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Why can't we use the power set as the sigma-algebra for a continuous uniform distribution?

We cannot use the power set as the sigma-algebra for a continuous uniform distribution on [0,1][0,1] because there is no function defined on all subsets of [0,1][0,1] that can simultaneously satisfy the length property, countable additivity, and translation invariance (assuming the axiom of choice).

Conditions

  • The sample space is Ω=[0,1]\Omega = [0,1].
  • We assume the axiom of choice.
  • The measure must satisfy length, countable additivity, and translation invariance.

Reasoning, step by step

  1. Define the desired properties for a uniform measure μ\mu on [0,1][0,1]: μ((a,b])=b−a\mu((a,b]) = b-a, countable additivity, and translation invariance.
  2. Attempt to define such a measure on the power set P([0,1])\mathcal{P}([0,1]).
  3. Recognize the mathematical fact that no such function exists for all subsets when the axiom of choice is assumed.
  4. Conclude that the power set is not a suitable sigma-algebra for this continuous setting.
  5. Restrict attention to a smaller sigma-algebra, such as the Borel sigma-algebra, to resolve the issue.

Example

The video states as a fact that there is no function μ\mu defined on all A⊆[0,1]A \subseteq [0,1] which satisfies the three properties of a uniform distribution, at least not if we assume the axiom of choice.

Common misconceptions

  • Assuming that because the power set works for discrete spaces, it must also work for continuous spaces.
  • Believing that translation invariance and countable additivity can always be satisfied for any subset of real numbers.

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