uniform convergence of $f_n = n \chi_{[1/n, 2/n]}$ 
$f_n = n \chi_{[1/n, 2/n]}$. Show that $f_n$ converge uniformly to $0$. 

$f_n = n$ if $1/n \le x \le 2/n$, and $0$ otherwise. 
But, I am a little bit confused here.
When $n$ goes to infinity,  
$f_n \to \infty$ if $0\le x\le 0 \implies x=0$? 
How can we deal with this case? 

Another question is that for $\delta > 0$, $f_n$ is uniformly convergent on the complement of $[0, \delta]$. However, there does not exists a set of measure $0$ on the complement of which $f_n$ is uniformly convergent.

I think that in this case, when $n \to \infty$, 
since $x=0$ is not considered, we can conclude that $f_n$ converge uniformly to $0$. But, since the measure of $[0, \delta]$ is $\delta$, such set does not exists. Am I understanding correct? 
 A: 
Show that $f_n$ converges uniformly to $0$.

It does not converge uniformly to zero, because $\sup_x |f_n(x)|=n$ does not converge to $0$.

When $n\to\infty$, $f_n\to \infty$ if ... $x=0$.

This is not correct. $f_n(0)=0$ for all $n$, so $f_n(0)\to 0$.

For $\delta>0$, $f_n\to 0$ uniformly on the complement of $[0,\delta]$. 

This is correct. 

There does not exist a set of measure zero on the complement of which $f_n$ is uniformly convergent. 

Also correct. 
A: The given sequence of functions converges pointwise and almost uniformly but NOT uniformly to $f \equiv 0$ on $\mathbb{R}$.
Notice that if $x \in \mathbb{R} \setminus (0,2]$ then $f_n(x)=0$ for every $n \in \mathbb N$. 
Suppose $x \in (0,2]$. By the Archimedean Property there is a positive integer $N$ so that $Nx>2$. Hence $f_n(x)=0$ whenever $n \geq N$. Therefore, $f_n(x) \to 0$ pointwise.
Select $\hat{\varepsilon}=1$, and let $N \in \mathbb N$. Select $x=\frac{3}{2(N+1)}$. So $N+1>N$ but $|\,f_{N+1}(x)-0|=N+1\geq1$. Therefore, the sequence of functions $\{\, f_n\}_{n=1}^\infty$ does not converge uniformly to $0$.
I will leave the almost uniform convergence to the question asker. Hint: Given $\delta>0$, find $N \in \mathbb N$ so that $N\delta>2$.
