$C_0(\mathbb R)$ is the vector space such that all f in $C_0(\mathbb R)$ is continuous function such that $\lim_{x\to \pm\infty}f(x) = 0$.
$\Vert f\Vert_{\sup}:= \sup \{|f(x)| : x \in \mathbb R\}$
$\Vert f\Vert_\infty := \textrm{ess}\sup\ |f|$ and $\textrm{ess}\sup f = \inf \{a\in \mathbb R : \mu(f^{-1}(a,\infty])=0\}$
Then, My book said that for any $f \in C_0(\mathbb R)$, two norms of $f$ are same. I tried to show $\leqq$ and $\geqq$ to show the equality. But, it was too hard for me, because I couldn't know how to use the fact that $f(x) \to 0$ as $x$ goes to $-\infty$ or $\infty$.
Could someone help me to prove this?
On the one hand, for all $x\in\mathbb{R}$, we have $|f(x)| \leq \|f\|_{\sup}$. So $x\notin |f|^{-1}(\|f\|_{\sup},\infty]$ and $\mu(|f|^{-1}(\|f\|_\sup,\infty)) = 0$, which means that $\|f\|_\infty \leq \|f\|_\sup$.
On the other hand, suppose by contradiction that $\|f\|_\infty < \|f\|_\sup$. Then there exists some $a<\|f\|_\sup$ such that $$\mu(|f|^{-1}(a,\infty]) = 0,$$ which means that $|f(x)|\leq a$ almost everywhere, and thus $|f(x)|\leq a$ for all $x\in\mathbb{R}$ since $f$ is continuous. This is contradictory to the definition of $\|f\|_\infty$.