Let $f_{n}=n \chi_{[\frac{1}{n},\frac{2}{n}]} $.
a) Show that the sequence $(f_{n})$ converges everywhere to the 0-function and converges in measure.
b) Show that the sequence $(f_{n})$ has the property that if $\delta > 0$, then it is uniformly convergent on the complement of the set $[0,\delta]$. However, show that there does not exist a set of measure zero, on the complement of which $(f_{n})$ is uniformly convergent.
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We first show point-wise convergence. Choose $x\in \Bbb R$ , then if it is outside of $(0,2]$ then certainly $f_n(x)=0$ for each $n$ so we are done. If $x\in (0,2)$ then choose $n\in \Bbb N$ such that $\frac{2}{n}<x$ then for each $m≥n$ we have $f_m(x)=0$ , again we are done.
Now we show convergence in measure. So let $1≥\epsilon >0$, hence $m(\{x : |f_n(x)|≥\epsilon\})≤\frac{1}{n}\rightarrow 0$ as $n\rightarrow \infty$ . Note also that for $\epsilon >1$ we have $m(\{x : |f_n(x)|≥1\})≥m(\{x : |f_n(x)|≥\epsilon\})$ . Hence $\{f_n\}_n $ converges in measure to $0$ function.
Now let $\delta >0$ and choose $n$ such that $\frac{2}{n}<\delta$ then for $k,m≥n$ we have $|f_k(x)-f_m(x)|=0$ for each $x\in \Bbb R-[0,\delta]$. Hence $\{f_n\}_n$ converges uniformly on $\Bbb R-[0,\delta]$ for each $ \delta >0$.
Now let $S$ be a subset of $\Bbb R $ having Lebesgue measure zero. Choose any $n\in \Bbb N$ and since $S$ does not contains any open interval we have an $ x\in [\frac{1}{2n},\frac{2}{2n}]\cap S^c$ with $|f_{2n}(x)-f_{3n}(x)|≥n>\frac{1}{2}$. Therefore $\{f_n\}_n$ does not converges uniformly on $S^c=\Bbb R-S$