Essential range
Encyclopedia
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'. The essential range can be defined for measurable
real or complex-valued functions on a measure space.
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 z such that the inverse image of each ε-neighbourhood of z 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.
2. The essential range of an essentially bounded 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, Vε, of w that doesn’t intersect A; then f^(-1)(Vε) 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.
Every bounded complex-valued function defined on (X, μ) is essentially bounded.
Proof:
If |f| is bounded, then |f| < a for some a > 0 so that if g = |f|, then g−1(a, ∞) is empty and therefore has measure 0. This implies that the set S = { x | μ(g−1((x, ∞))) = 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∞(μ) is compact if μ is an non-negative 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 (wn) 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 Vε under f has positive measure. First of all, choose N such that n > N => wn belongs to Vε. Since Vε is open and since wN+1 belongs to Vε, we may choose a δ-neighbourhood, Vδ about wN+1 that is contained in Vε. Since wN+1 belongs to S, the inverse image of Vδ under f has positive measure. Since Vδ is a subset of Vε, f^(-1) (Vδ) is a subset of f^(-1) (Vε). Noting that f^(-1) (Vδ) has positive measure, it follows that f^(-1) (Vε) 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 = |f|. Therefore, if w is a complex number such that |w| > a, and K = {complex numbers z | |z| > a}, then there is a p-neighbourhood, Vp, 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) (Vp) had positive measure, so would f^(-1) (K) since f^(-1) (Vp) is a subset of f^(-1) (K); a contradiction. Therefore, f^(-1) (Vp) 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.
Applications of the theorems and additional notes
1. Note that the essential range of a function always lies within a closed ball in R2 of radius equal to the essential supremum of the function.
2. An essentially bounded function is intuitively a function that is unbounded on a set of measure 0, i.e. a negligible set in a measure-theoretic sense. A bounded function is basically a function that is unbounded on the empty set (which is not mathematically precise but gives the basic idea). Since the empty set has measure 0, one can believe that every bounded function is essentially bounded. This fact is proven in theorem 1.
3. Note that the proof of theorem 2 is largely dependent on the fact that non-negative additive measures are monotone.
Mathematics
Mathematics is the study of quantity, space, structure, and change. Mathematicians seek out patterns and formulate new conjectures. Mathematicians resolve the truth or falsity of conjectures by mathematical proofs, which are arguments sufficient to convince other mathematicians of their validity...
, particularly measure theory, the essential range of a function
Function (mathematics)
In mathematics, a function associates one quantity, the argument of the function, also known as the input, with another quantity, the value of the function, also known as the output. A function assigns exactly one output to each input. The argument and the value may be real numbers, but they can...
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'. The essential range can be defined for measurable
Measurable function
In mathematics, particularly in measure theory, measurable functions are structure-preserving functions between measurable spaces; as such, they form a natural context for the theory of integration...
real or complex-valued functions on a measure space.
Terminology and useful facts
- Throughout this article, the ordered pair (X, μ) will denote a measure space with non-negative additive measureMeasure- Legal :* Measure of the Church of England is a law passed by the General Synod and the UK Parliament equivalent of an Act* Measure of the National Assembly for Wales, a law specific to Wales passed by the Welsh Assembly between 2007 and 2011...
μ.
- One property of non-negative 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, ∞) and let S = { x | μ(ƒ −1((x, ∞))) = 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 = |f| less than infinity, f is said to be essentially bounded.
- The vector spaceVector spaceA vector space is a mathematical structure formed by a collection of vectors: objects that may be added together and multiplied by numbers, called scalars in this context. Scalars are often taken to be real numbers, but one may also consider vector spaces with scalar multiplication by complex...
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 spaceBanach spaceIn mathematics, Banach spaces is the name for complete normed vector spaces, one of the central objects of study in functional analysis. A complete normed vector space is a vector space V with a norm ||·|| such that every Cauchy sequence in V has a limit in V In mathematics, Banach spaces is the...
. This Banach space is often referred to as L∞(μ) and is an Lp spaceLp spaceIn mathematics, the Lp spaces are function spaces defined using a natural generalization of the p-norm for finite-dimensional vector spaces...
.
Formal definition
Let f be a complex valued function defined on a measure space, (X, μ) that also belongs to L∞(μ). Then the essential range of f is defined to be the set:- S = { complex numbers z | μ({ x : abs(f(x) - z) < ε }) > 0 for all ε > 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 z such that the inverse image of each ε-neighbourhood of z 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.
Properties and examples
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 an essentially bounded function f is always compact
Compact space
In mathematics, specifically general topology and metric topology, a compact space is an abstract mathematical space whose topology has the compactness property, which has many important implications not valid in general spaces...
. The proof is given in the next section.
3. The essential range, S, of a function is always a subset of the closure
Closed set
In geometry, topology, and related branches of mathematics, a closed set is a set whose complement is an open set. In a topological space, a closed set can be defined as a set which contains all its limit points...
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, Vε, of w that doesn’t intersect A; then f^(-1)(Vε) 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.
Theorems
Theorem 1Every bounded complex-valued function defined on (X, μ) is essentially bounded.
Proof:
If |f| is bounded, then |f| < a for some a > 0 so that if g = |f|, then g−1(a, ∞) is empty and therefore has measure 0. This implies that the set S = { x | μ(g−1((x, ∞))) = 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∞(μ) is compact if μ is an non-negative 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
Sequence
In mathematics, a sequence is an ordered list of objects . Like a set, it contains members , and the number of terms is called the length of the sequence. Unlike a set, order matters, and exactly the same elements can appear multiple times at different positions in the sequence...
in S converges to an element in S. Let (wn) 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 Vε under f has positive measure. First of all, choose N such that n > N => wn belongs to Vε. Since Vε is open and since wN+1 belongs to Vε, we may choose a δ-neighbourhood, Vδ about wN+1 that is contained in Vε. Since wN+1 belongs to S, the inverse image of Vδ under f has positive measure. Since Vδ is a subset of Vε, f^(-1) (Vδ) is a subset of f^(-1) (Vε). Noting that f^(-1) (Vδ) has positive measure, it follows that f^(-1) (Vε) 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 = |f|. Therefore, if w is a complex number such that |w| > a, and K = {complex numbers z | |z| > a}, then there is a p-neighbourhood, Vp, 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) (Vp) had positive measure, so would f^(-1) (K) since f^(-1) (Vp) is a subset of f^(-1) (K); a contradiction. Therefore, f^(-1) (Vp) 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.
Applications of the theorems and additional notes
1. Note that the essential range of a function always lies within a closed ball in R2 of radius equal to the essential supremum of the function.
2. An essentially bounded function is intuitively a function that is unbounded on a set of measure 0, i.e. a negligible set in a measure-theoretic sense. A bounded function is basically a function that is unbounded on the empty set (which is not mathematically precise but gives the basic idea). Since the empty set has measure 0, one can believe that every bounded function is essentially bounded. This fact is proven in theorem 1.
3. Note that the proof of theorem 2 is largely dependent on the fact that non-negative additive measures are monotone.