Show that ${F : F ⊆ Y, f ^{-1}[F] ∈ Σ}$ is a σ-algebra

Question

I have two sets X and Y be and let Σ be a σ-algebra of subsets of X.

Let f : X → Y be a function.

How do I show that show that $\{F : F ⊆ Y, f^{-1}[F] ∈ \Sigma\}$ is a σ-algebra of subsets of Y .

Answer 1

Hint. Use the properties of the $f^{-1}[\cdot]$-operation, namely that $$ \def\ii#1{f^{-1}\left[#1\right]} \ii{Y \setminus F} = X \setminus \ii F, \ii{\bigcup_n F_n} = \bigcup_n \ii{F_n} $$ and the fact that $\Sigma$ is a $\sigma$-algebra.

Answer 2

Apply the definition of $\sigma$-algebra to show that $\Sigma_Y = \{F \subseteq Y \mid f^{-1}[Y] \in \Sigma\}$ is a $\sigma$-algebra on Y. But this is clear (nevertheless, you still have to show the details!), as $f^{-1}$ distributes over complements as well as over arbitrary intersections and unions.

Answer 3

Hint: Let's abuse notation and call $f^{-1}(\Sigma)$ that set. You must verify that:

• $\varnothing \in f^{-1}(\Sigma)$;

• $(A_n)_{n \geq 1}\subseteq f^{-1}(\Sigma) \implies \bigcup_{n \geq 1} A_n \in f^{-1}(\Sigma);$

• $A \in f^{-1}(\Sigma) \implies Y \setminus A \in f^{-1}(\Sigma).$

Use that $f^{-1}(\varnothing) = \varnothing$, and that inverse images preserve complements and unions.