Let $X,Y$ be two given non empty sets and let $\mathcal{F}$ be a $\sigma$-algebra on $Y$.
Let $f : X \rightarrow Y$ be a given function.
Define
$$\mathcal{G} = \lbrace \ f^{-1}(A)\ ; \ A\in\mathcal{F} \ \rbrace $$
- Prove that $\mathcal{G}$ is a $\sigma$-algebra on $X$.
- If $X=Y=\mathbb{R}$, and $\mathcal{F}=\mathcal{B}(\mathbb{R})$, let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a constant function, $f(x)=\alpha \ \forall x \in \mathbb{R}$.
Describe $\mathcal{G}$.
Thank you for your help.
To prove that $\mathcal{G}$ is a $\sigma$-algebra, we only need to prove that it is closed under complements and countable unions, and that it includes $X$ and $\varnothing$. The inverse image of $Y$ is $X$ and the inverse image of $\varnothing$ is $\varnothing$, so the last criteria are satisfied. Complement-closure is proved by the fact that $(f^{-1}(A))^c=f^{-1}(A^c)$, along with the fact that if $A$ is in $\mathcal{F}$, so is its complement -- because $\mathcal{F}$ is a $\sigma$-algebra. Finally, closure under countable unions is proved by the fact that $f^{-1}(A) \cup f^{-1}(B) = f^{-1}(A \cup B)$ -- by induction, this holds for all unions, up to countably infinite ones. We also know that if a countable collection of sets is in $\mathcal{F}$, so is the union of its members -- again, by $\mathcal{F}$ being a $\sigma$-algebra. Therefore, $\mathcal{G}$ is closed under countable unions, and is a $\sigma$-algebra.
The $\sigma$-algebra $\mathcal{G}$ in the second part of the question is trivial -- Given $A$ in $\mathcal{F}$, either $A$ contains $x$, or it does not. If it does, its inverse image is the whole set $X$, if it does not, it is $\varnothing$. So $\mathcal{G}$ is the minimal $\sigma$-algebra generated by $\{X\}$: $\left \{ X, \varnothing \right \}$.