Sigma Algebra produced by a Function

Last updated Nov 1, 2022

Let $(X, \mathcal{M})$ be a Measure Space, $Y$ a Set and $f : X \to Y$ a Function. Then $$\sigma^{*}(f) = {E \subset Y : f^{-1}(E) \in \mathcal{M}}$$ is a Sigma Algebra. It is called the Sigma Algebra produced by a Function.

We check the definition of a Sigma Algebra:

1. $f^{-1}(Y) = X \in \mathcal{M}$ because $\mathcal{M}$ is a Sigma Algebra. Therefore $Y \in \sigma^{*}(f)$.
2. Suppose $E \in \sigma^{}(f)$. Because $f^{-1}(E^{C}) = f^{-1}(E)^{C} \in \mathcal{M}$, we have that $E^{C} \in \sigma^{}(f)$.
3. Suppose $({E}{n}){n=1}^{\infty} \subset \sigma^{}(f)$. Then $f^{-1}(\bigcup\limits_{n=1}^{\infty} E_{n}) = \bigcup\limits_{n=1}^{\infty} f^{-1}(E_{n}) \in \mathcal{M}$. Therefore, $\bigcup\limits_{n=1}^{\infty} E_{n} \in \sigma^{}(f)$

Therefore $\sigma^{*}(f)$ is a Sigma Algebra on $Y$. $\blacksquare$

1. $f$ is by construction $(\mathcal{M}, \sigma^{*}(f))$-measureable.

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