Suppose that $f \in \mathcal{L}^\infty(X, \mathcal{M}, \mu)$ and that $||f||_\infty > 0$. Show that $||f||_\infty = \sup\{a : \mu(\{x : |f(x)| > a\}) > 0\}$.
My approach has been as follows. Let $A = \{a \ge 0 : \mu(\{x : |f(x)| > a\}) = 0\}$ and $B = \{a : \mu(\{x : |f(x)| > a\}) > 0\}$. So, $\inf A = ||f||_\infty$.
I've shown that $A = [||f||_\infty, \infty)$ and $B = (-\infty, ||f||_\infty)$, and it is then easy to see that $\sup B = \inf A = ||f||_\infty$, which shows the desired result.
My question is, why do we need the assumption that $||f||_\infty > 0$? It seems that this proof should work for the case when $||f||_\infty = 0$, which leads me to believe that I have made an error.
Hints only if possible! Thank you.