How to show that a sequence converges pointwise. Hmmm...I am almost embarrassed to ask this question, but I'll ask anyway. How do I show that the sequence defined by $f_n(x) = n^{1/p}\chi_{[0,1/n]}$ ,$1\le  p \lt \infty$  and $x\in [0,1]$ converges pointwise to $0$.
 A: It still doesn't, I'm afraid. We have that $f_n(0)=n^{1/p}\chi_{[0,\frac{1}{n}]}(0)=n^{1/p}$ for all $n$, and hence
$$\lim_{n\to\infty}f_n(0)=\lim_{n\to\infty} n^{1/p}= \infty.$$
However, for any $x&lt0$, we have that $f_n(x)=\chi_{[0,\frac{1}{n}]}(x)=0$ for all $n$, and for any $x>0$ there is some $m$ for which $x>\frac{1}{m}$, hence $x\notin [0,\frac{1}{n}]$ for all $n>m$, hence $\chi_{[0,\frac{1}{n}]}(x)=0$ for all $n>m$.
Thus the pointwise limit is the function to the extended reals
$$f(x)=\begin{cases}\infty &\text{ if }x=0,\\0 &\text{ if }x\neq0. \end{cases}$$

If it's as Asaf is proposing, and the original question was actually about $f_n(x)=x^{1/p}\chi_{[0,\frac{1}{n}]}$, then just note that $$f_n(0)=0\cdot \chi_{[0,\frac{1}{n}]}(0)=0\cdot 1=0$$
for all $n$, so that you do have 
$$\lim_{n\to\infty}f_n(0)=\lim_{n\to\infty} 0= 0.$$
Combined with my earlier argument for $x\neq0$, this then shows that the pointwise limit of $f_n(x)=x^{1/p}\chi_{[0,\frac{1}{n}]}$ is the zero function.
