TITLE: Definition of a measure subspace
QUESTION [6 upvotes]: The measure theory notes (John K. Hunter) state:
Definition 1.9 For set $X$ and $\sigma$-algebra $\mathcal{A}$ on set X, a measure $\mu$ on the measurable space $(X, \mathcal{A})$ is a function such that:

$\mu(\emptyset) = 0$
It is countably additive. In other words, if $\lbrace A_{i} \in \mathcal{A} : i \in \mathbb{N} \rbrace$ is a countable disjoint collection of sets in $\mathcal{A}$, then

$$\mu \left ( \bigcup_{i = 1}^{\infty} A_{i} \right) = \sum_{i = 1}^{\infty} \mu(A_{i})$$
Definition 1.10
If $(X, \mathcal{A}, \mu)$ is a measure space (a measurable space and a particular measure) and $E \subset X$ is a measurable subset, then the measurable subspace $(E, {\mathcal{A}|}_{E}, {\mu|}_{E})$ is defined by restricting $\mu$ to $E$ where:

${\mathcal{A}|}_{E} = \lbrace A \cap E : A \in \mathcal{A} \rbrace$
${\mu|}_{E}(A \cap E) = \mu(A \cap E)$

My question
From the answer in Show that subspace measure is a measure, I get that the measure subspace is indeed a measure space, except that the answer seems not to mention that $E$ need to be a "measurable subset".
My interpretation of that "$E$ is a measurable subset of $X$" is that:

$E$ is a subset of $X$
there exists a collection of subsets of $E$ that forms a $\sigma$-algebra.

Why does $E$ really have to be a "measurable subset" of $X$ instead of just a subset of $X$?

REPLY [3 votes]: If $E\notin\mathcal A$ then $E\cap A\notin\mathcal A$ for $A\in\mathcal  A$, so $\mu(E\cap A)$  is undefined.