The norm of essentially bounded functions Let $(X, \mathscr{A}, \mu)$ be a finite measure space, and let $f$ be a $\mathscr{A}$- measurable real- or complex- valued function on $X$.
Show that if $f\in \mathscr{L}^\infty$ then $\|f\|_\infty = \lim_{p \rightarrow+\infty} \|f\|_p$.
In my course literature $\|f\|_\infty$ is defined as the infimum of those non-negative numbers $M$ such that $\{x\in X : |f(x)| > M\}$ is locally $\mu$-null. 
But locally $\mu$- null seems to be equivalent with $\mu$-null subsets when the measure space is $\sigma$-finite? 
I know that $f \in \mathscr{L}$  for $p \in [1,+\infty)$ and that $\sup \{|f|_p : 1 \leq p < \infty\}$ is finite.
But how do I know that this converge? 
Thanks 
 A: I'll show it for the special case $\mu(X) = 1$. You should be able to derive the general case from that. First of all, we have
$$\lVert f\rVert_p = \left(\int_X \lvert f\rvert^p\, d\mu\right)^{1/p} \leqslant \left(\int_X \lVert f\rVert_\infty^p\, d\mu\right)^{1/p} = \lVert f\rVert_\infty$$
since $\mu(X) = 1$. Hence $\limsup\limits_{p\to\infty} \lVert f\rVert_p \leqslant \lVert f\rVert_\infty$.
Now we need to show that $\liminf\limits_{p\to\infty} \lVert f\rVert_p \geqslant \lVert f\rVert_\infty$. For $\lVert f\rVert_\infty = 0$ that is trivially satisfied, so let's assume $M := \lVert f\rVert_\infty > 0$. For $0 < \varepsilon < M$, let $A_\varepsilon = \{x \in X : \lvert f(x)\rvert \geqslant M - \varepsilon\}$. Then
$$\begin{align}
\lVert f\rVert_p &= \left(\int_X \lvert f\rvert^p\,d\mu\right)^{1/p}\\
&\geqslant \left(\int_{A_\varepsilon} \lvert f\rvert^p\,d\mu\right)^{1/p}\\
&\geqslant \left(\int_{A_\varepsilon} (M-\varepsilon)^p\,d\mu\right)^{1/p}\\
&= (M - \varepsilon)\cdot \mu(A_\varepsilon)^{1/p} \to M - \varepsilon,
\end{align}$$
hence $\liminf\limits_{p\to\infty} \lVert f\rVert_p \geqslant \lVert f\rVert_\infty - \varepsilon$. Since $\varepsilon$ was arbitrary, the proposition follows.
