Theorem 2.1.
is nonsquare if and only if
(a)
;
(b)
is nonsquare.
In order to prove the theorem, we give a lemma.
Lemma 2.2.
If
is nonsquare, then for any
, we have
Proof.
Case 1.
If
, then
or
Case 2.
If
, then
or
This implies
. This completes the proof.
Proof of Theorem 2.1.
-
(a)
Necessity. Suppose that
, then there exist
and
such that
. Pick
such that
is not a null set. Since
, there exist sequence
and disjont subsets
of
such that
Therefore, if we define
, then for any
, we have
This yields
Hence,
. But
, and we deduce that
. Moreover, we have
and
, a contradiction with nonsquareness of
.
If (b) is not true, then there exist
such that
. Pick
such that
. Put
Then we have
It is easy to see
. We know that
Hence, we have
It is easy to see
, a contradiction!
Sufficiency. Suppose that there exists
such that
We will derive a contradiction for each of the following two cases.
Case 1.
. Let
. Hence, we have
Since
, we have
. Hence,
. This implies
, a contradiction!
Case 2.
. By Lemma 2.2, without loss of generality, we may assume that there exists
such that
,
and
. Therefore,
Since
, we have
. Hence,
. This implies
, a contradiction!
Theorem 2.3.
is locally uniformly nonsquare if and only if
(a)
;
(b)
is locally uniformly nonsquare.
In order to prove the theorem, we give a lemma.
Lemma 2.4.
If
is locally uniformly nonsquare, then
(a)For any
,
, we have
(b)If
, then
, where
Proof.
-
(a)
Since
is locally uniformly nonsquare, we have
and
, where
and
In fact, since
is locally uniformly nonsquare, we have
Case 1.
If
, then
or
Case 2.
If
, then
or
Therefore, we get, the following inequality
holds.
(b1) Suppose that
, where
. Then there exist
and subsequence
of
, such that
. By definition of
, there exist
such that
We will derive a contradiction for each of the following two cases.
Case 1.
. Since
, there exists
such that
. Therefore,
This implies
, a contradiction!
Case 2.
. Without loss of generality, we may assume
, where
. Since
, there exists
such that
. Therefore, we have
This implies
Similarly, we have
Therefore, we have
This implies
, a contradiction! Hence,
.
(b2) Suppose that
, where
. Then there exist
and subsequence
of
, such that
. Since
, then there exist
such that
, whenever
. By definition of
, there exist
such that
Therefore, we have
whenever
. Since
, there exists
such that
, where
Hence, we have
This implies
which contradict (2.32). Hence,
.
Combing (b1) with (b2), we get
. This completes the proof.
Proof of Theorem 2.3.
Necessity. By Theorem 2.1,
. If (b) is not true, then there exist
,
such that
and
as
. Pick
such that
. Put
Then we have
It is easy to see
. We know that
Moreover, we have
,
. By the dominated convergence theorem, we have
It is easy to see
,
as
. By Lemma 1.1, we have
and
as
, a contradiction with locally uniform nonsquareness of
.
Sufficiency. Suppose that there exist
,
such that
as
. We will derive a contradiction for each of the following two cases.
Case 1.
There exist
,
such that
, where
. Put
We have
This implies
. Hence,
. We define a function
on
, where
. By Lemma 2.4, we have
-a.e on
. Let
-a.e on
, where
is simple function. Hence,
is
-measurable. By Lemma 2.4, we have
-a.e on
. Then
is
-measurable. Using
we get that there exists
such that
, where
Let
,
,
. It is easy to see
,
and
. If
, by Lemma 2.4, we have
Without loss of generality, we may assume that there exists
such that
Moreover, for any
, we have
Hence, if
, then
Let
. Then
. Therefore,
By Lemma 1.1, we have
,
as
. This is in contradiction with
.
Case 2.
For any
,
, there exists
such that
whenever
. By the Riesz theorem, without loss of generality, we may assume that
a.e on
. Using
we get that there exist
such that
, where
Since
is
-function, we can choose
such that 
. Since 
a.e on
, by the Egorov theorem, there exists
such that
whenever
, where
,
. Next, we will prove that if
, then
In fact, we have
Moreover, we have
By (2.54) and (2.55), we have
This shows that if
, then
It is easy to see
. Therefore,
for
large enough. By Lemma 1.1, we have
, 
as
, which contradicts
, for
large enough. This completes the proof.
Corollary 2.5.
The following statements are equivalent:
(a)
is locally uniformly nonsquare if and only if
is nonsquare;
(b)
is locally uniformly nonsquare.