Show that the sequence converges on the complement of the set $[0,\delta]$

Question

I'm stuck on this problem from Bartle; the problem says

Show that the sequence $f_{n}=n\chi_{[1/n,2/n]}$ 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.

I have no idea how to solve the second statement. For the first I know that the sequence converges (pointwise) to 0 and does not converge in $L^{p}(\mathbb{R},\mathcal{B},\lambda)$.

I found that the results holds if $\delta>2$ by showing that $\sup\limits_{x\in [0,\delta]^{c}}|f_{n}(x)|=0$, but I cannot see why $f_{n}$ converges uniformly on $[0,\delta]^{c}$ when $0<\delta\leq 2$.

I will appreciate any hint, thanks

Answer 1

For the first part note that whenever $n > 2/\delta$, then $f_n = 0$ on $[0,\delta]^c$.

For the second part, suppose $Z$ is a set of measure $0$. If $f_n$ converges uniformly on $Z^c$, then it necessarily converges to $0$, its pointwise limit. Uniform convergence to $0$ implies that $f_n$ converges to $0$ in $L^1$, yet $\int_{Z^c}f_n = \int f_n = 1$ for every $n$, contradiction.

Answer 2

Let $A\subset \mathbb R$ where the measure of $\mathbb R$ \ $A$ is zero.

For every $n\in \mathbb N$ the measure of $A\cap [1/n,2/n]$ is $1/n$, so $A\cap [1/n,2/n]\ne \phi$, so there exists $x_n\in [1/n,2/n]$ with $f_n(x_n)=n.$

Since $(f_n)_n$ converges point-wise to $0$ on $\mathbb R,$ uniform convergence of $(f_n)_n$ on $A$ is equivalent to $\lim_{n\to \infty} \|f_n\|_A=0,$ where $\|f_n\|_A=\sup \{|f_n(x)|:x\in A\}.$ But $\|f_n\|_A=f_n(x_n)=n.$

Actually the above argument, with some obvious modifications, applies to any dense $A\subset \mathbb R.$