How to prove essential supremum is a norm Let $f$ be a measure function on $X$. If there exists an $M>0$ such that: 
$\mu(\{t\in X: |f(t)|>M\})=0$, we say $f$ is essentially bounded. The infimum of all such $M$ is called the essential supremum of $|f|$. It is written as $ ||f||_{\infty}= {\rm ess \;sup}|f|$
How to prove essential supremum is a norm on $(X, \Gamma, \mu)$- a $\sigma$-finite measure space.

Lemma: Let $M' = \|f\|_{\infty}$, then
$$
\mu(\{t \in X : |f(t)| > M'\}) = 0
$$
Proof :
For each $n\in \mathbb{N}$, $M' + 1/n$ is not a lower bound for the set of $M$'s satisfying
$$
\mu(\{t \in X : |f(t)| > M\}) = 0
$$
Hence, there exists $M_n < M'+1/n$ such that
$$
\mu(\{t \in X : |f(t)| > M_n\}) = 0
$$
Now note that
$$
|f(t)| > M' \Leftrightarrow \exists n\in \mathbb{N} \text{ such that } |f(t)| > M_n
$$
This proves the lemma.

In other words,
$$
|f(t)| \leq M' \text{ almost everywhere}
$$
From this you see that


*

*If $\|f\|_{\infty} = 0$, then $f=0$ almost everywhere.


Note that for any $\alpha \neq 0$,
$$
|\alpha f(t)| \leq M \Leftrightarrow |f(t)| \leq M/|\alpha|
$$
This should help you prove


*

*$\|\alpha f\|_{\infty} = |\alpha|\|f\|_{\infty}$

*Finally the triangle inequality is obvious because
$$
|f(t) + g(t)| \leq |f(t)| + |g(t)| \quad\forall t\in X
$$
