Intuition behind subspace $\sigma$-Algebra Let $(X,\Sigma)$ be a measurable space and $D \subseteq X$. The subspace $\sigma$-Algebra of subsets of $D$, $$σ_D = \{E \cap D : E \in \Sigma\}$$
What is the intuition behind this definition? What is it trying to say? 
 A: It's analogous to the subspace topology. It turns out that this definition
gives a very "natural" definition of $\Sigma$ restricted to $D$. The definition is the right one for the structure induced by $\Sigma$ on $D\subseteq X$. Here are a couple of reasons why.

If $\Sigma$ is the Borel $\sigma$-algebra of a topology $\tau$ on a space $X$, and $D\subseteq X$, then the set $\Sigma_D$ as you define it is equal to the Borel $\sigma$-algebra of the subspace topology $\tau_D$ on $D$. The open sets of $\tau_D$ are exactly the sets $U\cap D$ for $U\in\tau$.

The subspace topology on $D$ has another description: it's the smallest topology on $D$ such that the inclusion $D \hookrightarrow X$ is continuous. Similarly, it's not hard to see that

$\Sigma_D$ is the smallest $\sigma$-algebra on $D$ such that the inclusion $i_D\colon D \hookrightarrow X$ is measurable — that is,  $\text{for all $E\in\Sigma, i_D^{-1}(E) \in \Sigma_D$.} $

It's immediate from the definitions that $i_D\colon (D,\Sigma_D)\to (X,\Sigma)$ is measurable, because $i_D^{-1}(E) = E\cap D$.
Suppose now that M is $\sigma$-algebra $M$ on $D$ such that $i_D\colon (D,M)\to (X,\Sigma)$ is measurable. Then for every $E\in\Sigma, $ we have $i_D^{-1}(E) = E\cap D \in M$. Thus $\Sigma_D\subseteq M$.

A concrete example: Suppose $X = \Bbb N$, $\Sigma = \mathcal{P}(X)$, and $D = $ the even integers. Do you see that $\Sigma_D = \{E\cap D\mid E\in \mathcal{P}(X)\} = \mathcal{P}(D)$?
A: Perhaps too short for an answer, but the intuition behind it, is that you want to focus on $D$, so you cut everything else off, and you keep only everything's part in $D$, i.e $D\cap E$ for all $E$'s.
