Theorems
Theorem 1
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 a 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−(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.
Read more about this topic: Essential Range