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In Egoroff theorem, the hypothesis that $ m (E) <\infty $ is essential. Construct an example of measurable functions $ f_n: \mathbb{R} \rightarrow \mathbb {R} $ that converge to the null function with the following property: if $ F \subset \mathbb{R} $ and $ m (R \backslash F) <\infty $ then $ \{f_n \}$ not converges uniformly on $ F $

I have not idea how to do this! I'm thinking for days but to no avail! I'm terrible with examples ...

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    $\begingroup$ $f_n=\chi_{[n,n+1]}$. $\endgroup$ Feb 3 '14 at 14:33
  • $\begingroup$ @DavidMitra I will try to show that it satisfies all of these properties of the exercise. It is not obvious to me. $\endgroup$
    – Felipe
    Feb 3 '14 at 14:37
  • $\begingroup$ Even easier, take $f_n=\chi_{[n,\infty)}$. $\endgroup$ Feb 3 '14 at 14:42
  • $\begingroup$ @DavidMitra But... you do not need to fix a set $F$ such that all works? $\endgroup$
    – Felipe
    Feb 3 '14 at 15:06
  • $\begingroup$ Yes. Fix $F$ and show that for any $n$ there is an $x\in F$ and $N\ge n$ with $f_N(x)=1$. This will show that $(f_n)$ does not converge uniformly to $0$ on $F$. $\endgroup$ Feb 3 '14 at 15:09
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Let $f_n:\mathbb{R}\rightarrow\mathbb{R}$ where $$ f_n(x)= \left \{ \begin{array}{ll} 1 & \textrm{ if } x\in[n,\infty) \\ 0 & \textrm{ if } x\not\in[n,\infty) \end{array} \right. $$ Let $F\subset \mathbb{R}$, $m(\mathbb{R}\backslash F)<\infty$ a closed set.

For each $n\in \mathbb{N}$ exists $x\in F$ and $N\geq n$ such that $f_N(x)=1$. This works because F is very large and always lets us find these elements (in $[N,\infty)$)

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