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I'm trying to construct a finite measure defined on Borel sigma algebra of $[0,1]$ such that every nonempty set has nonzero measure.

Can anybody give me a help to find out such measure?

Thanks in advance.

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  • $\begingroup$ Your title says "non-negative", your question says "nonzero". Which did you mean? $\endgroup$ – Ben Voigt Sep 11 '15 at 21:51
  • $\begingroup$ Non-zero. I edited. Thanks. $\endgroup$ – Guldam Sep 12 '15 at 4:44
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Such a measure doesn't exist.

Assume that $\mu([0, 1]) = M < \infty$. Then the cardinality of $T_n = \{x \in [0, 1] \mid \mu(\{x\}) > \frac{1}{n}\}$ is at most $Mn$ and therefore finite. Since every nonempty set is supposed to have a positive measure, the identity $[0, 1] = \bigcup \limits_{n \in \mathbb{N}} T_n$ holds. But this is impossible, since $[0, 1]$ is uncountable and the countable union of finite sets is at most countable.

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    $\begingroup$ It even follows, that such a measure can't be $\sigma$-finite at all. $\endgroup$ – Dominic Michaelis Sep 11 '15 at 19:19
  • $\begingroup$ +1, but: Possibly not the best choice to use $M$ as a real number and $M_n$ as sets. $\endgroup$ – Thomas Andrews Sep 11 '15 at 19:36
  • $\begingroup$ @ThomasAndrews You are right, I've changed $M_n$ to $T_n$. Thanks for the suggestion! $\endgroup$ – Dominik Sep 11 '15 at 19:38
  • $\begingroup$ Quite illuminating. Thanks! $\endgroup$ – Guldam Sep 11 '15 at 19:46

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