I have this (I thought easy) problem:Let K be a compact Hausdorff space and μ a regular Borel probability measure on K. Show that: supp ⁡ μ = ⋂ { A ⊆ K : A is closed and μ ( A ) = 1 }Here, supp ⁡ μ = { x ∈ K : ∀ open U ∋ x , μ ( U ) &gt; 0 }, the so-called topological support of μ.Clearly, calling the RHS of the above L, if x ∈ K ∖ L then there is an open neighbourhood of x disjoint from L, which by measure monotonicity implies the measure of that neighbourhood is 0 as μ ( L ) = 1 , μ ( K ∖ L ) = 0, so x ∉ supp ⁡ μ. Likewise, if x ∉ supp ⁡ μ, then one neighbourhood of x is null, so there is at least one closed set disjoint from x with measure 1. Thus x is not in the intersection of such closed sets, x ∉ L. Easily we have supp ⁡ μ = L.Why did the textbook assume regularity?

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I have this (I thought easy) problem:Let   K be a compact Hausdorff space and   μ a regular Borel probability measure on   K. Show that:  supp  ⁡  μ  =  ⋂  {  A  ⊆  K  :  A   is closed and   μ  (  A  )  =  1  }Here,   supp  ⁡  μ  =  {  x  ∈  K  :  ∀   open   U  ∋  x  ,    μ  (  U  )  &amp;gt;  0  }, the so-called topological support of   μ.Clearly, calling the RHS of the above   L, if   x  ∈  K  ∖  L then there is an open neighbourhood of   x disjoint from   L, which by measure monotonicity implies the measure of that neighbourhood is 0 as   μ  (  L  )  =  1  ,    μ  (  K  ∖  L  )  =  0, so   x  ∉  supp  ⁡  μ. Likewise, if   x  ∉  supp  ⁡  μ, then one neighbourhood of   x is null, so there is at least one closed set disjoint from   x with measure 1. Thus   x is not in the intersection of such closed sets,   x  ∉  L. Easily we have   supp  ⁡  μ  =  L.Why did the textbook assume regularity?

aletantas1x · Accepted Answer

I believe you&#039;re correct that regularity isn&#039;t needed to prove the identity. However, I also believe you&#039;ve inadvertently managed to slip an assumption of regularity into your argument for       x  ∉  L  ⇒    x  ∉  supp  ⁡  μ     when you assert that       μ  (  L  )  =  1    , which doesn&#039;t seem to be necessarily true in general. I suspect the reason why your textbook assumes regularity is precisely to ensure that       μ  (  L  )  =  1    .Nevertheless, if       x  ∉  L then       x  ∉  A     for some closed       A     with       μ  (  A  )  =  1    . Therefore       x  ∈  K  ∖  A    ,       K  ∖  A     is open, and       μ  (  K  ∖  A  )  =  0    , so your proof can be made to work without assuming      μ  (  L  )  =  1    .

I have this (I thought easy) problem: Let K be a compact Hausdorff space and μ<!-- μ

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