2.11 Markov Processes on Metric Spaces 195
Example 11.4 Let S ={m/(m + 1): m = 0, 1,...}∪{1}. Define p(m/
(m + 1), {(m + 1)/(m + 2)}) = 1 ∀m = 0, 1, 2,...,and p(1, {0}) = 1.
Then there does not exist any invariant probability, although S is com-
pact (Exercise). This is, of course, the case of an extremely degener-
ate transition probability given by a deterministic dynamical system:
f (m/m + 1) = (m + 1)/(m + 2) (m = 0, 1, 2,...), f (1) = 0. One may,
however, easily perturb this example to construct one in which the
transition probability is not degenerate. For example, let p(m/
(m +
1), {(m + 1)/(m + 2)}) = θ, p(m/(m + 1); {(m + 2)/(m + 3)}) = 1 −
θ (m = 0, 1, 2,...), p(1, {0}) = θ, p(1, {1/2}) = 1 − θ, for some θ ∈
(0, 1). One may show that there does not exist an invariant probability in
this case as well (Exercise).
It may be noted that the results on the existence and uniqueness of
an invariant probability in Section 2.9 did not require topological as-
sumptions. However, the assumptions were strong enough to imply the
convergence of the sequence (1/n)
n−1
r=0
p
(r)
(x, B) to a probability π (B)
uniformly for all B ∈ S. The existence of an invariant probability may be
demonstrated without the assumption of uniform convergence, as shown
below.
Proposition 11.1 Let (S, S) be a measurable state space. If for some
x ∈ S, there exist a probability measure π
x
and a sequence 1 ≤ n
1
<
n
2
< ··· such that
1
n
k
n
k
−1
r=0
p
(r)
(x, B) →
k→∞
π
x
(B) ∀B ∈ S, (11.10)
then π
x
is an invariant probability for p. If, in addition, π
x
does not
depend on x, then it is the unique invariant probability.
Proof. By standard measure theoretic arguments for the approximation
of bounded measurable functions by the so-called simple functions of
the form
m
i=1
a
i
1
A
i
(A
1
,...,A
m
disjoint sets in S, and a
1
, a
2
,...,a
m
nonzero reals), (11.10) may be shown to be equivalent to
1
n
k
n
k
−1
r=0
(T
r
f )(x) →
k→∞
"
f (y)π
x
(dy) ∀ f ∈ B(S). (11.11)
Now follow the same line of argument as used in the proof of Theorem
11.1, but for all f ∈ B(S).