Measurable Function is Simple Function iff Finite Image Set

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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

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