Measurability on subsets. Given a measure space $(X, \mathcal{M})$ we say that $f : X \to \mathbb{R}$ is $\mathcal{M}$-measurable if $f$ is ($\mathcal{M}, \mathcal{B}_\mathbb{R}$) measurable where $\mathcal{B}_\mathbb{R}$ is the Borel $\sigma$-algebra on $\mathbb{R}$.
Given $E \in \mathcal{M}$, we say that $f$ is measurable on $E$ if $f^{-1}(B) \cap E \in \mathcal{M}$ for all Borel $B$.
How do I see that this condition is equivalent to saying that $f$ restricted to $E$ is $M_E$ measurable where $E = \{F \cap E : F \in \mathcal{M}\}$?
 A: Hint: For any $E\subseteq X$, not necessarily belonging to ${\cal M}$, $\sigma$-algebra in $E$, ${\cal M}_{E} = \{E\cap A| A \in {\cal M}\} = i^{-1}\left({\cal M}\right)$, makes $\left(E, {\cal M}_{E}\right)$ a measurable space and the natural injection $i:\left(E, {\cal M}_{E} \right) \hookrightarrow \left(X, {\cal M} \right)$ a measurable function. 
Note that, if $E\in  {\cal M}$, then ${\cal M}_{E} = \{A| A \in {\cal M}, A\subseteq E\}$, so ${\cal M}_{E}\subseteq {\cal M}$.
Given $f: X \to \mathbb{R}$ an arbitrary function, we say $f$ is measurable on $E$ when $f|_{E} = f\circ i : \left(E, {\cal M}_{E} \right) \to \left(\mathbb{R}, {\cal B}_\mathbb{R}\right)$ is measurable. By definition this means $$\left(f|_{E}\right)^{-1}\left(B \right) \in {\cal M}_{E}$$
for all $B\in {\cal B}_\mathbb{R}$, which is equivalent to:
$$  f^{-1}\left(B \right)\cap E \in {\cal M}_{E}, $$
for all $B\in {\cal B}_\mathbb{R}$, as
$$ \left(f|_{E}\right)^{-1}\left(B \right) = (f\circ i)^{-1}\left(B \right) = i^{-1}\left(f^{-1}\left(B \right) \right) = f^{-1}\left(B \right)\cap E.$$
Note that, if $f: \left(X, {\cal M}\right) \to \left(\mathbb{R}, {\cal B}_\mathbb{R}\right)$ is measurable, then $f|_{E} : \left(E, {\cal M}_{E} \right) \to \left(\mathbb{R}, {\cal B}_\mathbb{R}\right)$ is automatically measurable.
