This is a sufficient condition for something that resembles “Weak convergence”: for all
Suppose that a.e.\ and that
,
. If
, we have
for all
,
. Note that the result is false if
.
Proof:
(Case: , where
is the domain of integration).
We may assume ,
,
otherwise the result is trivially true. Also, by Fatou’s Lemma,
Let . Since
, so
and there exists
such that for any measurable subset
with
,
.
Since a.e.\ (
is finite a.e.\ since
), by Egorov’s Theorem there exists closed
such that
and
converge uniformly to
on
. That is, there exists
such that for
,
for all
.
Then for ,
Since is arbitrary, this means
.
(Case: ). Error: See correction below.
Define , where
is the ball with radius
centered at the origin. Then
, so there exists
such that for
,
.
Since on
, by Monotone Convergence Theorem,
Thus there exists such that for
,
.
Then for ,
so that .
(Show that the result is false if ).
Let . Then
a.e., where
. Note that
,
so that
. Similarly,
.
However if ,
for all
but
.
Correction for the case :
Define , where
is the ball with radius
centered at the origin.
Since on
, by Monotone Convergence Theorem,
Thus there exists such that
.
Since , by the finite measure case there exists
such that for
,
So for ,
so that .
One thought on “Sufficient condition for “Weak Convergence””