In mathematics, particularly measure theory, the essential range of a function is intuitively the 'non-negligible' range of the function. One way of thinking of the essential range of a function is the set on which the range of the function is most 'concentrated'. Note that the essential range is only defined for complex valued functions belonging to a particular L^p space.

• Throughout this article, the ordered pair (X, μ) will denote a measure space with additive measure μ.

• One property of additive measures is that they are monotone; that is if A is a subset of B, then μ(A) <= μ(B) if μ is additive.

• Let f be a function from a measure space (X, μ) to [0, + infinity) and let S = {x | μ(f^(-1) (x, +infinity)) = 0}. The essential supremum of f, is defined to be the infimum of S. If S is empty, the essential supremum of f is defined to be infinity.

• If f is a function such that the essential supremum of g = abs(f) less than infinity, f is said to be essentially bounded.

• The vector space of all essentially bounded functions with the norm of a function defined to be its essential supremum, forms a complete metric space with the metric induced by its norm. Mathematically, this means that the collection of all essentially bounded functions form a Banach space. This Banach space is often referred to as L^infinity(μ) and is an L^p space.

Let f be a complex valued function defined on a measure space, (X, μ) that also belongs to L_infinity(μ). Then the essential range of f is defined to be the set:

S = {complex numbers z | μ({x: abs (f(x) - w) < e}) > 0 for all e > 0}

Note that: Another description of the essential range of a function is as follows:

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

The above description of the essential range is equivalent to the formal definition of the essential range and will therefore be used throughout this article.

1. Every complex-valued function defined on the measure space (X, μ) whose absolute value is bounded, is essentially bounded. A proof is provided in the next section.

2. The essential range of a function f is always compact. The proof is given in the next section.

3. The essential range, S, of a function is always a subset of the closure of A where A is the range of the function. This follows from the fact that if w is not in the closure of A, there is a ε-neighbourhood, V_ε, of w that doesn’t intersect A; then f^(-1)(V_ε) has 0 measure which implies that w cannot be an element of S.

4. Note that the essential range of a function may be empty even if the range of the function is non-empty. If we let Q' be the set of all rational numbers and let T be the power set of Q, then (Q, T, m) form a measurable space with T the sigma algebra on Q, and m a measure defined on Q that maps every member of T onto 0. If f is a function that maps Q onto the set of all points with rational co-ordinates that lie within the unit circle, then f has nonempty range (clearly). The essential range of f however is empty for if w is any complex number and V any ε-neighbourhood of w, then f^(-1) (V) has 0 measure by construction.

5. Example 4 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.

Theorem 1

Every essentially bounded complex-valued function defined on (X, μ) whose absolute value is bounded, is essentially bounded.

Proof

If abs(f) is bounded, then abs(f) < a for some a > 0 so that if g = abs(f), then g^(-1) (a, +infinity) is empty and therefore has measure 0. This implies that the set S = {x | μ(g^(-1) (x, +infinity)) = 0} is nonempty so that the essential supremum of g is less than infinity. Therefore, f is essentially bounded.

Theorem 2

The essential range of a complex valued function, f, defined on a measure space (X, μ) that belongs to L^infinity(μ) is compact if μ is an additive measure.

Proof

Let S denote the essential range of the function in question. By the Heine-Borel theorem, it suffices to show that S is closed and bounded. To show that S is closed, we will show that every convergent sequence in S converges to an element in S. Let (w_n) be a convergent sequence of points in S and let w be its limit. Let V be an ε-neighbourhood of w; we will prove that the inverse image of V_ε under f has positive measure. First of all, choose N such that n > N => w_n belongs to V_ε. Since V_ε is open and since w_N+1 belongs to V_ε, we may choose a δ-neighbourhood, V_δ about w_N+1 that is contained in V_ε. Since w_N+1 belongs to S, the inverse image of V_δ under f has positive measure. Since V_δ is a subset of V_ε, f^(-1) (V_δ) is a subset of f^(-1) (V_ε). Noting that f^(-1) (V_δ) has positive measure, it follows that f^(-1) (V_ε) has positive measure. Since ε was arbitrary, it follows that w belongs to S and S is closed.

Note that since f is essentially bounded, there exists a such that g^(-1) (a, +infinity) has 0 measure where g = abs(f). Therefore, if w is a complex number such that abs(w) > a, and K = {complex numbers z | abs(z) > a}, then there is a p-neighbourhood, V_p, of w that is contained in K (since K is open). Note that g^(-1) (a, +infinity) = f^(-1) (K) so that f^(-1) (K) has 0 measure. If f^(-1) (V_p) had positive measure, so would f^(-1) (K) since f^(-1) (V_p) is a subset of f^(-1) (K); a contradiction. Therefore, f^(-1) (V_p) has 0 measure so that w cannot be an element of S. This shows that S is a subset of the complement of K so that S is bounded.

• Essential supremum
• Additive measure

• Walter Rudin (1974). Real and Complex Analysis (2nd edition ed.). McGraw-Hill. ISBN 978-0070542341. {{cite book}}: |edition= has extra text (help)