Suppose ($X,\cal M, \mu$) is not complete. Show that there is a sequence {$f_n$} of measurable functions on $X$ that converges pointwise a.e. on $X$ to a function $f$ that is not measurable.


1 Answer 1


Pick a non-measurable subset $E$ of a null set, and let $f_n=0$ for all $n$. Then $f_n\to\chi_E$ a.e.

  • $\begingroup$ Harald, thank you for your response. So, since I'm showing that there exists a sequence that satisfies these conditions, I can just pick any non-measurable E. For showing purposes, should I actually pick a specific set? Or just say "non-measurable"? $\endgroup$
    – Jake Casey
    Commented Mar 17, 2013 at 21:01
  • 3
    $\begingroup$ You can't pick a specific set, because you don't know one. The hypothesis of non-completeness only guarantees the existence of one, it does not tell you how to specify one. So just say “non-measurable”. $\endgroup$ Commented Mar 18, 2013 at 6:36
  • $\begingroup$ I guess you mean a non-measurable subset of a null set. A null set must be measurable. $\endgroup$
    – Hua
    Commented Feb 5, 2017 at 17:02
  • $\begingroup$ @Hua I had in mind the following definition: A null set is a subset of a measurable set whose measure is zero. I think this is (a least slightly) better than assuming a priori that null sets are measurable. With this definition, a measure is complete iff all null sets are indeed measurable, for example. It is possible, though, that this convention is not universally followed, so I'll let your edit stand, in order to avoid the ambiguity. $\endgroup$ Commented Feb 7, 2017 at 8:24

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