Let X denote the set of equivalence classes of Lebesgue measurable subsets A ⊂ [ 0 , 1 ] under the equivalence relation: A ∼ B iff μ ( A Δ B ) = 0.If [ A ] , [ B ] ∈ X, set d ( [ A ] , [ B ] ) = μ ( A Δ B ).Now, how can we prove that (X, d) is a separable metric space?It is straightforward to show that d ( [ A ] , [ B ] ) = μ ( A Δ B ) induces a metric space, but how can we show there exists a countable subset of X which is dense in X?

Question

Let   X denote the set of equivalence classes of Lebesgue measurable subsets   A  ⊂  [  0  ,  1  ] under the equivalence relation:  A  ∼  B iff   μ  (  A  Δ  B  )  =  0.If   [  A  ]  ,  [  B  ]  ∈  X, set   d  (  [  A  ]  ,  [  B  ]  )  =  μ  (  A  Δ  B  ).Now, how can we prove that (X, d) is a separable metric space?It is straightforward to show that   d  (  [  A  ]  ,  [  B  ]  )  =  μ  (  A  Δ  B  ) induces a metric space, but how can we show there exists a countable subset of   X which is dense in   X?

Meadow Knox · Accepted Answer

This comes either from the regularity of the Lebesgue measure or from its definition. For any Lebesgue mesurable set   A and any   ε  &amp;gt;  0, there is a finite union of intervals       I    1  ,...,       I    n   such that  μ  (  A  Δ  (  ⋃      I    j    )  )  &amp;lt;  ε  .We may approximate each of these intervals with intervals with rational endpoints. So the countable set you are looking for is the set of subsets of [0,1] which are finite unions of intervals with rational endpoints.

Let X denote the set of equivalence classes of Lebesgue measurable subsets A ⊂<!

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