Royden real analysis 問題求助

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這個引理
Let E be a bounded measurable set of real numbers.
Suppose there is a bounded, countably infinite set of real numbers Λ for which the collection of translates of E,
{λ + E} λ∈Λ is disjoint. Then m( E) = o.

將 Λ 條件分別改為以下三種 ,結論仍然成立嗎
1. Λ 是 有界可數有限
2. Λ 是 有界不可數無限
3. Λ 是 無界可數無限

這是課本的題目,只是想不出來該如何去推,會的人麻煩幫個忙

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关键词：Analysis Analysi alysis Analys Royden collection infinite numbers 如何

are you sure there is no typo?
1. how come you have a set of real numbers that is "unbounded and (countably) finite"? please note that finiteness implies boundedness. are you talking about "unbounded and countably infinite set \Lambda"?
2. what is the difference between "Condition 3. Λ 是 有界可數無限" and "a bounded, countably infinite set of real numbers Λ"? they should be the same except that one is in Chinese and the other in English, right? so do you really mean "bounded and uncountably infinite"?

I am assuming that you are talking about Lebesgue measure (denoted by m as you have suggested). (Actually the lemma should hold for any Lebesgue-Stieltjes measure on R, i.e. bounded on compact subsets and translation invariant.) The lemma itself is easy to show, and if you read the proof, you should notice that boundedness is necessary given countably infinity. So if Condition 1, the conclusion is not true. Indeed, consider E=(0,1), the unit interval in R, where E is bounded measurable with m(E)=1 \neq 0. Let \Lambda = Z, the set of all integers. Obviously \Lambda is unbounded and countably infinite, while \lambda+E = (\lambda, \lambda+1) are disjoint for distinct \lambda's. Thus, replacing the original condition with Condition 1 does not yield the same conclusion.
Let's look at Condition 3 first. It is easy that Condition 3 implies the original condition. Simply pick a countably infinite subset A of \Lambda (which always exists), and A is automatically bounded. Also by the property of \Lambda, the collection {a+E: a \in A} is pairwise disjoint as well. Hence by the lemma, (existence of) A implies m(E)=0.
Finally, let's show that Condition 2 implies Condition 3. Suppose that \Lambda is uncountable, then there is one n>0 such that the intersection of (-n,n) and \Lambda is uncountable. Otherwise, if each intersection of (-n,n) and \Lambda were countable, then \Lambda would be countable as a countable union of countable sets, contradicting the assumption (Condition 2). Hence Condition 3 is true given Condition 2. From above, m(E)=0.

In summary, only Condition 1 does not yield the same conclusion.

KevinOu 发表于 2012-12-18 02:46
are you sure there is no typo?
1. how come you have a set of real numbers that is "unbounded and (c ...

非常抱歉,我在編輯的時候反而給他打相反了,1 2都是有界的 只有3是無界的
1. Λ 是 有界可數有限
2. Λ 是 有界不可數無限
3. Λ 是 無界可數無限


Λ = Z 是條件3的,這個我看懂了, 非常感謝你的幫忙

CrazyDavis 发表于 2012-12-18 07:07
非常抱歉,我在編輯的時候反而給他打相反了,1 2都是有界的 只有3是無界的
1. Λ 是 有界可數有限[/backco ...

If that is the case, then only Condition 2 yields the same conclusion, since you can always find a countably infinite subset as mentioned in a previous post. (Indeed, all you need is "infinity" here, or in other words, "as many as you wish", no matter "countably" or not. Please note that "countable infinity" is the "minimal infinity", so "all infinities" "contain" countable infinity.)

For Condition 1, m(E) may not be zero. This is rather trivial and I believe you know it (what if \Lambda is "the simplest finite" set?)

恩, 條件1 與課本上的證明差不多, 只差在原本引理可以使用矛盾證法證出m(E) = 0

條件2的意思看懂了,可是我不明白為什麼最後可以得出結論m(E) = 0.

Maybe I should have listed the arguments line by line instead of clustering them together. Let me be clear.

E is a fixed bounded measurable set of reals.

Assume that \Lambda is
i) bounded
ii) (uncountably) infinite
iii) such that the collection of sets {a+E: a \in \Lambda} is pairwise disjoint

Then one can find a subset A of \Lambda such that A is countably infinite.

But A, as a subset of the bounded set \Lambda, must be bounded, too.

Also, {a+E: a \in A} is pairwise disjoint, since {a+E: a \in A} is a sub-collection of {a+E: a \in \Lambda}, which is pairwise disjoint.

Now, A satisfies the condition of the Lemma:
i) A is bounded;
ii) A is countably infinite;
iii) {a+E: a \in A} is pairwise disjoint.

Apply the Lemma to A (A is \Lambda in the Lemma), m(E)=0. This completes the proof (using the Lemma).


If you think twice what countable summation from one to infinity really means, you will find that the proof of the Lemma relies on the concept "as many as possible" but not on countability. In other words, you only need infinity (given boundedness) to prove the original lemma.

A final comment of the original lemma:
a) boundedness of E and \Lambda is to uniformly control for any natural number n,
   n*m(E)
= sum m(a_i + E) for 1<=i<=n             [by translation-invariant]
= m(union of {a_i + E} for 1<=i<=n)    [by (finite) additivity]
<= m(\Lambda + E) = C (a constant)    [by monotonicity and boundedness]
< \infinity

b) infinity (no matter countable or not) makes "taking n to infinity" valid for the following:
    m(E) <= C/n    [inequality from above]

KevinOu 发表于 2012-12-18 11:34
Maybe I should have listed the arguments line by line instead of clustering them together. Let me be ...

總算清楚了,非常感謝你的熱心幫忙~