Editing Outer measure (section)

==Outer measures==
Given a set <math>X,</math> let <math>2^X</math> denote the [[Power set|collection of all subsets]] of <math>X,</math> including the [[empty set]] <math>\varnothing.</math> An '''outer measure''' on <math>X</math> is a [[set function]]
<math display=block>\mu: 2^X \to [0, \infty]</math>
such that 
* {{em|null empty set}}: <math>\mu(\varnothing) = 0</math>
* {{em|countably subadditive}}: for arbitrary subsets <math>A, B_1, B_2, \ldots</math> of <math>X,</math><math display=block>\text{if } A \subseteq \bigcup_{j=1}^\infty B_j \text{ then } \mu(A) \leq \sum_{j=1}^\infty \mu(B_j).</math>
Note that there is no subtlety about infinite summation in this definition. Since the summands are all assumed to be nonnegative, the sequence of partial sums could only diverge by increasing without bound. So the infinite sum appearing in the definition will always be a well-defined element of <math>[0, \infty].</math> If, instead, an outer measure were allowed to take negative values, its definition would have to be modified to take into account the possibility of non-convergent infinite sums.

'''An alternative and equivalent definition.'''<ref>The original definition given above follows the widely cited texts of Federer and of Evans and Gariepy. Note that both of these books use non-standard terminology in defining a "measure" to be what is here called an "outer measure."</ref> Some textbooks, such as Halmos (1950) and Folland (1999), instead define an outer measure on <math>X</math> to be a function <math>\mu: 2^X \to [0, \infty]</math> such that
* {{em|null empty set}}: <math>\mu(\varnothing) = 0</math>
* {{em|monotone}}: if <math>A</math> and <math>B</math> are subsets of <math>X</math> with <math>A \subseteq B,</math> then <math>\mu(A) \leq \mu(B)</math>
* for arbitrary subsets <math>B_1, B_2, \ldots</math> of <math>X,</math><math display=block>\mu\left(\bigcup_{j=1}^\infty B_j\right) \leq \sum_{j=1}^\infty \mu(B_j).</math>
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| '''Proof of equivalence.'''
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|Suppose that <math>\mu</math> is an outer measure in sense originally given above. If <math>A</math> and <math>B</math> are subsets of <math>X</math> with <math>A \subseteq B,</math> then by appealing to the definition with <math>B_1 = B</math> and <math>B_j = \varnothing</math> for all <math>j \geq 2,</math> one finds that <math>\mu(A) \leq \mu(B).</math> The third condition in the alternative definition is immediate from the trivial observation that <math>\cup_j B_j \subseteq \cup_j B_j.</math>

Suppose instead that <math>\mu</math> is an outer measure in the alternative definition. Let <math>A, B_1, B_2, \ldots</math> be arbitrary subsets of <math>X,</math> and suppose that 
<math display=block>A \subseteq \bigcup_{j=1}^\infty B_j.</math>
One then has
<math display=block>\mu(A) \leq \mu\left(\bigcup_{j=1}^\infty B_j\right) \leq \sum_{j=1}^\infty\mu(B_j),</math>
with the first inequality following from the second condition in the alternative definition, and the second inequality following from the third condition in the alternative definition. So <math>\mu</math> is an outer measure in the sense of the original definition.
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