# Understanding the indicator function counterexample to Egorov's Theorem

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?

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