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Can someone give an example why a non-complete measure is a problem? Or what are some theorems that require a measure to be complete?

user53970
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  • my area of studies uses stochastic processes. I just asked http://math.stackexchange.com/questions/539733/measurability-question-with-regard-to-a-stochastic-process See if you understand this. – Lost1 Oct 25 '13 at 23:29
  • The wikipedia page provides a motivation for a measure to be complete. – Julien Oct 25 '13 at 23:31
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    Without completeness, your theorems would have to refer to sets that are subsets of measure 0. That's awkward. – KCd Oct 25 '13 at 23:31
  • http://math.stackexchange.com/questions/147381/why-do-we-essentially-need-complete-measure-space –  Oct 25 '13 at 23:32
  • Thank you all for the quick answers. @Lost1, I have studied stochastic processes but I need a very intense revision to relearn all the definitions that I have forgotten in your example. – user53970 Oct 25 '13 at 23:37
  • @julien, the problem with that example at the wikipedia page is that it is just the monotonicity of measures but applied to two dimensional cases. – user53970 Oct 25 '13 at 23:39
  • @KCd Could you provide an example of such a theorem? (Preferably a simple one) Does the lack of completeness change anything substantially or merely make the statements more verbose? – paperskilltrees May 09 '25 at 02:04
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    @paperskilltrees if $f_n \to f$ everywhere on $X$ and each $f_n$ is measurable on $X$, then so is $f$. But what if we only know $f_n \to f$ almost everywhere on $X$ and each $f_n$ is measurable on $X$: is $f$ measurable on $X$? We are assuming there is $N\subset X$ with measure $0$ s.t. $f_n \to f$ on $X-N$, so $f|_{X-N}$ is measurable, but we have no idea whether the set of $x \in N$ s.t. $f_n(x) \to f(x)$ is even in the $\sigma$-algebra for the measure. If the measure is complete, then an almost everywhere limit of measurable functions is measurable, but otherwise we can't say that. – KCd May 09 '25 at 19:09
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    @paperskilltrees see also https://mathoverflow.net/questions/11554/whats-the-use-of-a-complete-measure. – KCd May 09 '25 at 19:10
  • @KCd Thanks! But can't we always work around this by replacing $f$ with some measurable $\tilde{f}$ such that $\tilde{f}|{X-N}=f|{X-N}$? Relaxing everywhere to almost everywhere convergence expands the set of suitable $f_n$ and the set of suitable $f$, which causes inconvenience. This inconvenience can be overcome in two ways: (1) by choosing measurable representatives, or (2) by considering a completion of the measure, in which all candidates are now measurable. I have difficulties thinking of a situation when option (2) is necessary because option (1) cannot work. Does this make sense? – paperskilltrees May 09 '25 at 20:06
  • @paperskilltrees we don't relax convergence everywhere to convergence almost everywhere just for the sake of extra generality: in an $L^p$-space its elements can only be regarded as functions known almost everywhere, so convergence almost everywhere rather than convergence everywhere is the more relevant pointwise limit concept in $L^p$-spaces. – KCd May 09 '25 at 20:19
  • @KCd Agree. My point was that we can choose a measurable $\tilde{f}\in{f: f_n \to f \textrm{ almost everywhere on } X}$, e.g. by defining it as a (measurable) pointwise limit $f|_{X-N}$ on $X-N$ and as $0$ on $N$, which makes it measurable overall. Not having a guarantee that any choice of $f$ is measurable and having the ability to choose a measurable $\tilde{f}$ are the two sides of the same coin. So I wonder if there are situations in which we are unable to choose a measurable a.e. limit for a given incomplete measure, and thereby are forced to consider its completion. – paperskilltrees May 09 '25 at 21:43

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