MATH 394: Probability I
Supplementary Notes
Department of Mathematics
University of Washington
Summer 2026
Arman Jahangiri
1
1 Continuity of Probability
Motivation
In probability, we often describe complicated events as limits of simpler events. For example,
suppose we have events
A
1
, A
2
, . . . , A
n
, . . .
and we want to understand the probability of an event obtained by taking an infinite union or an
infinite intersection. The key fact is that the probability function P(·) behaves continuously under
limits of certain sequences of events, in a precise mathematical sense. This is called the continuity
of probability, which we will prove here. Before that, we revisit the probability axioms.
Probability Space
A probability space is a triple
(S, F, P),
where
(1) Sample space.
S is the set of all possible outcomes of the experiment.
(2) Sigma-algebra.
F is a collection of subsets of S, called events, satisfying:
(a)
S ∈ F.
(b) If A ∈ F, then
A
c
∈ F.
(c) If
A
1
, A
2
, . . . , A
n
, . . . ∈ F,
then
∞
[
n=1
A
n
∈ F.
A collection satisfying these three properties is called a σ-algebra (or sigma-field).
(3) Probability measure.
P : F → [0, 1] is a function that assigns a probability to every event in F. Such a function is
called a probability measure and satisfies the following axioms.
(A1) Nonnegativity:
P(A) ≥ 0, for every A ∈ F.
(A2) Normalization:
P(S) = 1.
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(A3) Countable additivity:
If
A
1
, A
2
, . . . , A
n
, . . .
are pairwise disjoint events, then
P
∞
[
n=1
A
n
!
=
∞
X
n=1
P(A
n
).
Increasing and Decreasing Sequences of Events
Definition 1 (Increasing sequence of events). A sequence of events (A
n
)
n≥1
is called increasing
if
A
1
⊆ A
2
⊆ · · · ⊆ A
n
⊆ · · · .
Figure 1: An increasing sequence of events, A
1
⊆ A
2
⊆ · · · ⊆ A
n
⊆ · · · , whose limit is
∞
[
n=1
A
n
.
In this case, we write
A
n
↑
∞
[
n=1
A
n
,
and we define
lim
n→∞
A
n
:=
∞
[
n=1
A
n
.
So A
n
↑ lim
n→∞
A
n
means that the events are getting larger and their limiting event is the
union.
Definition 2 (Decreasing sequence of events). A sequence of events (A
n
)
n≥1
is called decreasing
if
A
1
⊇ A
2
⊇ · · · ⊇ A
n
⊇ · · · .
In this case, we write
A
n
↓
∞
\
n=1
A
n
,
where
lim
n→∞
A
n
:=
∞
\
n=1
A
n
.
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Figure 2: A decreasing sequence of events, A
1
⊇ A
2
⊇ · · · ⊇ A
n
⊇ · · · , whose limit is
∞
\
n=1
A
n
.
So A
n
↓
T
∞
n=1
A
n
means that the events are getting smaller and their limiting event is the
intersection.
Continuity of the probability function from Below
Theorem 1 (Continuity from below). Suppose (A
n
)
n≥1
is a sequence of events in F, and
A
1
⊆ A
2
⊆ · · · ⊆ A
n
⊆ · · · .
Then
P
lim
n→∞
A
n
= lim
n→∞
P(A
n
).
Equivalently, if A
n
↑ A, then
P(A
n
) ↑ P(A).
Proof. Define disjoint pieces
B
1
= A
1
, B
n
= A
n
\ A
n−1
for n ≥ 2.
These are disjoint as for any m < n, we have
B
n
∩ B
m
= (A
n
− A
n−1
) ∩ (A
m
− A
m−1
) = A
n
∩ A
c
n−1
∩ A
m
∩ A
c
m−1
Since m < n, we have A
m
⊆ A
n
, and A
m−1
⊆ A
n−1
, and therefore, A
c
n−1
⊆ A
c
m−1
. This implies
that A
n
∩ A
m
= A
m
and A
c
n−1
∩ A
c
m−1
= A
c
n−1
. Now, since n − 1 ≤ m, A
n−1
⊆ A
m
and therefore,
A
m
∩ A
c
n−1
= ∅. Hence, we have shown
A
n
∩ A
c
n−1
∩ A
m
∩ A
c
m−1
= A
m
∩ A
c
n−1
= ∅
So B
n
∩ B
m
= ∅.
