Let f ( x ) = sup r &gt; 0 μ ( B r ( x ) ) m ( B r ( x ) ) . Prove it is B ( R d ) measurable. μ is a measure on B ( R d ) such that μ ( B ( R d ) ) is finite, and m is the Lebesgue measure.My thoughts: suppose x ∈ f &lt; a , then there exists ϵ s.t. ( 1 + ϵ ) f ( x ) &lt; a. So given some δ &gt; 0 for any x ′ ∈ B δ ( x ) we have μ ( B r ( x ′ ) ) m ( B r ( x ′ ) ) ≤ μ ( B r + δ ( x ) ) m ( B r ( x ′ ) ) = μ ( B r + δ ( x ) ) m ( B r ( x ) ) = μ ( B r + δ ( x ) ) m ( B r + δ ( x ) ) ( r + δ r ) d ≤ f ( x ) ( r + δ r ) d but I need δ to be uniform in order to prove { f &lt; a } is open and I don't know how to treat it when r is very small.

Question

Let   f  (  x  )  =      sup          r      &amp;gt;      0                  μ      (              B        r            (      x      )      )              m      (              B        r            (      x      )      )      . Prove it is       B    (            R        d    ) measurable.   μ is a measure on       B    (            R        d    ) such that   μ  (      B    (            R        d    )  ) is finite, and m is the Lebesgue measure.My thoughts: suppose   x  ∈      f    &amp;lt;    a  , then there exists   ϵ s.t.   (  1  +  ϵ  )  f  (  x  )  &amp;lt;  a. So given some   δ  &amp;gt;  0 for any       x    ′    ∈      B          δ        (  x  ) we have            μ      (              B        r            (              x        ′            )      )              m      (              B        r            (              x        ′            )      )        ≤            μ      (              B                  r          +          δ                    (      x      )      )              m      (              B        r            (              x        ′            )      )        =            μ      (              B                  r          +          δ                    (      x      )      )              m      (              B        r            (      x      )      )        =            μ      (              B                  r          +          δ                    (      x      )      )              m      (              B                  r          +          δ                    (      x      )      )        (            r      +      δ        r        )    d    ≤  f  (  x  )  (            r      +      δ        r        )    d  but I need   δ to be uniform in order to prove   {  f  &amp;lt;  a  } is open and I don&#039;t know how to treat it when r is very small.

Kayley Jackson · Accepted Answer

Fix an   r  &amp;gt;  0.                    x        →                                            μ              (                              B                                  r                                            (              x              )              )                                      m              (                              B                                  r                                            (              x              )              )                                          is a lower semi-continuous function. Then, f as the supremum of a class of lower semi-continuous functions, is still a lower semi-continuous function, and hence a Borel function.

Let f ( x ) = <munder> <mo movablelimits="true" form="prefix">sup <mrow

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