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Why does Egorov's theorem not hold in the case of infinite measure? It turns out that, for example, $f_n = \chi_{[n,n+1]}x$ does not converge nearly uniformly, that is, it does not converge on E such that for a set F m(E\F) < $\epsilon$. Is this simply true because it takes on the value 1 for each n but suddenly hits 0 when n ---> infinity?

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2 Answers 2

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Set $\forall n\geq1: f_n:[0,\infty[\to\{0,1\}, f_n:=\chi_{[n-1,n]}$. Then $f_n\to0$ pointwise on $\mathbb{R}$. Suppose $\exists F\subseteq\mathbb{R}: f_n\stackrel{u.}{\to}0$ on $F$, i.e. that

$$\forall \epsilon>0,\exists N,\forall n\geq N,\forall x\in F: |f_n(x)|<\epsilon.$$

For $\epsilon:=1, \exists N,\forall n\geq N,\forall x\in F:|f_n(x)|<1$, so that $x\not\in[N,\infty[$. Thus $F\subseteq [0,N[$, and consequently $m(\mathbb{R}-F)\geq m([N,\infty[)=\infty$.


The moral of the story is then that for this example any set on which we have uniform convergence has to be of finite measure, and we cannot make the rest have arbitrarily small measure.

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  • $\begingroup$ Should it be $f_n := \chi_{[n-1, n)}$? $\endgroup$
    – user529632
    Commented Dec 3, 2022 at 2:35
  • $\begingroup$ @Note You can use that sequence as well. $\endgroup$
    – Alp Uzman
    Commented Dec 4, 2022 at 23:22
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$f_n$ converges pointwise to the zero function on $\mathbb{R}$ (here $E = \mathbb{R})$. However there doesn't exist a set of finite measure $F$ such that $f_n$ converges uniformly on $\mathbb{R} \setminus F$. To see this note that for large enough $n$, $f_n$ will take both the values $0$ and $1$ on $\mathbb{R} \setminus F$.

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  • $\begingroup$ So we have that the largest possible set we can take away is of measure zero? $\endgroup$
    – user71443
    Commented Apr 11, 2013 at 21:58
  • $\begingroup$ The definition states that if you have a sequence of functions in a set of finite measure $E$ which converge pointwise to $f$ almost everywhere. Then for every $\epsilon >0$ you can find a measurable set $F \subset E$ such that $m(F) < \epsilon$ and $f_n$ converges uniformly to $f$ on $E \setminus F$. $\endgroup$
    – chango
    Commented Apr 12, 2013 at 9:17
  • $\begingroup$ Here your hypothesis of finite measure fails since $m(\mathbb{R}) = \infty$ and the conclusion fails as well because for any set of finite measure $F$, your function won't converge uniformly to the zero function. $\endgroup$
    – chango
    Commented Apr 12, 2013 at 9:20
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    $\begingroup$ I don't think the important conclusion of Egorov's Theorem is that we are guaranteed uniform convergence on a set of finite measure (though to be sure this is what we get from the theorem), rather it's that we can make the complement of such a set arbitrarily small. For the infinite measure case this is precisely what fails. $\endgroup$
    – Alp Uzman
    Commented Oct 17, 2016 at 12:27

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