Talk:Tight measure

"On a locally compact space $X$ such measures are also outer regular" – really? Even if the space is not separable, like (0,1) times a uncountable discrete space? --Boris Tsirelson (talk) 09:37, 30 November 2012 (CET)

No, if the $\sigma$-algebra generated by the compact sets does not coincide with the Borel $\sigma$-algebra, then there are counterexamples. I missed that the references claiming this fact without assuming separability of the space assume anyway finiteness of the measure. Camillo (talk) 13:05, 30 November 2012 (CET)