Next, we claim (please show) that for every n ≥ 1,
A
n
=
n
[
k=1
B
k
.
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Also,
∞
[
n=1
A
n
=
∞
[
n=1
B
n
.
Therefore, by countable additivity,
P
∞
[
n=1
A
n
!
= P
∞
[
n=1
B
n
!
=
∞
X
n=1
P(B
n
).
On the other hand, for each fixed n,
P(A
n
) = P
n
[
k=1
B
k
!
=
n
X
k=1
P(B
k
).
Taking the limit as n → ∞, we get
lim
n→∞
P(A
n
) = lim
n→∞
n
X
k=1
P(B
k
) =
∞
X
k=1
P(B
k
).
Thus,
lim
n→∞
P(A
n
) = P
∞
[
n=1
A
n
!
.
Remark 1 (Analogy with left-continuity of functions). Suppose f : R → R is continuous from the
left at a point a. If
a
n
↑ a,
meaning
a
1
≤ a
2
≤ · · · ≤ a
n
≤ · · · and a
n
→ a,
then
f(a
n
) → f(a).
Equivalently,
lim
n→∞
f(a
n
) = f
lim
n→∞
a
n
= f(a).
Continuity from below for probability says something very similar. If
A
n
↑ A,
then
P(A
n
) → P(A).
Equivalently,
lim
n→∞
P(A
n
) = P( lim
n→∞
A
n
)
In this sense, probability is continuous along increasing sequences of events. The role of “ap-
proaching from the left” is played by the inclusions
A
1
⊆ A
2
⊆ · · · ⊆ A
n
⊆ · · · ⊆
∞
[
i=1
A
i
.
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Continuity of the probability function from Ab ove
Theorem 2 (Continuity from above). Suppose (A
n
)
n≥1
is a sequence of events in F, and
A
1
⊇ A
2
⊇ · · · ⊇ A
n
⊇ · · · .
Then
P
∞
\
n=1
A
n
!
= lim
n→∞
P(A
n
).
Equivalently, if A
n
↓ A, then
P(A
n
) ↓ P(A).
Proof. Let
A =
∞
\
n=1
A
n
.
Since A
n
↓ A, the complements satisfy
A
c
1
⊆ A
c
2
⊆ · · · ⊆ A
c
n
⊆ · · · .
Also, by De Morgan’s law,
∞
[
n=1
A
c
n
=
∞
\
n=1
A
n
!
c
= A
c
.
By continuity from below,
P(A
c
) = P
∞
[
n=1
A
c
n
!
= lim
n→∞
P(A
c
n
).
Using complements,
P(A
c
n
) = 1 − P(A
n
),
and
P(A
c
) = 1 − P(A).
Therefore,
1 − P(A) = lim
n→∞
(1 − P(A
n
)) .
Hence,
P(A) = lim
n→∞
P(A
n
).
Remark 2 (Analogy with right-continuity of functions). Suppose f : R → R is continuous from the
right at a point a. If
a
n
↓ a,
meaning
a
1
≥ a
2
≥ · · · ≥ a
n
≥ · · · and a
n
→ a,
then
f(a
n
) → f(a).
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Equivalently,
lim
n→∞
f(a
n
) = f
lim
n→∞
a
n
= f(a).
Continuity from above for probability says the analogous statement for decreasing sequences of
events. If
A
n
↓ A,
then
P(A
n
) → P(A).
Here, the role of “approaching from the right” is played by
A
1
⊇ A
2
⊇ · · · ⊇ A
n
⊇ · · · ⊇ A.
“‘latex
Properties of a Cumulative Distribution Function
Let X be a random variable on the probability space
(S, F, P).
The cumulative distribution function, or CDF, of X is the function
F : R → [0, 1]
defined by
F (x) = P(X ≤ x).
More precisely,
F (x) = P({ω ∈ S : X(ω) ≤ x}).
