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http://planetmath.org/anysigmafinitemeasureisequivalenttoaprobabilitymeasure http://planetmath.org/anysigmafinitemeasureisequivalenttoapr...

• panic 11y ago
  Yes, but the probability measure they construct doesn't give "equal probability" in the sense that Xcelerate probably meant. Going back to the plane example, if we use the unit squares as our sequence A1, A2, ..., we'll get P(A1) = 1/2, P(A2) = 1/4, P(A3) = 1/8, and so on. This sums to 1 as you'd expect, but it's not uniform!
• apoklasdjasd 11y ago
  Note that the OP desired that "each point has an equal probability of being chosen". The infinite plane being infinite, this requirement does not formally make sense, but a natural way to interpret it is to ask for invariance under isometries (or translation-invariance). As the post that you replied to indicates, this is not possible for the Euclidean plane. The measure constructed by the lemma that you cite does not have this property.
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  • alsdkalsjdalkd 11y ago
    Lebesgue measure on R^n is sigma-finite.
• enupten 11y ago
  Yes, I think the above is a fairly useless digression. Thank you for making me think more about this.