Is there an integrable g that dominates all fn's, $f_n(x)=\frac{\chi_{[0,n]}(x)}{n^p}$?

Question

Is there an $ g\in L_1([0,\infty]) $that dominates all $f_n$'s, $f_n(x)=\frac{\chi_{[0,n]}(x)}{n^p}$, $n\geq 1, 0 < p <\infty$? If exists how can I find it?

I did "a proof" for p=1 there is $x\in[0,n]$, $f_n=\frac{1}{n}$,

$\frac{1}{n}\leq g(x)$

Hence $$\infty=\sum_{n=1}^{\infty}\frac{1}{n} \leq \sum |g_n(x)|\leq \int|g_n(x)|d\lambda$$

Therefore there's no $g\in L_1([0,\infty])$ that dominates all $f_n$.

The problem is, this is just for the case p=1, I have no idea what happens for $p>1$ and I need a rigorous solution to this problem.

Thanks in advance. I need some help

Answer 1

For any $p>0$ we have $|\frac {\chi_{[0,1]}} {n^{p}}| \leq \chi_{[0,1]}$ and $\chi_{[0,1]}$ is integrable.

Answer for the revised question: For $p>1$ let $g(x)=\frac 1 {j^{p}}$ on $[j,j+1)$, $0,1,2,...$. Then $\int g(x) dx =\sum_j \frac 1 {j^{p}} <\infty$ and $f_n(x) \leq g(x)$ for all $n$ and $x$: If $ 0 \leq x \leq n$ then there exists $j \in \{1,2...,n-1\}$ such that $j \leq x <j+1$. Hence $f_n(x)=\frac 1 {n^{p}} \leq \frac 1 {j^{p}} =g(x)$.