Sigma Algebra induced by Function

Last updated Nov 1, 2022

Let $X$ and $(Y, \mathcal{N})$ be Measure Spaces. Let $f: X \to Y$ be a Function. Then the sigma algebra induced by $f$ is $$\sigma(f) := {f^{-1}(E) : E \in \mathcal{N}}$$

1. We should check that $\sigma(f)$ is indeed a Sigma Algebra

   Proof: We check the criteria for being a Sigma Algebra

   1. $Y \in \mathcal{N}$ and $f^{-1}(Y) = X \in \sigma(f)$.
   2. Suppose $A \in \sigma(f)$. Then there exists $E \in \mathcal{N}$ so $f^{-1}(E) = A$. Since $\mathcal{N}$ is Sigma Algebra we know $E^{C} \in \mathcal{N}$. Thus $$A^{C} = f^{-1}(E)^{C} = f^{-1}(E^{C}) \in \sigma(f)$$
   3. Suppose $(A_n){n=1}^{\infty} \subset \sigma(f)$. Then there exists $(E_n){n=1}^{\infty} \subset \mathcal{N}$ such that $f^{-1}(E_{n}) = A_{n}$. Then $\bigcup\limits_{n \in \mathbb{N}} E_{n} \in \mathcal{N}$. $$\bigcup\limits_{n \in \mathbb{N}} f^{-1}(E_{n}) = f^{-1}\Big(\bigcup\limits_{n \in \mathbb{N}} E_{n} \Big) \in \sigma(f)$$ $\blacksquare$