Proving that $\|f\|_\infty=\sup\{M:\mu(\{x:|f(x)|>M\})>0\}$

Question

I need show that identity $\|f\|_\infty=\sup\{M:\mu(\{x:|f(x)|>M\})>0\}$.

I know that $\|f\|_\infty=\inf\{c>0:\mu(\{x:|f(x)|>c\})=0\}$ and $ \|f\|_{\infty}=\inf\{C\geq 0;\,\,|f(x)| \leq C \text{ almost everywhere}\}. $

But i don't but I do not know how to relate these three definitions

Answer 1

Actually it's a trivial proof by definition of $\sup$ and $\inf$.

Let define $$K := \sup\{M:\mu(\{x:|f(x)|>M\})>0\}$$ Then $K$ is an upper bound of $$A :=\{M:\mu(\{x:|f(x)|>M\})>0\}$$ so on the one hand we have for all $c>K$ that $$\mu(\{x:|f(x)|>c\})=0$$ On the other hand by monotony of the measure $\mu$ we have for $c < K$ that $$\mu(\{x:|f(x)|>c\})>0$$ So it follows for all $c > 0$ that $$\mu(\{x:|f(x)|>c\})=0 \Rightarrow c \ge K$$ So $K$ is a lower bound of $$B = \{c>0:\mu(\{x:|f(x)|>c\})=0\}$$ and additionally for each $\varepsilon >0$ it has to hold $K+\varepsilon \in B$ otherwise $K+\varepsilon$ would be an upper bound of $A$ so $K$ is not just a lower bound of $B$ but the least one… hence it holds $$K = \inf B = \inf\{c>0:\mu(\{x:|f(x)|>c\})=0\}$$ And we are done.

Answer 2

The final two conditions are exactly equivalent.

$|f(x)| \leq C \,\,a.e \iff \mu(\{x \mid f(x)>C\})=0$, which is the definition of almost everywhere.

Reconciling this with the first definition is just ensuring that these two things are equal, heuristically: "one collection ends where the other begins."

Basically, $C:=\sup\{M \mid \mu(\{x \mid |f(x)|>M\})>0\}$ implies that $C$ is a lower bound for $\{M \mid \mu(\{x\mid f(x)>M\})=0\}$, and it is also not hard to see that this is the greatest lower bound, since it is essentially the $\sup$ of the complement.