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mathematic: ½ð±Ò+1, ¡ïÓаïÖú 2015-12-07 06:48:01
mathematic: ½ð±Ò+1, ¡ïÓаïÖú 2015-12-07 07:03:46
Edstrayer: ½ð±Ò+5, LatexÓ¦Öú 2015-12-08 00:16:02
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Let , then for any, there exists and such that

However, for all , by the property of countably additivity of the measure, we have that

Therefore, for any which directly implies that

Since is arbitrary, put and we then complete our first proof.
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2Â¥: Originally posted by junefi at 2015-12-06 09:22:41
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Let \sup{E_1}(A)=\alpha, then for any \varepsilon>0, there exists {B_n} and \bigcup\limits_n {{B_n}}  \subset A such that
\alpha  - \varepsilon  & ...

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2Â¥: Originally posted by junefi at 2015-12-06 09:22:41
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Let \sup{E_1}(A)=\alpha, then for any \varepsilon>0, there exists {B_n} and \bigcup\limits_n {{B_n}}  \subset A such that
\alpha  - \varepsilon  & ...

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3Â¥: Originally posted by mathematic at 2015-12-07 06:46:47
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2Â¥: Originally posted by junefi at 2015-12-06 09:22:41
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Let \sup{E_1}(A)=\alpha, then for any \varepsilon>0, there exists {B_n} and \bigcup\limits_n {{B_n}}  \subset A such that
\alpha  - \varepsilon  & ...

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