An example of a function in $L^1[0,1]$ which is not in $L^p[0,1]$ for any $p>1$ Title says most of it. Could you help me find an example?
It is easy obviously to show a function that would not be in $L^p[0,1]$ for a specific $p$ (say $(1/x)^{1/p}$, but I can't see how it would be done for all $p$. The reason I'm asking is because we proved in class that $L^p[0,1]$ is nowhere dense as a subset of $L^1[0,1]$, so there must be some $L^1[0,1]$ like this..
Thanks :)
Added: thanks for all the comments. there was some missing parts about how to use convergence theorems that i couldn't complete my own so i'd love assistance :)

Take
$$f(x) = a_n \quad\text{ if }\quad x\in\left(\frac{1}{2^n},\frac{1}{2^{n-1}}\right]$$
for $n = 1, 2, \dots$ and for some $a_n$.
You'll get
$$\int_0^1 f(x)\, dx = \sum_{n=1}^{\infty}\frac{a_n}{2^n}$$
$$\int_0^1 f^p(x)\, dx = \sum_{n=1}^{\infty}\frac{a_n^p}{2^n}$$
Then choose the sequance $a_n$ so that the first sum is convergent, while the second one is divergent for any $p>1$. For example
$$a_n = \frac{2^n}{n^2}.$$
$$\sum_{n=1}^{\infty}\frac{a_n}{2^n} = \sum_{n=1}^{\infty}\frac{1}{n^2}<\infty$$
$$\sum_{n=1}^{\infty}\frac{a_n^p}{2^n} = \sum_{n=1}^{\infty}\frac{2^{n(p-1)}}{n^{2p}} = \infty$$
