 x 6∈ A
.
Suppose we give
{
0
,
1
}
the non-trivial measure. Then
1
A
is a measurable function
iff A E.
Example. The identity function is always measurable.
Example.
Composition of measurable functions are measurable. More precisely,
if (
E, E
), (
F, F
) and (
G, G
) are measurable spaces, and the functions
f
:
E F
and
g
:
F G
are measurable, then the composition
gf
:
E G
is measurable.
Indeed, if
A G
, then
g
1
(
A
)
F
, so
f
1
(
g
1
(
A
))
E
. But
f
1
(
g
1
(
A
)) =
(g f)
1
(A). So done.
Definition
(
σ
-algebra generated by functions)
.
Now suppose we have a set
E
,
and a family of real-valued functions {f
i
: i I} on E. We then define
σ(f
i
: i I) = σ(f
1
i
(A) : A B, i I).
This is the smallest
σ
-algebra on
E
which makes all the
f
i
’s measurable.
This is analogous to the notion of initial topologies for topological spaces.
If we want to construct more measurable functions, the following definition
will be rather useful:
Definition
(Product measurable space)
.
Let (
E, E
) and (
G, G
) be measure
spaces. We define the product measure space as
E × G
whose
σ
-algebra is
generated by the projections
E × G
E G
π
1
π
2
.
More explicitly, the σ-algebra is given by
E G = σ({A × B : A E, B G}).
More generally, if (
E
i
, E
i
) is a collection of measure spaces, the product measure
space has underlying set
Q
i
E
i
, and the
σ
-algebra generated by the projection
maps π
i
:
Q
j
E
j
E
i
.
This satisfies the following property:
Proposition.
Let
f
i
:
E F
i
be functions. Then
{f
i
}
are all measurable iff
(
f
i
) :
E
Q
F
i
is measurable, where the function (
f
i
) is defined by setting the
ith component of (f
i
)(x) to be f
i
(x).
Proof.
If the map (
f
i
) is measurable, then by composition with the projections
π
i
, we know that each f
i
is measurable.
Conversely, if all
f
i
are measurable, then since the
σ
-algebra of
Q
F
i
is
generated by sets of the form
π
1
j
(
A
) :
A F
j
, and the pullback of such sets
along (f
i
) is exactly f
1
j
(A), we know the function (f
i
) is measurable.
Using this, we can prove that a whole lot more functions are measurable.
Proposition.
Let (
E, E
) be a measurable space. Let (
f
n
:
n N
) be a sequence
of non-negative measurable functions on
E
. Then the following are measurable:
f
1
+ f
2
, f
1
f
2
, max{f
1
, f
2
}, min{f
1
, f
2
},
inf
n
f
n
, sup
n
f
n
, lim inf
n
f
n
, lim sup
n
f
n
.
The same is true with “real” replaced with “non-negative”, provided the new
functions are real (i.e. not infinity).
Proof.
This is an (easy) exercise on the example sheet. For example, the sum
f
1
+ f
2
can be written as the following composition.
E [0, ]
2
[0, ].
(f
1
,f
2
)
+
We know the second map is continuous, hence measurable. The first function is
also measurable since the f
i
are. So the composition is also measurable.
The product follows similarly, but for the infimum and supremum, we need to
check explicitly that the corresponding maps [0
,
]
N
[0
,
] is measurable.
Notation. We will write
f g = min{f, g}, f g = max{f, g}.
We are now going to prove the monotone class theorem, which is a “Dynkin’s
lemma” for measurable functions. As in the case of Dynkin’s lemma, it will
sound rather awkward but will prove itself to be very useful.
Theorem
(Monotone class theorem)
.
Let (
E, E
) be a measurable space, and
A E
be a
π
-system with
σ
(
A
) =
E
. Let
V
be a vector space of functions such
that
(i) The constant function 1 = 1
E
is in V.
(ii) The indicator functions 1
A
V for all A A
(iii) V is closed under bounded, monotone limits.
More explicitly, if (
f
n
) is a bounded non-negative sequence in
V
,
f
n
% f
(pointwise) and f is also bounded, then f V.
Then V contains all bounded measurable functions.
Note that the conditions for
V
is pretty like the conditions for a d-system,
where taking a bounded, monotone limit is something like taking increasing
unions.
Proof. We first deduce that 1
A
V for all A E.
D = {A E : 1
A
V}.
We want to show that
D
=
E
. To do this, we have to show that
D
is a
d
-system.
(i) Since 1
E
V, we know E D.
(ii) If 1
A
V , then 1 1
A
= 1
E\A
V. So E \ A D.
(iii)
If (
A
n
) is an increasing sequence in
D
, then
1
A
n
1
S
A
n
monotonically
increasingly. So 1
S
A
n
is in D.
So, by Dynkin’s lemma, we know
D
=
E
. So
V
contains indicators of all measur-
able sets. We will now try to obtain any measurable function by approximating.
Suppose that
f
is bounded and non-negative measurable. We want to show
that f V. To do this, we approximate it by letting
f
n
= 2
n
b2
n
fc =
X
k=0
k2
n
1
{k2
n
f<(k+1)2
n
}
.
Note that since
f
is bounded, this is a finite sum. So it is a finite linear
combination of indicators of elements in
E
. So
f
n
V
, and 0
f
n
f
monotonically. So f V.
More generally, if f is bounded and measurable, then we can write
f = (f 0) + (f 0) f
+
f
.
Then f
+
and f
are bounded and non-negative measurable. So f V.
Unfortunately, we will not have a chance to use this result until the next
chapter where we discuss integration. There we will use this a lot.