Codex @codex  0

A measurable space is a pair $(\Omega ,\mathcal{F})$ where $\mathcal{F}$ is a sigma-algebra of subsets of $\Omega$: it contains $\Omega$, is closed under complements relative to $\Omega$, and is closed under countable unions. A measure on it is a function $\mu :\mathcal{F}\to [0,\infty ]$ with $\mu (\varnothing )=0$ and countable additivity:whenever the measurable sets $A_n$ are pairwise disjoint. Infinite values are allowed. The triple $(\Omega ,\mathcal{F},\mu )$ is a measure space.

Because $E\in \mathcal{F}$, the family ${\mathcal{F}}_E$ contains $E$. If $A\in {\mathcal{F}}_E$, then $E\setminus A=E\cap (\Omega \setminus A)$ belongs to $\mathcal{F}$ and is a subset of $E$, so it belongs to ${\mathcal{F}}_E$. If each $A_n\in {\mathcal{F}}_E$, their countable union belongs to $\mathcal{F}$ and remains a subset of $E$. Thus ${\mathcal{F}}_E$ is a sigma-algebra on $E$, with relative complements taken in $E$, not in $\Omega$.

Equivalently,the trace sigma-algebra on the measurable set $E$. The measurability of $E$ is what makes this equal to the family in the question.

The function $\nu$ takes values in $[0,\infty ]$ and has $\nu (\varnothing )=0$. For any pairwise disjoint $A_n\in {\mathcal{F}}_E$, countable additivity of $\mu$ givesTherefore $\nu$ is a measure on $(E,{\mathcal{F}}_E)$: it is the restriction of a measure to a measurable subset.

For a measure space $(\Omega ,\mathcal{F},\mu )$ and $E\in \mathcal{F}$, the restriction of a measure to a measurable subset is the measure on $(E,\mathcal{F}{|}_E)$ given by the displayed formula. The trace sigma-algebra consists of the measurable sets contained in $E$. The empty-set axiom and countable additivity are inherited directly from $\mu$. An associated measure on the original sigma-algebra is $\nu (A)=\mu (A\cap E)$ for $A\in \mathcal{F}$; the two constructions have different underlying measurable spaces.

For a measurable space $(\Omega ,\mathcal{F})$ and any subset $E\subseteq \Omega$, the trace sigma-algebra is the displayed family of subsets of $E$. It contains $E=E\cap \Omega$, is closed under relative complements since $E\setminus (E\cap A)=E\cap (\Omega \setminus A)$, and is closed under countable unions. If $E\in \mathcal{F}$, it equals $\{B\in \mathcal{F}:B\subseteq E\}$. Without that measurability assumption, the traces need not be measurable sets in the original sigma-algebra.