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Sep 27, 2012 at 12:58 comment added Alexander Pruss I stand corrected about my comment above. I kind of forgot that I also required $\mu((a,b])=b-a$. (My initial thinking about the problem only required $\mu([a,b))=b-a$. I still don't know the answer to that one, either.)
Sep 27, 2012 at 3:36 history edited Joel David Hamkins CC BY-SA 3.0
Fixed error in comment about rho
Sep 26, 2012 at 22:49 history edited Joel David Hamkins CC BY-SA 3.0
added 22 characters in body
Sep 26, 2012 at 22:46 comment added Joel David Hamkins Sean, I agree, and I have edited.
Sep 26, 2012 at 22:43 history edited Joel David Hamkins CC BY-SA 3.0
Simplified the argument via Sean's comment; added 45 characters in body
Sep 26, 2012 at 22:18 comment added Sean Eberhard Perhaps more simply, if $a<b$ then $$\mu(\lbrace a\rbrace) + (b-a) = \mu(\lbrace a\rbrace\cup(a,b]) = \mu([a,b]) = \mu([a,b)\cup\lbrace b\rbrace) = (b-a) + \mu(\lbrace b\rbrace).$$
Sep 26, 2012 at 21:51 history answered Joel David Hamkins CC BY-SA 3.0