Essential range of a continous function. Good evening everyone.
I am struggeling understanding the essential range of a function, defined as $$\lambda\in\text{ess Ran}(f)\Leftrightarrow \forall \epsilon>0 \mu\{x \,: \, |f(x)-\lambda|<\epsilon\}\neq 0$$
I know I am not the only one who has difficulties with the concept of essential supremum, and I have read through all the questions about essential range that can be found on this site, but I did not get sufficiently wiser.
My question is: given any continuous function $f:[0,1]\to \mathbb{C}$, where we let the measure be the Lebesgue measure (this doesn't matter?), what can we say about its essential range?
I am inclined to think that it does not have any essential range, my intuition behind this is that when the function is continuous there will only be one point such that $|f(x)-\lambda|<\epsilon$, and the Lebesgue measure of a point is zero. But I don't know if there is any correctness in this idea, and I wouldn't know exactly how to prove it, any help would be greatly appreciated.
 A: If $(\Omega,\mu)$ is a measure space, then you can define $L^2(\Omega,\mu)$, and a natural operator on $L^2(\Omega,\mu)$ is a multiplication operators $M_f$ defined by $M_f g = fg$. The function $f$ must be measurable or this will not make sense in general. If the function $f$ is bounded, then $M_f$ is bounded on $L^2(\Omega,\mu)$. Otherwise, $M_f$ may be an unbounded operator. The spectrum of $M_f$ is the essential range of $f$. If $\lambda$ is not in the essential range of $f$, then $M_f - \lambda I$ has a bounded inverse on $L^2$ given by the multiplication operator $M_{1/(f-\lambda)}$. The essential range is a good way to describe the spectrum of $M_f$, which you cannot describe directly in terms of the range, and which must depend on the measure.
A: If $f:[0,1]\to \mathbb C$ is continuous, then the essential range of $f$ is equal to the range of $f.$
Proof: Let $z\in f([0,1]).$ Let $\epsilon>0.$ Then $f^{-1}(D(z,\epsilon))$ is open in $[0,1]$ by continuity, and it is nonempty because $z\in f([0,1]).$ Every nonempty open subset of $[0,1]$ has positive Lebesgue measure. This implies $z$ is in the essential range. (This is essentially what Rob Arthan was doing in a comment.)
Now suppose $z\notin f([0,1]).$ Note that $f([0,1])$ is compact by the continuity of $f.$ Thus the distance from $z$ to $f([0,1])$ is some positive number $r.$ It follows that $D(z,r/2)\cap f([0,1])=\emptyset.$ Therefore $z$ is not in the essential range.
Thus the essential range is precisely the range, and we're done. 
