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

Roland Manning

Roland Manning

Answered

2022-06-26

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).

Answer & Explanation

trajeronls

trajeronls

Expert

2022-06-27Added 21 answers

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 > 0 N 1 m , n N { x X : | f m ( x ) f n ( x ) | < 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 ) | < 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 "natural" examples of Borel sets outside of Π 4 0 is known in the literature.
Micaela Simon

Micaela Simon

Expert

2022-06-28Added 3 answers

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.

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