Proving a sigma algebra Let $(X, \mathcal A)$ be a measurable space and $M\subset X$ an arbitrary set:
Show: $\mathcal A_M$:{$A\cap M: A\in \mathcal A$} is a sigma algebra on M
So I got:
$A\cap M \rightarrow A\in \mathcal A\,, A\cap M\in \mathcal A$ (How I can show that $M\in \mathcal A$? wrong way) can I say $A\cap M\in \mathcal A$ so $A\in \mathcal A$ and $M\in \mathcal A$? Also where would I get the empty set from?
$A\in\mathcal A \to A\subset X\; and\; M\subset X$ also $(A\cap M)^c=A^c\cup M^c$ Where do I go from here?
Edit so I can say because $\mathcal A $ is a sigma algebra and A belong to it then so does the A compliment so the intersection of A and M compliment will belong in the sigma algebra. Still  not sure where to go with this information.
Let $A_1\cap M_1,...$ be a countable sequence of elements of A_M. The union $\cup^\infty_{k=1}(A_k\cap M)=A_1\cup A_2\cup...\cup M$, is in $A_M$ not sure how to explain.
If anyone can fully explain how to do everything correctly it would be most appreciated.
 A: Let $\iota: M\to X$ denote the inclusion function prescribed by $x\mapsto x$ for $x\in M$.
Then $\iota^{-1}(A)=A\cap M$ for every $A\subseteq X$ so that: $$\mathcal A_M=\{\iota^{-1}(A)\mid A\in\mathcal A\}\subseteq\mathcal P(M)\tag1$$
So proving that $\mathcal A_M$ is a $\sigma$-algebra boils down to proving that:

*

*$M=\iota^{-1}(X)$

*$\iota^{-1}(A)^c=\iota^{-1}(A^c)$

*$\bigcup_{n=1}^{\infty}\iota^{-1}(A_n)=\iota^{-1}\left(\bigcup_{n=1}^{\infty}A_n\right)$
This because we already know that $\mathcal A$ is a $\sigma$-algebra.

Edit (to make things more clear)
The 3 bullets make it easy to prove that $\mathcal A_M$ is a $\sigma$-algebra. The first tells us immediately that $M\in\mathcal A_M$ because $M =\iota^{-1}(X)$ and $X\in\mathcal A$.
With the second it can be shown that $\mathcal A_M$ is closed under complementation. If $B\in\mathcal A_M$ then according to $(1)$ we have $B=\iota^{-1}(A)$ for some $A\in\mathcal A$. Then $B^c=(\iota^{-1}(A))^c=\iota^{-1}(A^c)$ and from $A\in\mathcal A$ it follows that $A^c\in\mathcal A$. Then $B^c\in\mathcal A_M$ according to $(1)$.
On a similar way it can be shown that $\mathcal A_M$ is closed under the formation of countable unions by means of the statement behind the third bullet.
A: For the complements it is required that $M\setminus (A\cap M) \in\mathcal A_M$ whenever $A\in\mathcal A$. Note that
$$ M\setminus (A\cap M) = A^c\cap M \in\mathcal A_M.$$
A: To long for a comment - but a comment:
In addition to the above hint, a clarification. You do NOT need to show that $M \cap A\in {\cal A}$, or $M\in {\cal A}$ in particular. For instance, take $X= \{1,0\}$, ${\cal A} =\{\emptyset, X\}$, and $M=\{0\}$. $M$ is certainly not in $\cal A$.
The question is analogous to the question about the induced, 'relative topology' on a subset $M$: show that if ${\cal T}$ is a topology on $X$, then $ \{A\cap M\mid A\in {\cal T}\}$ forms a topology on $M$. ($M$ need NOT be open!)
