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I am trying to find an example that shows that Egorov's Theorem fails when the measure of the set under study is not finite. I found this example here:

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 $m(\mathbb{R}-F)\geq m([N,\infty[)=\infty$.

There is one thing I don't understand about this proof. I understand how the author got to $F\subseteq [0,N[$. How does this imply that $m(\mathbb{R}-F)\geq m([N,\infty[)=\infty$? I thought that $F\subseteq [0,N[ \implies F^c \subseteq [N, \infty) \implies m\{F^c\} \le m\{[N, \infty)\}$, where the last implication follows from the fact that measure preserve order, but this conclusion is almost the opposite of what the proof concluded. Where did I go wrong?

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$A\subseteq B$ implies $B^{c}\subseteq A^{c}$.

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  • $\begingroup$ Wow, much simpler than I thought. Just out of curiosity, is there a name for this result? $\endgroup$ Apr 23, 2021 at 20:24
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    $\begingroup$ I think I never ever see there is any title to that. $\endgroup$
    – user284331
    Apr 23, 2021 at 20:25

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