lp spaces - convergence of $L^p $ norm

If I define $|f|_{L^\infty}= \lim_{n\to \infty} |f|_{L^n}$. How can I prove that this limit is esssup $|f|$?

Answer

The main reason to choose $\text{ess}\sup\vert f\vert$ over $\sup \vert f\vert$ is that "functions" in $L^p$ are in fact equivalence classes of functions: $f\sim g$ if $\{x:f(x)\neq g(x)\}$ has measure zero. By construction of the Lebesgue integral, for all $1\leq p<\infty$ we have $\|f\|_p=\|g\|_p$ if $f\sim g$; we would like $\|f\|_\infty$ to have the same property. $\sup\vert f\vert$ won't work because we can have $f\sim g$ but $\sup\vert f\vert \neq \sup\vert g\vert$, i.e. two functions in the same equivalence class will have different norm. Since $\text{ess}\sup\vert f\vert$ "ignores" sets of measure zero, we will have $\text{ess}\sup\vert f\vert=\text{ess}\sup\vert g\vert $ if $f\sim g$ and hence the norm $\|\cdot\|_\infty$ will be well defined on our equivalence classes.

Edit: I guess the question changed as I was writing this. This is more the reason for $\text{ess}\sup$ rather than the proof requested.