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这里，我们使用Carathéodory’s criterion来证明这个结论。勒贝格可测， 我们需要证明下面的等式对所有的。接下来我们就需要证明。

本证明是一个简要的证明。这两个结论这里没有直接给出。因为如果给出这两个结论的证明就需要更多篇幅。其中1和3需要读者自行查找资料证明。NOTE: 所谓集合的处处收敛，是说得他们的特征函数处处收敛.

现在我们可以来讨论，如果两个集合不相交且两个集合都是闭集的时候，并不能说明两个集合不可分。也就是说即便两个集合都是闭集且不相交，他们也可能不可分。或者我们用一种不太严谨的说法，两个集合的并集没有缝隙，那么就说明两个集合不可分。而这个缝隙的数学含义就是。本文就构造出两个例子，一个是在有理数集合里面构造出两个集合， 一个是在。如果一个集合的聚点都属于这个集合本身吗，那么这个集合是一个闭集。也就是说这两个例子都说明，即便两个都是闭集且交集为空，这两个集合依然可能是不可分。同理，这两个集合也是闭集，且交集为空集。

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Let X be a topological vector space. If A⊂XA\subset XA⊂X and B⊂XB\subset XB⊂X, then A‾+B‾⊂A+B‾\overline{A}+\overline{B}\subset \overline{A+B}A+B⊂A+B​.Take a∈A‾a\in \overline{A}a∈A, b∈B‾b\in \overline{B}b∈B; Let WWW be a neighborhood of a+ba+ba+b.There ar

a subset of a zero(0) Lebesgue measure set is Lebesgue measurable

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Definition Suppose SSS is an ordered set, E⊂SE\subset SE⊂S, and EEE isbounded above. Supose there exists an α∈S\alpha\in Sα∈S with the followingproperties:α\alphaα is an upper bound of EEEIf γ<α\gamma<\alphaγ<α then γ\gammaγ is not an uppe

As all we know that Expanding natural numbers N\mathbb{N}N a little bityields integers Z\mathbb{Z}Z. Similarily, Expanding integersZ\mathbb{Z}Z a little bit generates Q\mathbb{Q}Q. But, what about realnumbers R\mathbb{R}R ? How to get real numbers R\mat

What is the foundation of differential calculus? We will discover itfrom one of the most important theorem in differential calculus whichcalled Mean-Value theorem. In order to prove this theorem, we only needseveral steps.THEOREM 1—The Extreme Value

UTF8gbsn多约束的拉格朗日乘子问题.f(x)h1(x)=0⋮hm(x)=0\left. \begin{aligned} \quad & f(x)\\ \quad& h_1(x)=0\\ & \quad \quad \vdots\\ & h_m(x)=0 \end{aligned} \right.​f(x)h1​(x)=0⋮hm​(x)=0​假设这个问题的解是x∗x^{*}x∗.

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