Definition 1:Consider the measure space ( X , Σ , μ ).A function f : X → R is said to be measurable if, for every α ∈ R , { x ∈ X : f ( x ) &gt; α } ∈ Σ.The definition I knew was thatDefinition 2 :A function is measurable if the pre-image of every measurable set is measurable.How come these definitions are equivalent?I suspect it has something to do with the fact that half-lines are generators of the Borel sigma algebra, but I haven't been able to prove it.I guess the implication 2 -&gt; 1 is trivial. If a function is measurable, half lines being borel sets, are measurable sets on R , therefore their preimages are measurable.Can someone tell me if I am on the right track and complete the proof?

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Definition 1:Consider the measure space   (  X  ,  Σ  ,  μ  ).A function   f  :  X  →      R   is said to be measurable if, for every       α    ∈      R    ,  {  x  ∈  X  :  f  (  x  )  &amp;gt;  α  }  ∈  Σ.The definition I knew was thatDefinition 2 :A function is measurable if the pre-image of every measurable set is measurable.How come these definitions are equivalent?I suspect it has something to do with the fact that half-lines are generators of the Borel sigma algebra, but I haven&#039;t been able to prove it.I guess the implication 2 -&amp;gt; 1 is trivial. If a function is measurable, half lines being borel sets, are measurable sets on       R  , therefore their preimages are measurable.Can someone tell me if I am on the right track and complete the proof?

pralkammj · Accepted Answer

Let       H   denote the collection of half lines.We know they generate the Borel sigma-algebra, i.e.   σ  (      H    )  =      B  , so the preimages of measurable sets are      f          −      1        (      B    )  =      f          −      1        (  σ  (      H    )  )  .We also know       f          −      1        (      H    )  ⊆  Σ, which implies  σ  (      f          −      1        (      H    )  )  ⊆  Σ  .Thus, if we show                              (                      ∗                    )                                      f                      −            1                          (        σ        (                  H                )        )        =        σ        (                  f                      −            1                          (                  H                )        )        ,            then we are done, since we would then have       f          −      1        (      B    )  ⊆  Σ. Try showing this last claim yourself.

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