Clearly, the pointwise limit of a sequences of measurable functions is measurable: $$f_n\text{ measurable}\implies f\text{ measurable}\quad(f_n\to f\text{ pointwise})$$ (Especially, this holds true for uniform limit of a sequence.)

On the other hand this fails for nets in general, e.g.: $$f_{E\in\mathcal{B}([0,1]):E\subseteq V}:=\chi_E:\quad f_E\text{ measurable},f\text{ nonmeasurable}$$ with a Vitali set $V$ of the unit interval $[0,1]$.

But what about the uniform limit of a net: $$f_\lambda\text{ measurable}\implies f\text{ measurable}\quad(f_\lambda\to f\text{ uniformly})$$ This, especially, would imply that the bounded Borel functions form a Banach space under the supremum norm...

  • $\begingroup$ well I mean uniform convergence implies pointwise convergence $\endgroup$
    – M A Pelto
    Sep 28, 2014 at 0:46
  • $\begingroup$ @Pelto: Yes, sure, but pointwise convergence of nets doesn't imply measurability. $\endgroup$ Sep 28, 2014 at 1:49
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    $\begingroup$ To see that the bounded Borel functions form a Banach space, you do not need the measurability of (uniform) limits of nets, just that of sequences. $\endgroup$
    – PhoemueX
    Sep 28, 2014 at 7:45
  • $\begingroup$ @PhoemueX: Ah right I forgot ^^ thanks! $\endgroup$ Sep 28, 2014 at 13:02

1 Answer 1


Choose $\{ f_{\alpha_{n}}\}_{n=1}^{\infty}$ such that $\sup |f_{\alpha_{n}}-f| < 1/n$.


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