Measurable Function is Simple Function iff Finite Image Set
Theorem
Let $\struct {X, \Sigma}$ be a measurable space.
Let $f: X \to \R$ be a measurable function.
Then $f$ is a simple function if and only if its image is finite:
- $\card {\Img f} < \infty$
Corollary
Let $f: X \to \R$ be a measurable function.
Then $f$ has a standard representation.
Proof
Necessary Condition
Suppose that $f$ is a simple function, and that:
- $\ds \forall x \in X: \map f x = \sum_{i \mathop = 1}^n a_i \map {\chi_{S_i} } x$
Since each of the $\chi_{S_i}$ is a characteristic function, it can take only the values $0$ and $1$.
Thus each summand can take two values.
It follows immediately that $f$ can take at most $2^n$ different values.
The conclusion follows from Simple Function is Measurable.
$\Box$
Sufficient Condition
Suppose that the image of $f$ is finite.
Call the distinct values $f$ attains $y_1, \ldots, y_n$.
For brevity, denote $\set {f = a}$ to mean $\set {x \in X: \map f x = a}$ (compare Set Definition by Predicate).
Define for each $i$ with $1 \le i \le n$:
- $B_i := \set {f = y_i}$
From Characterization of Measurable Functions $(2)$ and $(4)$, and Sigma-Algebra Closed under Intersection we obtain that:
- $\set {f = y_i} = \set {f \ge y_i} \cap \set {f \le y_i} \in \Sigma$
Furthermore, since the $y_i$ are distinct, the $B_i$ are necessarily disjoint.
It follows that:
- $(1): \quad \map f x = \ds \sum_{i \mathop = 1}^n y_j \map {\chi_{B_j} } x$
As the $B_i$ are measurable, $f$ is shown to be a simple function.
$\blacksquare$
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $8.7 \ \text{(iii)}$