Essential range

In mathematics, particularly measure theory, the essential range, or the set of essential values, of a function is intuitively the 'non-negligible' range of the function: It does not change between two functions that are equal almost everywhere. One way of thinking of the essential range of a function is the set on which the range of the function is 'concentrated'.

Formal definition
Let $$(X,{\cal A},\mu)$$ be a measure space, and let $$(Y,{\cal T})$$ be a topological space. For any $$({\cal A},\sigma({\cal T}))$$-measurable $$f:X\to Y$$, we say the essential range of $$f$$ to mean the set
 * $$\operatorname{ess.im}(f) = \left\{y\in Y\mid0<\mu(f^{-1}(U))\text{ for all }U\in{\cal T} \text{ with } y \in U\right\}.$$

Equivalently, $$\operatorname{ess.im}(f)=\operatorname{supp}(f_*\mu)$$, where $$f_*\mu$$ is the pushforward measure onto $$\sigma({\cal T})$$ of $$\mu$$ under $$f$$ and $$\operatorname{supp}(f_*\mu)$$ denotes the support of $$f_*\mu.$$

Essential values
We sometimes use the phrase "essential value of $$f$$" to mean an element of the essential range of $$f.$$

Y = C
Say $$(Y,{\cal T})$$ is $$\mathbb C$$ equipped with its usual topology. Then the essential range of f is given by
 * $$\operatorname{ess.im}(f) = \left\{z \in \mathbb{C} \mid \text{for all}\ \varepsilon\in\mathbb R_{>0}: 0<\mu\{x\in X: |f(x) - z| < \varepsilon\}\right\}.$$

In other words: The essential range of a complex-valued function is the set of all complex numbers z such that the inverse image of each ε-neighbourhood of z under f has positive measure.

(Y,T) is discrete
Say $$(Y,{\cal T})$$ is discrete, i.e., $${\cal T}={\cal P}(Y)$$ is the power set of $$Y,$$ i.e., the discrete topology on $$Y.$$ Then the essential range of f is the set of values y in Y with strictly positive $$f_*\mu$$-measure:
 * $$\operatorname{ess.im}(f)=\{y\in Y:0<\mu(f^\text{pre}\{y\})\}=\{y\in Y:0<(f_*\mu)\{y\}\}.$$

Properties

 * The essential range of a measurable function, being the support of a measure, is always closed.
 * The essential range ess.im(f) of a measurable function is always a subset of $$\overline{\operatorname{im}(f)}$$.
 * The essential image cannot be used to distinguish functions that are almost everywhere equal: If $$f=g$$ holds $$\mu$$-almost everywhere, then $$\operatorname{ess.im}(f)=\operatorname{ess.im}(g)$$.
 * These two facts characterise the essential image: It is the biggest set contained in the closures of $$\operatorname{im}(g)$$ for all g that are a.e. equal to f:
 * $$\operatorname{ess.im}(f) = \bigcap_{f=g\,\text{a.e.}} \overline{\operatorname{im}(g)}$$.


 * The essential range satisfies $$\forall A\subseteq X: f(A) \cap \operatorname{ess.im}(f) = \emptyset \implies \mu(A) = 0$$.
 * This fact characterises the essential image: It is the smallest closed subset of $$\mathbb{C}$$ with this property.
 * The essential supremum of a real valued function equals the supremum of its essential image and the essential infimum equals the infimum of its essential range. Consequently, a function is essentially bounded if and only if its essential range is bounded.
 * The essential range of an essentially bounded function f is equal to the spectrum $$\sigma(f)$$ where f is considered as an element of the C*-algebra $$L^\infty(\mu)$$.

Examples

 * If $$\mu$$ is the zero measure, then the essential image of all measurable functions is empty.
 * This also illustrates that even though the essential range of a function is a subset of the closure of the range of that function, equality of the two sets need not hold.
 * If $$X\subseteq\mathbb{R}^n$$ is open, $$f:X\to\mathbb{C}$$ continuous and $$\mu$$ the Lebesgue measure, then $$\operatorname{ess.im}(f)=\overline{\operatorname{im}(f)}$$ holds. This holds more generally for all Borel measures that assign non-zero measure to every non-empty open set.

Extension
The notion of essential range can be extended to the case of $$f : X \to Y$$, where $$Y$$ is a separable metric space. If $$X$$ and $$Y$$ are differentiable manifolds of the same dimension, if $$f\in$$ VMO$$(X, Y)$$ and if $$\operatorname{ess.im} (f) \ne Y$$, then $$\deg f = 0$$.