# Thread: uniform convergence

1. ## uniform convergence

Let $f_n= n \chi_{[\frac{1}{n},\frac{2}{n}]} \,$ on $\mathbb{R}$ for all $n \in \mathbb{N}$ and $f=0$ on $\mathbb{R}.$
Show that ther does not exist a set of measure zero ,on the complement of which $(f_n)$ is uniformly convergent.

2. Originally Posted by problem
Let $f_n= n \chi_{[\frac{1}{n},\frac{2}{n}]} \,$ on $\mathbb{R}$ for all $n \in \mathbb{N}$ and $f=0$ on $\mathbb{R}.$
Show that ther does not exist a set of measure zero ,on the complement of which $(f_n)$ is uniformly convergent.
Your functions $f_n$ are converging pointwise to the zero function. If you had some set $E$ as above, then you'd need, for any $\varepsilon>0$, that there exists some $N_\varepsilon$ so that $|f_n|<\varepsilon$ on $E$ for all $n>N_\varepsilon$. But if $E$ has measure zero, then the complement contains points arbitrarily near zero, i.e. points $x$ such that $x\in[\frac1n,\frac2n]$ for $n$ as large as you like, which contradicts uniform convergence.