Proving a sigma algebra

Question

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.

Edit 2 Method 1.

1. $\emptyset =\emptyset \cap M \in \mathcal A_M,\mathcal A_M\neq \emptyset $

2. $A\cap M \in \mathcal A_M, M$\ $(A\cap M)=(A^c\cap M)\in \mathcal A_M$ Because $\mathcal A$ is closed under compliment $\mathcal A_M$ is closed under compliment
3. Suppose $A\cap M \subset \mathcal A_M$ is a given sequence . $\bigcup^\infty_{n=1}A_n\cap M=(\bigcup^\infty_{n=1}A_n)\cap M \in\mathcal A_M $ Since $\mathcal A$ is closed under countable union $\mathcal A_M$ is also closed under countable union.

Method 2 $T:M\to X, T^{-1}(A)=A\cap M\; for\; every\; A\subset X $ so that $\mathcal A_M=\{T^{-1}(A)\mid A\in\mathcal A\}\subseteq\mathcal P(M) $

1. $T(M)=X\implies T^{-1}(X)=M \in \mathcal A\implies M\in \mathcal A_M$
2. $A\subset M $ Let $x\in X$ $x\in T^{-1}(A)^c \iff x\notin T^{-1}(A) \iff T(x)\notin (A)\iff f(x)\in A^c\iff x\in T^{-1}(A^c)$
3. $\bigcup^\infty_{n=1}T^{-1}(A_n)=T^{-1}\bigcup^\infty_{n=1}(A_n)$, then let $x\in X$.

$x\in \bigcup^\infty_{n=1}T^{-1}(A_n) \iff \exists_{n_0}\in N, x\in T^{-1}(A_{n_0}) \iff \exists_{n_0} N T(x)\in A_{n_0} \iff T(x)\in \bigcup^\infty_{n=1} A_n \iff x\in T^{-1}\bigcup^\infty_{n=1}(A_n)$

I tried to do it two ways with all of the suggestions and revision on my part, I would appreciate feedback

Answer 1

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.

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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.

Answer 2

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!)

Answer 3

For the complements it is required that $M\setminus (A\cap M) \in\mathcal A_M$ whenever $A\in\mathcal A$. Note that $$ M\setminus (A\cap M) = A^c\cap M \in\mathcal A_M.$$