Let we have $X=R$ the set of real numbers and $B_R$ is borel algebra and $2^X=P(X)$ How can I prove that $B_R$ is not equal to $P(X)$
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In other words you need to show there are subsets of $\mathbb R$ that are not Borel sets. I don't think that's particularly easy to do. What level class are you taking? – Gregory Grant Apr 02 '15 at 21:47
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There's an example on this wiki page of a set that's not Borel: http://en.wikipedia.org/wiki/Borel_set#Non-Borel_sets – Gregory Grant Apr 02 '15 at 21:48
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Have you searched the site for the many users who have asked exactly this before? – Asaf Karagila Apr 02 '15 at 22:17
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If extending pre-measures is fair game, it would suffice to show that there doesn't exist a measure on $P(\mathbb{R})$ which assigns to each interval its length. There's a fairly simple proof for that. – Jonathan Y. Apr 02 '15 at 22:21
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@Jonathan: Not even a week ago, http://math.stackexchange.com/questions/1209126/borel-sigma-algebra-not-containing-all-subsets-of-mathbbr and while I don't mind duplicates, searching for them is only part of the work I expect someone posting a question to put into the question. And here practically no work was put into the question. – Asaf Karagila Apr 02 '15 at 22:37
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@Jonathan: I found it by looking through the first three pages (set to 50 questions per page) of the measure theory tag. Not an unreasonable effort. – Asaf Karagila Apr 02 '15 at 22:49
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@AsafKaragila can we agree the question at least deserves a reference to an answer? Even if OP didn't quite present his/her attempts at solving it? – Jonathan Y. Apr 02 '15 at 22:52