Let be some measurable domain and f : E → R a measurable map. Let B ∈ Bor ⁡ ( R ) be a Borel set. Show that f − 1 ( B ) is measurable.I'm advised to define A = { A ∣ f − 1 ( A ) measurable }. Now A consists of sets whose preimage is measurable, and since f is continuous these sets are open. This collection seems to form a σ-algebra on R , but I'm confused about the construction here as it seems that A is the smallest σ-algebra containing open sets, but that would mean that it's equal to Bor ⁡ ( R )?

Question

Let    be some measurable domain and   f  :  E  →      R   a measurable map. Let   B  ∈  Bor  ⁡  (      R    ) be a Borel set. Show that       f          −      1        (  B  ) is measurable.I&#039;m advised to define       A    =  {  A  ∣      f          −      1        (  A  )   measurable  }. Now       A   consists of sets whose preimage is measurable, and since   f is continuous these sets are open. This collection seems to form a   σ-algebra on       R  , but I&#039;m confused about the construction here as it seems that       A   is the smallest   σ-algebra containing open sets, but that would mean that it&#039;s equal to   Bor  ⁡  (      R    )?

Freddy Doyle · Accepted Answer

Your proof will depend on your definition of &quot;measurable function&quot;.Let&#039;s say &quot;measurable function&quot; means       f          −      1        (  G  ) is measurable for all open sets   G  ⊆      R  .Then define       A    =  {  A  ⊆      R    ∣      f          −      1        (  A  )   measurable  }. Show that       A   contains all open sets and that       A   is a sigma-algebra. Conclude that       A    ⊇  Bor  ⁡  (      R    ).

Pattab · Accepted Answer

Note: you cannot prove       A    =  Bor  ⁡  (      R    ).

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