how to show that the set B δ = { x ∉ D g : ∃ y ∈ S such that d S ( x , y ) &lt; δ but d S ′ ( g ( x ) , g ( y ) ) &gt; ϵ }is measurable with respect to the Borel σ-algebra of S?I changed the notation a bit since the result is stated generally for an a.e. continuous mapping g : S → S ′ between two (separable, I presume) metric spaces, S and S ′ .

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how to show that the set      B          δ        =  {  x  ∉      D          g          :    ∃  y  ∈  S   such that       d          S        (  x  ,  y  )  &amp;lt;  δ   but       d                  S                  ′                      (  g  (  x  )  ,  g  (  y  )  )  &amp;gt;  ϵ  }is measurable with respect to the Borel   σ-algebra of   S?I changed the notation a bit since the result is stated generally for an a.e. continuous mapping   g  :  S  →      S          ′       between two (separable, I presume) metric spaces,   S and       S          ′      .

hopeloothab9m · Accepted Answer

If your space is separable then the distance function on it is measurable with respect to the Borel   σ-algebra on the product space. For a proof see Parthasarathy&#039;s measure and probability book. If we set   G  (  x  ,  y  )  =      d                  S        ′              (  g  (  x  )  ,  g  (  y  )  ) and   H  (  x  ,  y  )  =      d          S        (  x  ,  y  ) then both   G and   H are measurable functions from   S  ×  S to   R. Notice that   W  =      G          −      1        (  0  ,  δ  )  ∩      H          −      1        (  ϵ  ,  ∞  )  ∩      D    g    c    ×  S is measurable in   S  ×  S. Since   S is separable,   W  =  A  ×  B for measurable   A and   B in the borel algebra. Notice that the set A is exactly       B          δ      .

how to show that the set B <mrow class="MJX-TeXAtom-ORD"> δ

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