Let G = { G ⊆ R 0 + ∣ G denumerable or G ¯ denumerable } and f : R + ⟶ ( 0 , 1 ) : x ↦ { exp ⁡ ( − x ) x rational 0 else I am wondering if the function f is G − B ( ( 0 , 1 ) ) measurable, because as the e x p ( − x ) is continuous and Q is denumerable, normally all f − 1 ( B ( ( 0 , 1 ) ) ⊂ G for the exponential function, but I am lacking the final idea to finish the proof. My idea was to use the following: As B ( ( 0 , 1 ) ) is created by σ ( E ) with E =(0,1) then we can use E to proof that f − 1 ( E ) ⊂ G Any help is much appreciated.

Question

Let       G    =      {    G    ⊆                  R            0              +              ∣    G       denumerable or             G      ¯       denumerable   } and   f  :            R              +        ⟶  (  0  ,  1  )  :  x  ↦      {                            exp          ⁡          (          −          x          )                          x           rational                                       0                           else                         I am wondering if the function   f is       G    −      B    (  (  0  ,  1  )  ) measurable, because as the   e  x  p  (  −  x  ) is continuous and       Q   is denumerable, normally all       f          −      1            (                  B            (      (      0      ,      1      )        )    ⊂      G   for the exponential function, but I am lacking the final idea to finish the proof. My idea was to use the following: As       B    (  (  0  ,  1  )  ) is created by   σ  (      E    ) with       E  =(0,1) then we can use       E   to proof that      f          −      1        (      E    )  ⊂      G  Any help is much appreciated.

Isla Klein · Accepted Answer

For any set   E the inverse image       g          −      1        (  E  ) is the union of the set of all irrational numbers and a countable set or a subset of the set of rational numbers (depending on whether   0  ∈  E or not). Hence,   g is measurable.

Let <mrow class="MJX-TeXAtom-ORD"> <mi class="MJX-tex-caligraphic" mathvariant="script">G

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