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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.

ugjumb
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  • Ugjumb, $f(X)=Y$ is true only if $f$ is a surjective function (onto function). – Angelo Mar 06 '23 at 16:06
  • If it is not stated can I just say "assume that f is/let f be a surjective function..."? Or is there a better way I can prove nr.1 – ugjumb Mar 06 '23 at 16:08
  • You do not need $f$ is a surjective function because, even if $f(X)\neq Y$, $f^{-1}(Y)=X$ is always true. Actually, you do not need to write $f(X)=Y$. Please see other comments and answers. – Angelo Mar 06 '23 at 16:10
  • $f(X)=Y$ is not needed. Enough is $f^{-1}(Y)=X$ which does not need surjectivity. – drhab Mar 06 '23 at 16:11

2 Answers2

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  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.
S.L.
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  • Thank you! Can you maybe recommend some other tasks with sigma algebras? – ugjumb Mar 06 '23 at 16:20
  • Here's a nice one: Prove that the intersection of two sigma algebras is also a sigma algebra, and give a counter example to show that the union of two sigma algebras may not be a sigma algebra – S.L. Mar 06 '23 at 16:31
  • It truly is a nice one, but I already solved it. Usually I find proving sigma algebras with intersection or subset in the requirements, could the be a union or a difference instead? – ugjumb Mar 06 '23 at 16:35
1

Compare your efforts with this.

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 $\bigcup_{n=1}^{\infty}B_n\in\mathcal B$

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.

drhab
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