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Hmm. So the Riemann hypothesis requires Λ to be <= 0. Now they’ve showed that Λ >= 0. Since the intersection of the two ranges is non-measurable, probabilistica

Hmm. So the Riemann hypothesis requires Λ to be <= 0. Now they’ve showed that Λ >= 0. Since the intersection of the two ranges is non-measurable, probabilistically the Riemann hypothesis must be false! :^)

nimish 9y ago
{0} is measurable (It's a borel set) with the standard lebesgue measure and has measure 0. Of course, with some other measure it might not be measurable, but that would be a weird measure.
- RBerenguel 9y ago
  Been thinking for a while how such a measure would be defined, but my measure theory abilities are a bit rusty (heck, unused for 10 years already, time flies). I'd bet it's impossible (in \mbb{R})... and sadly I'll now keep thinking about it until I remember how to prove for sure it or fall asleep, thanks :D
  - deleted 9y ago
    [deleted]
  - vostok 9y ago
    The Lebesgue measure is not defined for all subsets of R. It is defined on the Lebesgue sigma algebra which is larger than the Borel sigma algebra. You could instead choose a sigma algebra that does not contain {0}. Trivially, you could choose the standard Lebesgue measure restricted to {{}, R}.
    - RBerenguel 9y ago
      Oh well, I was thinking of a measure where 0 was in some Borel set of it, otherwise it loses a bit the fun of it
      - vostok 9y ago
        In this case, 0 is contained in R which is a Borel set and is part of the sigma algebra on which the measure is defined.
        RBerenguel 9y ago
        I mean, one with a more rich sigma algebra.
- contravariant 9y ago
  You could identify 0 with some other point to get a topology where {0} is not a Borel set anymore. A somewhat more useful example would be to identify x with x+1 for all x (or equivalently use the topology on R induced by the map t -> e^2πit from R to S^1), which results in a topology where only periodic functions are continuous. For a complete measure you get the additional requirement that 0 isn't allowed to be contained in any null-set.
- tzahola 9y ago
  You’re right. I should have used the term null set.

nimish 9y ago

{0} is measurable (It's a borel set) with the standard lebesgue measure and has measure 0. Of course, with some other measure it might not be measurable, but that would be a weird measure.

RBerenguel 9y ago
Been thinking for a while how such a measure would be defined, but my measure theory abilities are a bit rusty (heck, unused for 10 years already, time flies). I'd bet it's impossible (in \mbb{R})... and sadly I'll now keep thinking about it until I remember how to prove for sure it or fall asleep, thanks :D
- deleted 9y ago
  [deleted]
- vostok 9y ago
  The Lebesgue measure is not defined for all subsets of R. It is defined on the Lebesgue sigma algebra which is larger than the Borel sigma algebra. You could instead choose a sigma algebra that does not contain {0}. Trivially, you could choose the standard Lebesgue measure restricted to {{}, R}.
  - RBerenguel 9y ago
    Oh well, I was thinking of a measure where 0 was in some Borel set of it, otherwise it loses a bit the fun of it
    - vostok 9y ago
      In this case, 0 is contained in R which is a Borel set and is part of the sigma algebra on which the measure is defined.
      - RBerenguel 9y ago
        I mean, one with a more rich sigma algebra.
contravariant 9y ago
You could identify 0 with some other point to get a topology where {0} is not a Borel set anymore. A somewhat more useful example would be to identify x with x+1 for all x (or equivalently use the topology on R induced by the map t -> e^2πit from R to S^1), which results in a topology where only periodic functions are continuous. For a complete measure you get the additional requirement that 0 isn't allowed to be contained in any null-set.
tzahola 9y ago
You’re right. I should have used the term null set.