let $f$ be a bounded measurable function on $[0,1]$, then $\lim_{p \rightarrow \infty}\|f\|_{p} = \|f\|_{\infty}$. Let $f$ be a bounded measurable function on $[0,1]$ with lebesgue measure $m$, then  $\lim_{p \rightarrow \infty}\|f\|_{p} = \|f\|_{\infty}$.
There are several solutions to more general versions of this on SE already, but I'm wondering if the assumptions that $f$ is bounded and measurable on $[0,1]$ admits a simpler solution?
Here is one such solution: Limit of $L^p$ norm
 A: The assumption that $f$ is bounded and measurable on a set $E$ with $m(E) = 1$ does simplify things somewhat. In this case, $\|f\|_p$ is monotonically increasing as a function of $p$. 
To see this, suppose that $1 < p < q < \infty$. Let $r$ be the number such that $1/p = 1/q + 1/r$. Then $1 < r < \infty$, and by this generalization of Hölder's inequality we have
$$\|f\|_p = \|f \chi_E\|_p \leq \|f\|_q \|\chi_E\|_r = \|f\|_q m(E)^{1/r} = \|f\|_q$$
Moreover, 
$$\|f\|_p^p = \int_E |f|^p \leq \|f\|_{\infty}^p m(E) = \|f\|_{\infty}^p$$
so
$$\|f\|_p \leq \|f\|_{\infty}$$
As $p \mapsto \|f\|_p$ is increasing and bounded, the limit $\lim_{p \to \infty}\|f\|_p$ exists and does not exceed $\|f\|_{\infty}$. It remains to show that the limit is in fact $\|f\|_{\infty}$.
If $\|f\|_{\infty} = 0$, there is nothing to prove. Otherwise, fix any $\epsilon$ satisfying $0 < \epsilon < \|f\|_{\infty}.$ Then $|f(x)| > \|f\|_{\infty} - \epsilon > 0$ on a set $M$ of positive measure. Consequently,
$$\int_E |f|^p \geq \int_M |f|^p \geq (\|f\|_{\infty} - \epsilon)^p m(M)$$
and therefore
$$\|f\|_p \geq (\|f\|_{\infty} - \epsilon)m(M)^{1/p}$$
Now $m(M) \leq m(E) = 1$, so $m(M)^{1/p} \uparrow 1$ as $p \to \infty$, and therefore we have
$$\lim_{p \to \infty}\|f\|_p \geq \|f\|_{\infty} - \epsilon$$
As this holds for arbitrarily small positive $\epsilon$, the result follows.
