Let B([0,1]) be the borel set on [0,1] and suppose that μ is a finite measure on B([0,1]). Then, define A = [ 0 , 1 ] ∩ { x ; μ ( { x } ) &gt; 0 } . I have shown that if A is countable, then [0,1]∖A is a separable metric space (because it is a subset of a separable metric space). Now if ( x n ) n is a dense sequence in [0,1]∖A, then I want to show that it is also dense in [0,1]. How can I prove it?

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Let B([0,1]) be the borel set on [0,1] and suppose that   μ is a finite measure on B([0,1]). Then, define   A  =  [  0  ,  1  ]  ∩      {    x    ;    μ    (          {      x      }        )    &amp;gt;    0    }  . I have shown that if A is countable, then [0,1]∖A is a separable metric space (because it is a subset of a separable metric space). Now if   (      x          n            )          n       is a dense sequence in [0,1]∖A, then I want to show that it is also dense in [0,1]. How can I prove it?

persstemc1 · Accepted Answer

If I=(a,b) is an open interval in [0,1] then I∖A is uncountable so in particular a non-empty open subset of [0,1]∖A and so contains some       x    n   etc.

Raul Walker · Accepted Answer

The complement of a countable subset of       R   is dense. Thus, [0,1]∖A is dense in [0,1].The rest follows from the transitivity of the denseness property.

Let B([0,1]) be the borel set on [0,1] and suppose that μ is a finite measure

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