Let $X$ be a set and $M$ be a sigma algebra of $X$. If $U \subset X$, is it true that the set $M_U = \{E \cap U|E \in M\}$ is a sigma algebra? I'm getting the countable union part, but not compliments:
$$(E\cap U)^C = E^C \cup U^C$$
I'm not sure what to do with $U^C$
On
If $X$ is a set, $\mathcal M$ a $\sigma$-algebra on it and $f:Y\to X$ is a function then the collection $f^{-1}(\mathcal M)$ can be shown to be a $\sigma$-algebra on $Y$.
Observe that in that situation $f^{-1}(E^{\complement})=f^{-1}(E)^{\complement}$.
You can apply that here by taking $Y=U$ and $f:U\to X$ the inclusion.
That will lead to $f^{-1}(\mathcal M)=\mathcal M_U$ here.
The complement of $E\subseteq U$ in universe $U$ is $E^{\complement}\cap U$.
This because you are focusing on a collection of subsets of $U$.
Well, you want to show that $\mathcal{M}_U$ is an $U$ $\sigma-$ algebra, thus you have to show that $$U \cap (E \cap U)^c \in \mathcal{M}_U$$ But $$U \cap (E \cap U)^c = U \cap (E^c \cup U^c) = U \cap E^c \in \mathcal{M}_U$$