P2464
Let $E \subseteq F$ such that $E \in \mathcal{F}$. Denote $E^{\complement} : = F \setminus E$. The sets $E$ and $E^{\complement}$ are disjoint with $E \cup E^{\complement} = F$, so additivity of measure implies
\begin{equation}
0 = \mu(F) = \mu(E \cup E^{\complement}) = \mu(E) + \mu(E^{\complement}) \geq \mu(E) \geq 0
\end{equation}
Hence, $\mu(E) = 0$. $\square$
