Here we prove the following important result.
Equation (
) is an informal way of stating
the required continuity assumption. To state it more
formally, let 
denote the ``law'' or
distribution of X. That is, if 
is a Borel set, the 
. Thinking
of X as a map from the underlying probability space
, then 
where P denotes
the underlying probability measure on . Thus,
it is common to write 
= 
. Then
we may restate () as
The proof of Theorem will be easy with
the following result.
Proof.
Suppose 
. Let 
be any subsequence.
Then 
also, i.e., 
,
as 
.
Therefore, for each 
, we can find an
such that
Let 
.
Thus, for any 
,
Thus,
which implies
Restated in terms of quantifiers,
and taking complements:
The latter implies 
.
Conversely, suppose 
does not converge in probability
to X, i.e. for some 
,
Since there is always a subsequence which has the lim sup as
its limit, there is a subsequence 
such that
Recall that convergence a.s. implies convergence in probability,
so if any subsequence satisfied 
,
then 
, but that can't happen in view of ().
It is clear then that no subsequence of 
can converge a.s. to
X.
We claim and 
implies 
. To see this, let
denote the underlying probability space, and let
Note that the set of such 
's has probability 1.
Then the sequence of vectors 
so by the definition of continuity
.
Since such 's
make up almost all of , the claim follows.
Now apply Theorem to complete the proof.
