Equivalence of definitions for $L_\infty$ norm via essential range and essential supremum I have the following two norms for mesurable complex valued functions $f \colon X\to \mathbb{C}$:
$$\|f \|_1 := \inf \big\{ c \geq 0 \colon \mu (\{ x \in X \colon |f(x)| > c \}) = 0 \big\}$$
and
$$\|f\|_2 := \sup \big\{ |\lambda| \colon \lambda \in r_{\mathrm{ess}}(f)\big\},$$
where
$$
r_{\mathrm{ess}}(f) := \big\{\lambda \in \mathbb{C} \colon \forall \varepsilon > 0: \mu(f^{-1}(B_\varepsilon(\lambda)) > 0 \big\},$$
where $B_\varepsilon(\lambda)$ denotes the open ball with radius $\varepsilon$ centered at $\lambda$.
It is easy to show, that $\|f\|_2 \leq \|f\|_1$ but I have not been able to prove the converse. Any suggestions? 
I think the problem boils down to showing that $\mu\big(\;f^{-1}(\mathbb{C} \setminus r_{\mathrm{ess}}(f))\big) = 0$.

EDIT
One possible proof could be like this:
We exhaust the set $S := \mathbb{C}  \setminus \overline{ B_{\|f\|_2}(0)}$ in two steps: 
First for all $m \in \mathbb{N}$ by annuli
$$
S_{m,n} := S_{\|f\|_2 + \frac{1}{m}, \|f\|_2 + n} := \bigg\{ \lambda \in \mathbb{C} \colon \|f\|_2 + \frac{1}{m} \leq |\lambda | \leq \|f\|_2 + n\bigg\}
$$
with $\mu(f^{-1}(S_{m,n})) = 0$, i.e. for
$$
S_m := \bigcup_{n \in \mathbb{N}} S_{m,n}
$$
we get  $\mu(f^{-1}(S_{m}))  = 0$.
Secondly, by considering
$$
S = \bigcup_{m \in \mathbb{N}} S_{m}
$$
we show the claim.
Proof.
Let $m,n$ be given. 
Then, for all $\lambda \in S_{m,n}$ there exists $\varepsilon_\lambda > 0$ such that 
$$\mu\bigg(\big\{f^{-1}\big(B_{\varepsilon_\lambda}(\lambda)\big)\big\}\bigg) = 0.$$
But $S_{m,n}$ is compact, i.e. we get $\lambda_1,\dots,\lambda_k$ such that
$$
S_{m,n} \subseteq \bigcup_{i = 1,\dots,k} B_{\varepsilon_{\lambda_i}}
(\lambda_i).
$$
Hence (assuming completeness of $\mu$) we have that $f^{-1}(S_{m,n})$ is a set of measure zero, as well as the countable union of all $S_{m,n}$ for $n \in \mathbb{N}$.
So, for all $m \in \mathbb{N}$ the sets $f^{-1}(S_m)$ are of measure zero and the fact that $S = \bigcup_{m \in \mathbb{N}}$ gives the claim: We have
$$\mu (\{ x \in X \colon |f(x)| > \|f\|_2 \}) = 0$$
i.e.
$\|f\|_2 \geq \|f\|_1$. $\square$
What do you think?
 A: 
I think the problem boils down to showing that $\mu\big(\;f^{-1}(\mathbb{C} \setminus r_{\mathrm{ess}}(f))\big) = 0$.

Strictly speaking, that is a bit more than needed. With $s = \lVert f\rVert_2$, showing
$$\mu\Bigl(f^{-1}\bigl(\mathbb{C}\setminus \overline{B_s(0)}\bigr)\Bigr) = 0\tag{1}$$
is what is precisely needed to show $\lVert f\rVert_1 \leqslant \lVert f\rVert_2$. And that is exactly what you correctly proved in your edit.
Let me add a proof of
$$\mu\bigl(f^{-1}(\mathbb{C} \setminus r_{\mathrm{ess}}(f))\bigr) = 0$$
though. By definition of the essential range, if $\lambda \notin r_{\text{ess}}(f)$, then there is an $\varepsilon_{\lambda} > 0$ such that $\mu\bigl(f^{-1}(B_{\varepsilon_{\lambda}}(\lambda)\bigr) = 0$. For $\lambda' \in U := B_{\varepsilon_{\lambda}}(\lambda)$, the ball $B_{\delta}(\lambda')$, where $\delta = \varepsilon_{\lambda} - \lvert \lambda' - \lambda\rvert > 0$, is contained in $U$, hence $\mu\bigl(f^{-1}(B_{\delta}(\lambda')\bigr) \leqslant \mu\bigl(f^{-1}(U)\bigr) = 0$, and so $\lambda' \notin r_{\text{ess}}(f)$. Thus $U \cap r_{\text{ess}}(f) = \varnothing$, and we have shown

The essential range of a measurable function is closed.

Choosing an $\varepsilon_{\lambda} > 0$ as above for every $\lambda \notin r_{\text{ess}}(f)$, we get an open cover
$$\mathscr{U} = \bigl\{ B_{\varepsilon_{\lambda}}(\lambda) : \lambda \in \mathbb{C}\setminus r_{\text{ess}}(f)\bigr\}$$
of $\mathbb{C}\setminus r_{\text{ess}}(f)$. Since $\mathbb{C}$ is second countable, this open cover has a countable subcover, so there is a sequence $(\lambda_n)_{n\in\mathbb{N}}$ such that
$$\mathbb{C}\setminus r_{\text{ess}}(f) = \bigcup_{n = 0}^{\infty} B_{\varepsilon_{\lambda_n}}(\lambda_n).$$
Then
$$f^{-1}\bigl(\mathbb{C}\setminus r_{\text{ess}}(f)\bigr) = \bigcup_{n = 0}^{\infty} f^{-1}\bigl( B_{\varepsilon_{\lambda_n}}(\lambda_n)\bigr)$$
is a countable union of $\mu$-null sets.
