Is this the correct way to prove a sigma algebra?

Question

Let $X, Y$ be non-empty sets, $\mathcal A$ a $\sigma$-algebra on $X$ and $f:X\to Y$ a mapping.
Prove that $\mathcal B:=\{B\subseteq Y : f^{−1}(B)\in\mathcal A\}$ is a $\sigma$-algebra on $Y$.

1. $ f^{-1}(Y)=X \in \mathcal A\implies Y\in \mathcal B$

2. Let $B\subseteq Y$,
   $f^{-1}(B)^c=f^{-1}(B^c)$,
   then $x\in f^{-1}(B)^c \iff x\notin f^{-1}(B) \iff f(x)\notin (B)\iff f(x)\in B^c\iff x\in f^{-1}(B^c)$

3. $\bigcup\limits^\infty_{n=1}(B_n)\in \mathcal B$

$(\forall n\in N):B_n\in \mathcal B \implies (\forall n\in N): f^{-1}(B)\in \mathcal A\implies \bigcup^\infty_{n=1}f^{-1}(B_n)=f^{-1}\bigcup^\infty_{n=1}(B_n)\in \mathcal A\implies \bigcup^\infty_{n=1}(B_n)\in \mathcal B$

I would appreciate if anyone can tell me if this proof is correct, if not, to point out my mistakes. Thanks in advance!

EDIT Can anyone also recommend similar exercises? I want to practice proving sigma algebras.

Answer 1

1. A small comment: we don't necessarily know that $f(X)=Y$ (e.g. $f:\mathbb{R}\to\mathbb{R}, x\mapsto x^2$), but since $f$ has domain $X$, we can be sure that $f^{-1}(Y)=X$, so your argument still holds there.
2. Looks good
3. Besides what looks like a small typo (I assume you meant $f^{-1}\bigcup_{n=1}^\infty (B_n)\in\mathcal{A}$ rather than $f^{-1}\bigcup_{n=1}^\infty (B_n)\mathcal{A}$, everything else looks alright.

Answer 2

Compare your efforts with this.

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In general: $$f^{-1}(Y)=X\tag1$$ So from $X\in\mathcal A$ it follows that $Y\in\mathcal B$

In general: $$f^{-1}(B^c)=f^{-1}(B)^c\tag2$$ So if $B\in\mathcal B$ then $f^{-1}(B^c)\in\mathcal A$ implying that $B^c\in\mathcal B$

In general: $$f^{-1}\left(\bigcup_{n=1}^{\infty}B_n\right)=\bigcup_{n=1}^{\infty}f^{-1}(B_n)\tag3$$

So if $B_n\in\mathcal B$ for $n=1,2,\dots$ then $f^{-1}(\bigcup_{n=1}^{\infty}B_n)\in\mathcal A$ and consequently $\bigcup_{n=1}^{\infty}B_n\in\mathcal B$

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Proved is now that $\mathcal B$ is a $\sigma$-algebra.

I recommend you to have a look at this. Actually you just proved one side of what is posed there.