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The Induced Sigma Algebra of a Borel Function is Finite iff it is Simple

Last updated Nov 1, 2022

# Statement

Let $X$ be a Set. Suppose $f: X \to \mathbb{R}$. Then $\sigma(f)$ with respect to $\mathcal{B}(\mathbb{R})$ is a Finite Set If and Only If $f$ is a Simple Function.

# Proof

$(\Rightarrow)$: Suppose $f$ is not simple. Then $f(X)$ is an Infinite Set. Since Singletons are Closed in the Real Numbers, they are in $\mathcal{B}(\mathbb{R})$. In other words, ${y} \in \mathcal{B}(\mathbb{R})$ for all $y \in f(X)$. Since $y \in f(X)$, we have that $f^{-1}({y}) \neq \emptyset$. On the other hand for $y, z \in f(X)$ so that $y \neq z$, we have $f^{-1}({y}) \cap f^{-1}({z}) = f^{-1}({y} \cap {z}) = f^{-1}(\emptyset) = \emptyset$, so each of our preimages are Nonempty Disjoint Sets (and are thus distinct). Therefore, we have infinitely many distinct sets in $\sigma(f)$. $\checkmark$

($\Leftarrow$): A Simple Function Induces a Finite Sigma Algebra. $\checkmark$ $\blacksquare$

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