= Solution
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>:
$$
\mu\left(\bigcup_{n=1}^{\infty}A_n\right)=\sum_{n=1}^{\infty}\mu(A_n)
$$
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 \b[$\mathcal F_E$ is a <sigma-algebra> on $E$], with relative <complements> taken in $E$, not in $\Omega$.
Equivalently,
$$
\mathcal F_E=\{E\cap B:B\in\mathcal F\},
$$
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 $\mu$ gives
$$
\nu\left(\bigcup_nA_n\right)=\mu\left(\bigcup_nA_n\right)
=\sum_n\mu(A_n)=\sum_n\nu(A_n).
$$
Therefore \b[$\nu$ is a <measure> on $(E,\mathcal F_E)$]: it is the <restriction of a measure to a measurable subset>.
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