Intuition behind the essential range We define the essential range in the following way. Let $F$ be a real-valued function on a measure space $\langle M , \mu \rangle$. We say $\lambda$ is in the essential range of $F$ if and only if $$\mu \{ m \ \vert \ \lambda - \epsilon < F(m) < \lambda + \epsilon \} > 0$$ for all $\epsilon >0$. Intuitively, should we think of the essential range as the range of $F$ seen by the measure? Does this at all relate to the essential supremum defined on $L^{\infty}?$
 A: Define the pushforward measure $F_*\mu$ on $\mathbb{R}$ by the formula
$$ F_*\mu(B):=\mu(F^{-1}(B)). $$
Using the language of probability (and assuming that $\mu$ is a probability measure to get a better grasp), $F_*\mu(B)$ tells you how much it is likely that $F$ (thought as a random variable) takes on a value in the set $B$.
The essential range is precisely the support of the measure $F_*\mu$ (this is immediate to check using the definitions). Any value $\lambda$ outside the essential range is essentially far from the real values which matter: there exists a neighborhood of $\lambda$ which is negligible according to $F_*\mu$.
As an exercise, try to show that $\|F\|_\infty$ (the essential supremum) coincides with $\sup|\lambda|$, as $\lambda$ varies in the essential range.
A: Consider the function $$f(t)=\begin{cases} 1,&\ t\ne0\\ 7,&\ t=0\end{cases}$$The maximum of the function is $7$, and the range is $\{1,7\}$. But the function takes the value $7$ at a single point, as opposed to everywhere else for $1$. From the point of view of measure theory, the value at $0$ is irrelevant. This is what the essential range captures: for the $f$ above, its essential range is $\{1\}$. 
And it is definitely related with the essential supremum: the essential supremum is the supremum of the essential range. 
