Suppose is a -compact Hausdorff space, is a regular Borel measure on , and is a Banach function space containing as a sublattice.
By denote the unique regular Borel measure such that for all .
Lemma.
If
is dense in
, then TFAE.
- is order continuous.
- For each positive function on , for any Borel set .
.
Let
be an element in
and
be a sequence in
convergent to
with respect to the norm on
, then
converges to
.
Assume that
converges to
in order (replaced by a subsequence otherwise),
then there is an increasing positive sequence
and a decreasing positive sequence
in
that are both convergent to
in order and satisfy
.
Since
is an order ideal of
,
converges to
with respect to
and thus with respect to
by the assumption in (2).
Therefore, By the arbitrariness of
,
Now we have hence the equation (1) holds for all .
Suppose
is a monotone order bounded sequence in
, then
increases or decreases to some
in
almost everywhere with respect to
.
Thanks to Dini’s Theorem,
converges to
with respect to the norm on
, i.e.,
is order continuous.
.
Suppose
,
is a compact subset of
, and
.
Let
be a sequence of open sets containing
such that
converges to 0. We can find a decreasing sequence
in
such that
for all
.
Replacing with this sequence, we can assume
decreases to
almost everywhere with respect to
. Hence,
decreases to
in order.
Since
is order continuous,
converges to
weakly, that is,
converges to
.
However,
decreases to
almost everywhere with respect to
, which implies that
converges to
, so
. As
is
-compact,
.
Ignore the old version.
Theorem. Suppose is a locally compact space, is a Dedekind complete Banach lattice which consists of Borel measurable functions on and contains as a subspace.
For any compact subset of , the inclusion map is continuous.
For any decreasing sequence in , converges to with respect to order is equivalent to converges to with respect to the norm on .
The norm on is order continuous if is dense in .
Proof.
(a) For each
,
Let , then for all . Suppose is a positive linear functional on , then is a positive linear functional on where is the inclusion map from to . Since there is a unique regular Borel measure on such that converges to by the Dominated Convergence Theorem. Hence, -converges to . By Dini’s Theorem, converges to 0 with respect to the norm on .
Suppose is a decreasing sequence in and is a sequence in such that . Let , then is a decreasing sequence in such that . So, there is a subsequence and a positive element such that . Since , converges to in order and so does . Hence converges to 0 in norm by (a). The proof ends with converging to in norm.