By definition we know that
$$\|f\|_{L^\infty(\Omega)}=\inf\{M:\mu \{ x: |f(x)|>M \}=0\}.$$
We note that for each pair $(\lambda, M)$ such that $\mu\{x:|f(x)|>M\}=0$ and $\mu\{x:|f(x)|>\lambda\}>0$ we have $\lambda < M$. Suppose the contrary, then $\lambda \geq M$. Therefore, $$\{x:|f(x)|>\lambda \} \subset \{x:|f(x)|>M \},$$ which implies that $$0<\mu \{x:|f(x)|>\lambda \} \leq\mu \{x:|f(x)|>M \}=0,$$ which is a contradiction. Thus, $$\sup \{\lambda: \mu \{x:|f(x)|>\lambda\}>0\} \leq \inf\{M:\mu\{x:|f(x)|>M\}=0\}.$$
My question: How to prove the other inequality, that is, how to show that $$\sup \{\lambda: \mu \{x:|f(x)|>\lambda\}>0\} \geq \inf\{M:\mu\{x:|f(x)|>M\}=0\}?$$
My attempt: Let $K$ be defined by $K=\sup \{\lambda: \mu \{x:|f(x)|>\lambda\}>0\}$. Suppose that $M \in (K,\infty)$ then $\mu\{x:|f(x)|>M\}=0$. Indeed, if we had $M>K$ with $\mu\{x:|f(x)|>M\}>0$ this would be a contradiction with the definition of upper bound. Hence, $$(K,\infty) \subset \{M:\mu\{x:|f(x)|>M\}=0\}$$ which implies that $$\inf\{M:\mu\{x:|f(x)|>M\}=0\} \leq \inf (K,\infty)= K=\sup \{\lambda: \mu \{x:|f(x)|>\lambda\}>0\}.$$
Is this proof correct?
