More Convergence Concepts
[16/31%]Prove the following simple claims.
Let denote a Cauchy sequence in , where . Show that is a Cauchy sequence in measure: for every and there is such that for every we have .
Let and suppose that , where . If , then and .
While convergence in pth mean implies convergence in measure, it is not the case that convergence in measure implies convergence in pthe mean.
It is possible for a sequence in to converge -a. e. to some but not in pth mean. That is, convergence -a. e. does not force convergence in th mean.
It is possible for a sequence in to converge in pth mean to zero, but converges at no point of .
It is possible to have functions such that but .
Let . There is a sequence in for all with for all , but fails to converge in th mean.
Let , and let denote a sequence in . Suppose that there is and with and . Then -a.e. on .
Suppose that is a sequence in , where . Suppose that with -a.e. on . If converges in measure to some , then and .
If are such that uniformly, then .
Here .
Let , where , and suppose that . Then either of the conditions -a.e. on and in measure is sufficient for the conclusion .
Consider the measure space , where here denotes the Borel subsets of and denotes Lebesgue measure restricted to . Let for each . Then each is in for , -a.e. on and in measure, but when , and hence weakly.
Definition. Consider a generic measure space , and let . Let . We will say that weakly [in ] iff
Here denotes the conjugate exponent of : for , and .
Consider the measure space . Let for every . Then each is in for , uniformly, but weakly in .
Consider the function . We have that for each , is bounded, and for every , while .
Consider the measure space , where here denotes the Borel subsets of and denotes Lebesgue measure restricted to . For each and , define . Then each is in , and weakly in . However, . In fact, we have and -a.e. on .
This exercise might be difficult.
Refer to the Vitali Convergence Theorem. Prove that conditions (i) and (ii) imply that and in pth mean when .
Vitali Convergence Theorem (finite-measure case). Let denote a measure space with , and let . Let denote a sequence of functions in . Let denote an -measurable function on such that -a.e. on and is finite -a.e. on . Then the condition and is equivalent to the dual conditions (i) for every there is [depending on ] with and for every , and (ii) for every there is [depending only on , not ] such that whenever we have for every .
Let denote a measure space with . Let denote a sequence in , and let denote an -measurable function that is finite -a.e. on with -a.e. on . For every , let . Make the following additional assumption (a): for every there is a set [depending on ] such that and for every .
Then the following four statements are logically equivalent:
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and
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If is a nonincreasing sequence of -sets with limit , then for every there is such that forces for every .
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For every there is such that forces for every .
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For every there is such that forces for every .
Vitali Convergence Theorem (finite-measure case). Let denote a measure space with , and let . Let denote a sequence of functions in . Let denote an -measurable function on such that -a.e. on and is finite -a.e. on . Then the condition and is equivalent to the dual conditions (i) for every there is with and for every , and (ii) for every there is such that whenever we have for every .
The following outline is useful. By the Vitali Convergence Theorem and Exercise 8.7.14, it is immediate that (i) and (iv) are logically equivalent.
We show that (i) implies (ii). Let . Find such that implies . For and , Minkowski's Inequality gives
Apply DCT to to show that for large enough , say . Then for every and , we have . For , observe for every that we have
and therefore DCT can be used to obtain (ii).
Now assume (ii), picking . For each , define . Then and on ; from this we have . Next, write for every . We have that and . Choose such that implies for every . For every we have . Therefore, if , then
For , we have , say, and hence , where . Use DCT to derive (iii) in direct fashion.
Now assume (iii). Pick such that forces for every . Argue that implies
Therefore we may deduce (iv) from (iii).
Assume everything in Exercise 8.7.15, including (a). Assume further that (ii) holds. Argue that (i) and (ii) of the Vitali Convergence Theorem follow as consequences.
Vitali Convergence Theorem (finite-measure case). Let denote a measure space with , and let . Let denote a sequence of functions in . Let denote an -measurable function on such that -a.e. on and is finite -a.e. on . Then the condition and is equivalent to the dual conditions (i) for every there is with and for every , and (ii) for every there is such that whenever we have for every .
The particular form of the Vitali Convergence Theorem as presented in this text and the previous two exercises [and the outlines of the solution] owe their existence and form to the most excellent presentation of Hewitt and Stromberg, pp. 203-205.