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 its image is finite:


 * $\card {\Img f} < \infty$

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.

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.