Show that $\bigcap_{p<\infty }\mathcal{L}^p([0,1])\neq \mathcal{L}^{\infty }([0,1])$

Question

I have an exercise that says

Show that $\bigcap_{p<\infty }\mathcal{L}^p([0,1])\neq \mathcal{L}^{\infty }([0,1])$

Im not sure if $p$ ranges from one or from zero, but from the context where this exercise appeared it seems to range from zero. Clearly if $f$ is essentially bounded then $\|f\|_p\leqslant \|f\|_\infty $ for any chosen $p\in(0,\infty )$.

Thus I want to show that there is some $f$ such that $\|f\|_p<\infty $ for each $p\in(0,\infty )$ but that it is not essentially bounded.

However Im not sure that the statement to be proved is true. Let $A_n:=\{x\in[0,1]:|f(x)|\in [n,n+1)\}$, then is clear that

$$ \sum_{n\geqslant 1}n\chi_{A_n}\leqslant f\implies \sum_{n\geqslant 1}n^p\lambda(A_n)\leqslant \int_{[0,1]}|f|^p \,\mathrm d \lambda $$ Thus if the statement of the exercise would be true we can assume WLOG that exists some $f$ such that $\mu(A_n)>0$ for each $n\in \Bbb N $ and that $\sum_{n\geqslant 1}n^p\lambda(A_n)<\infty $ for all $p\in(0,\infty )$, however its hard to believe that this is true and I dont know how to proceed further. Some help will be appreciated, thanks.

Answer 1

Take a disjoint sequence $(A_{n})$ of $[0,1]$ such that $|A_{n}|=1/2^{n}$, and consider $f(x)=\sum_{n=1}^{\infty}n\chi_{A_{n}}$.

Answer 2

Alternative solution: $$0\le \int_0^1 [\log (1/x)]^p dx = \int_0^\infty y^{p}e^{-y}dy = \Gamma(p+1) <\infty,$$ for any $p\in(0,\infty)$. (The rightmost integral actually converges for $\Re p > -1$)