A Simple Function Induces a Finite Sigma Algebra

Last updated Nov 1, 2022

Let $(Y, \mathcal{N})$ be a Measure Space and let $X$ be a Set. Suppose $f: X \to Y$ is a Simple Function. Then $\sigma(f)$ is a Finite Set.

Suppose $f$ is a Simple Function. Then $f(X) \subset Y$ is a Finite Set. Suppose $|f(X)| = n \in \mathbb{N}$. Let $E \in \mathcal{N}$ be a Measureable Set. Then $$\begin{align*} f^{-1}(E) &= f^{-1}(E) \cap X \\ &= f^{-1}(E) \cap f^{-1}(f(X))\\ &=f^{-1}(E \cap f(X)) \end{align*}$$ Since $E \cap f(X) \subset f(X)$ and $f(X)$ is finite, $E \cap f(X)$ has only $2^{n}$ possibilities. Thus, $|\sigma(f)| \leq 2^{n}$ and is thus a Finite Set. $\blacksquare$

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