Does the intersection of an infinite $\sigma$-algebra with the complement of an element in the $\sigma$-algebra create a new $\sigma$-algebra?

Question

I am wondering if I have an infinite $\sigma$-algebra $M$ such that it contains infinitely many sets, and if I then take an $E \in M$ and define $$M_1=\{F \cap E^c : F \in M\},$$ then can I say that $M_1$ is also a $\sigma$-algebra? Perhaps not for the same sets, but maybe for the sets of $E^c$?

I hope my question makes sense. I am new to this concept.

Thank you.

Answer 1

You are right, $M_1$ is a $\sigma$-algebra in $E^c$. This is because

1. $E^c = X \cap E^c$ (assuming $M$ is a $\sigma$-algebra in $X$).
2. If $F \in M_1$, then $F = G \cap E^c$ for some $G \in M$. So, $G^c \cap E^c \in M_1$ as well, and $(G \cap E^c) \cup (G^c \cap E^c) = E^c \implies M_1$ is closed under complements.
3. Let $\{ F_i \}_{i=1}^\infty$ be a countable subcollection of $M_1$. Then, there exists a countable subcollection $\{ G_i \}_{i=1}^\infty$ of $M$ such that $F_i = G_i \cap E^c$. We have that $G = \bigcup_{i=1}^\infty G_i \in M$, so $F = G \cap E^c \in M_1$. But, $$G \cap E^c = \left( \bigcup_{i=1}^\infty G_i \right) \cap E^c=\bigcup_{i=1}^\infty (G_i \cap E^c) = \bigcup_{i=1}^\infty F_i.$$ Hence, $M_1$ is closed under countable unions.

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As an aside, I do not see why you choose to describe $M_1$ as $$ M_1 = \{ F \cap E^c : F \in M \}, $$ because you might as well call $E^c$ as another element $\tilde{E}$ in the original $\sigma$-algebra $M$ (because a $\sigma$-algebra is closed under taking complements).

So, the essential result is that if you take a $\sigma$-algebra and take the intersection of all its members with some fixed member of the $\sigma$-algebra, then this gives rise to another $\sigma$-algebra.

Also note that this result is true regardless of the number of elements in $M$.