I have somehow confused myself with this fairly straightforward proof. We need to show that λ f is a measurable function on ( S , S ), i.e. that for any c ∈ R : { s ∈ S : f ( s ) ≤ c } ⊂ S. If λ ≠ 0, then the claim follows immediately from the measurability of f, namely as c / λ ∈ R it follows that { s ∈ S : f ( s ) ≤ c / λ } = { s ∈ S : λ f ( s ) ≤ c }But then, if λ = 0 , λ f = 0 and I am not sure how to convince myself that λ f is measurable.

Question

I have somehow confused myself with this fairly straightforward proof. We need to show that   λ  f is a measurable function on   (  S  ,      S    ), i.e. that for any   c  ∈      R    :  {  s  ∈  S  :  f  (  s  )  ≤  c  }  ⊂  S. If   λ  ≠  0, then the claim follows immediately from the measurability of   f, namely as   c      /    λ  ∈      R   it follows that   {  s  ∈  S  :  f  (  s  )  ≤  c      /    λ  }  =  {  s  ∈  S  :  λ  f  (  s  )  ≤  c  }But then, if   λ  =  0  ,  λ  f  =  0 and I am not sure how to convince myself that   λ  f is measurable.

Aryanna Caldwell · Accepted Answer

If   λ  =  0, then your function   λ  f  =  0 is identically 0. The inverse image of any measurable set containing 0 is   S and of any set not containing 0 is the empty set. Both of these are measurable.

aggierabz2006zw · Accepted Answer

f  :      S    →      R    measurable function.  g  :      R    →      R   defined by   g  (  x  )  =  λ  x ,   λ  ∈      R   , is continuous.Hence,   g  ∘  f  =  λ  f is measurable.

I have somehow confused myself with this fairly straightforward proof. We need to show that &#x0

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