In fact, for any B ⊆ R, we write {X ∈ B} instead of {ω ∈ S : X(ω) ∈ B}.
Proposition 1 (Basic properties of a CDF). Let F be the CDF of a random variable X. Then:
(1)
0 ≤ F (x) ≤ 1 for every x ∈ R.
(2)
lim
x→−∞
F (x) = 0.
(3)
lim
x→∞
F (x) = 1.
(4) F is increasing. That is, if x ≤ y, then
F (x) ≤ F (y).
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Proof. We prove each property separately.
(1) Proof that F (x) ∈ [0, 1].
Fix x ∈ R. By definition,
F (x) = P(X ≤ x).
The set {X ≤ x} is an event in F. Since P is a probability measure, every event has probability
between 0 and 1. Therefore,
0 ≤ P(X ≤ x) ≤ 1.
Hence,
0 ≤ F (x) ≤ 1.
(2) Proof that lim
x→−∞
F (x) = 0.
For each n ∈ N, define
A
n
= {X ≤ −n}.
As n increases, the number −n becomes smaller. Therefore, the events become smaller, and thus,
we have a decreasing chain of events.
A
1
⊇ A
2
⊇ A
3
⊇ · · · .
Reason: If ω ∈ A
n+1
, then X(ω) ≤ −(n + 1) < −n, so X(ω) ≤ −n. Thus, ω ∈ A
n
., and hence,
A
n+1
⊆ A
n
.
Now consider the intersection
∞
\
n=1
A
n
=
∞
\
n=1
{X ≤ −n}.
This event means that X is less than or equal to −n for every positive integer n. But no real
number can be less than or equal to every number
−1, −2, −3, . . . .
Since X is real-valued, this is impossible. Therefore,
∞
\
n=1
A
n
= ∅.
By Theorem 2,
P
∞
\
n=1
A
n
!
= lim
n→∞
P(A
n
).
Thus,
P(∅) = lim
n→∞
P(X ≤ −n).
Since P(∅) = 0, we get
lim
n→∞
F (−n) = 0.
This shows that along the sequence x = −n,
F (x) → 0.
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Since F is increasing, this implies the full limit
lim
x→−∞
F (x) = 0.
(3) Proof that lim
x→∞
F (x) = 1.
For each n ∈ N, define
B
n
= {X ≤ n}.
As n increases, the number n becomes larger. Therefore, the events become larger:
B
1
⊆ B
2
⊆ B
3
⊆ · · · .
Reason: If ω ∈ B
n
, then X(ω) ≤ n < n+1,, so X(ω) ≤ n+1. Thus, ω ∈ B
n+1
.. Hence, B
n
⊆ B
n+1
.
Now consider the union.
∞
[
n=1
B
n
=
∞
[
n=1
{X ≤ n}.
Since X is real-valued, for every outcome ω ∈ S, the number X(ω) is a real number. Therefore,
there exists some positive integer n such that
X(ω) ≤ n.
Hence every ω ∈ S belongs to at least one of the events B
n
. Therefore,
∞
[
n=1
B
n
= S.
By Theorem 1,
P
∞
[
n=1
B
n
!
= lim
n→∞
P(B
n
).
Thus,
P(S) = lim
n→∞
P(X ≤ n).
Since P(S) = 1, we get
lim
n→∞
F (n) = 1.
This shows that along the sequence x = n,
F (x) → 1.
Since F is increasing and bounded above by 1, this implies the full limit
lim
x→∞
F (x) = 1.
(4) Proof that F is increasing.
Let x, y ∈ R and suppose
x ≤ y.
We claim that
{X ≤ x} ⊆ {X ≤ y}.
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Indeed, if ω ∈ {X ≤ x}, then
X(ω) ≤ x.
Since x ≤ y, we have
X(ω) ≤ y.
Therefore,
ω ∈ {X ≤ y}.
So,
{X ≤ x} ⊆ {X ≤ y}.
By monotonicity of probability,
P(X ≤ x) ≤ P(X ≤ y).
Therefore,
F (x) ≤ F (y).
Hence F is increasing.
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