Sequence of measurable functions $f_n=n\mathcal X_{[\frac{1}{n},\frac{2}{n}]}$, uniform convergence

Question

For each $n \in \mathbb N$, let $f_n:[0,\infty) \to \mathbb R: f_n(x)=n\mathcal X_{[\frac{1}{n},\frac{2}{n}]}$. Show that there is no $E \subset [0,\infty)$ such that $|E|=0$ and $(f_n)_{n \geq 1}$ converges uniformly on $E^c$.

I could show that $f_n$ converges pointwise to the function $f \equiv 0$, and also that $f_n \rightrightarrows 0$ on all the intervals of the form $[a,\infty)$ with $a>0$. Since for each $f_n$, all $x \in [0,\frac{1}{n}) \cup (\frac{2}{n}, \infty)$ are in the set $\{x \in [0,\infty) : |f_n(x)|>\frac{1}{2}\}$, then $m(\{x \in [0,\infty) : |f_n(x)|>\frac{1}{2}\})=\infty$ so it is clear that $f_n \not \xrightarrow{m} 0$.

What I thought of is: if there is some null set $E$ such that $f_n \rightrightarrows 0$ on $E^c$, then $f_n \xrightarrow{m} 0$ on $E^c$, I could try to show that this cannot happen but I have no idea how to prove this.

Any suggestions and corrections would be greatly appreciated. Thanks in advance.

Answer 1

If $E$ is a set such that $f_n\to 0$ uniformly in $E^c$, then $E$ has a positive measure. Indeed, notice that $$\sup_{x\in E^c}f_n(x)=\begin{cases}n&\mbox{if }[1/n,2/n]\cap E^c\neq\emptyset, \\ 0&\mbox{otherwise}. \end{cases} $$ Therefore, we should have $[1/n,2/n]\cap E^c=\emptyset $ for some $n$. This implies that $$\frac 1n=\lambda\left(\left[\frac 1n,\frac 2n\right]\right)= \lambda\left(\left[\frac 1n,\frac 2n\right]\cap E\right)\leqslant \lambda(E).$$

Answer 2

Let $I = \overline {E^c} \cap [0,1]$ (the bar denotes topological closure). Being closed an bounded, $I$ is compact. Note that since $|E|=0$, then $\Big| I \cap [\frac 1 n, \frac 2 n] \Big|= \frac 1 n$. It is known that if $f_n \to f$ uniformly on a compact $X$, then $\int \limits _X f_n \to \int \limits _X f$. In your case, applying this on $I$ gives $1 \to 0$, contradiction, so the convergence cannot be uniform on any such set $E$.