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.