A basic result in real analysis is that any measurable function is an a.e. limit of a step function sequence (yet a pointwise limit of a simple function sequence), but the statement does not hold when the “a.e.” is replaced with everywhere. How to find a counterexample to the “everywhere” statement?I’ve tried to use the fact that a step function is different from a simple one in that it is continuous on the complement of a zero-measure set, then maybe apply the Egorov’s thm. Considering this we are motivated to choose an everywhere discontinuous characteristic function of some “bad” measurable set. But then I got stuck, since once a.e. is involved, it seems hard to dispense with it (so as to arrive at an counter argument).

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trajeronls · Accepted Answer

It is possible to prove that there exists a Borel-measurable function which is not the everywhere-limit of any step function.If   (      f    n    ) is a sequence of real-valued functions on a set   X, then its point of convergence is given by  E  =      ⋂          ε      ∈                        Q                          &amp;gt;          0                          ⋃          N      ≥      1            ⋂          m      ,      n      ≥      N        {  x  ∈  X  :      |        f    m    (  x  )  −      f    n    (  x  )      |    &amp;lt;  ε  }  .(Note that   E is precisely the set of all   x at which   (      f    n    (  x  )      )          n      ≥      1       is a Cauchy sequence in       R  .)Now, if   (      f    n    ) is any sequence of step functions, then   {  x  ∈  X  :      |        f    m    (  x  )  −      f    n    (  x  )      |    &amp;lt;  ε  } is a finite union of intervals. So, it is a       G          δ      -set and hence belongs to the class             Π        2          0       in the Borel hierarchy on       R  .From this, we know that   E is an       G          δ      σ      δ      -set, and so, it lies in the class             Π        4          0      . Since we know that there exists a Borel set   B which is not in the class             Π        4          0      , the indicator function             1              B       is Borel-measurable but cannot be an everywhere-limit of any sequence of step functions.Although I have little expertise in the descriptive set theory, it seems that no &quot;natural&quot; examples of Borel sets outside of             Π        4          0       is known in the literature.

Micaela Simon · Accepted Answer

I believe I have an answer. Note that step functions are all Borel-measurable, and a limit of Borel-measurable is also Borel-measurable. So take the characteristic function of a Lebesgue-measurable but non-Borel-measurable set and we have an example.

A basic result in real analysis is that any measurable function is an a.e. limit of a step function

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