[proofplan] Use the defining closure properties of a sigma-algebra: it contains the ambient set, is closed under complements, and is closed under countable unions. [/proofplan] [step:Verify the empty set and countable intersections] Since $\Omega\in\mathcal F$ and $\mathcal F$ is closed under complements, \begin{align*} \varnothing=\Omega^c\in\mathcal F. \end{align*} If $A_n\in\mathcal F$ for all $n$, then $A_n^c\in\mathcal F$ for all $n$, and \begin{align*} \bigcap_{n=1}^{\infty}A_n=\left(\bigcup_{n=1}^{\infty}A_n^c\right)^c. \end{align*} The right-hand side belongs to $\mathcal F$, so $\bigcap_n A_n\in\mathcal F$. [/step] [step:Verify complements of unions and intersections] Since each $A_n^c\in\mathcal F$ and sigma-algebras are closed under countable unions, \begin{align*} \bigcup_{n=1}^{\infty}A_n^c\in\mathcal F. \end{align*} Applying countable intersection closure to the sequence $(A_n^c)$ gives \begin{align*} \bigcap_{n=1}^{\infty}A_n^c\in\mathcal F. \end{align*} [/step] [step:Verify set difference and symmetric difference] If $A,B\in\mathcal F$, then \begin{align*} A\setminus B=A\cap B^c\in\mathcal F. \end{align*} Also \begin{align*} A\triangle B=(A\setminus B)\cup(B\setminus A), \end{align*} and both terms on the right are in $\mathcal F$. Hence $A\triangle B\in\mathcal F$. [/step]