Measurable Function is Simple Function iff Finite Image Set

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\left({X, \Sigma}\right)$ be a measurable space.

Let $f: X \to \R$ be a measurable function.


Then $f$ is a simple function iff its image is finite:

$\# \operatorname{Im} \left({f}\right) < \infty$


Corollary

Every simple function $f: X \to \R$ has a standard representation.


Proof

Necessary Condition

Suppose that $f$ is a simple function, and that:

$\displaystyle \forall x \in X: f \left({x}\right) = \sum_{i \mathop = 1}^n a_i \chi_{S_i} \left({x}\right)$

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 $\left\{{f = a}\right\}$ for $\left\{{x \in X: f \left({x}\right) = a}\right\}$ (compare Set Definition by Predicate).


Define for each $i$ with $1 \le i \le n$:

$B_i := \left\{{f = y_i}\right\}$

From Characterization of Measurable Functions $(2)$ and $(4)$, and Sigma-Algebra Closed under Intersection we obtain that:

$\left\{{f = y_i}\right\} = \left\{{f \ge y_i}\right\} \cap \left\{{f \le y_i}\right\} \in \Sigma$


Furthermore, since the $y_i$ are distinct, the $B_i$ are necessarily disjoint.

It follows that:

$(1): \quad f \left({x}\right) = \displaystyle \sum_{i \mathop = 1}^n y_j \chi_{B_j} \left({x}\right)$

and as the $B_i$ are measurable, $f$ is shown to be a simple function.

$\blacksquare$


Proof of Corollary

Applying the main theorem to a simple function yields the representation $(1)$ which is of the required form.

$\blacksquare$


Sources