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Suppose $X$ is compact metric space. Let $A$ be the smallest set of complex functions containing all continuous functions such that:

If $f_n \in A$ are uniformly bounded and $f_n \to f$ pointwise then $f \in A$.

Is it true that $A$ is equal to the set of all Borel functions? If not, what is relation between these two? What assumptions are needed to get this?

Answers as well as references will be appreciated.

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  • $\begingroup$ Check Kechris' Classical Descriptive Set Theory, 11.6 and 11.7. I think this should give you the union of all continuous and all bounded Borel functions. You can also generalize to any metric space $X$. To get all Borel functions, drop the boundedness requirement $\endgroup$ Apr 22, 2016 at 23:01
  • $\begingroup$ Thank you, this is precisely what I was looking for. If you make it an answer, I will accept it. $\endgroup$
    – Blazej
    Apr 23, 2016 at 8:41

1 Answer 1

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In Kechris' Classical Descriptive Set Theory, 11.6 and 11.7, it is proved that that for every metrizable $X$, the class of (real, complex) Borel functions is the smallest class containing the continuous functions and it is closed under taking pointwise limits. It is also noted that bounded real Borel functions are generated by bounded continuous functions and uniformly bounded pointwise limits.

Taking this two into account, the set you describe will result in the union of all continuous and all bounded complex Borel functions on $X$. This follows from the real case by considering real and imaginary parts, and that unbounded continuous functions can't be used in any uniformly bounded limit to get new functions.

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