Past exam of the integration course of the University of Oxford / 2015 / 1 / a

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.