A Doubt From Shiryayev's Probability Let $(\Omega,\mathcal{F},P)$ be a probability space and $\mathcal{A}$ be an algebra of subsets of $\Omega$ such that $\mathcal{F}=\sigma(\mathcal{A})$ (this says that $\mathcal{F}$ is the smallest sigma algebra generated by $\mathcal{A}$). Show that given any $\epsilon>0$ and $B\in \mathcal{F}$, there exists a set $A\in \mathcal{A}$ such that
\begin{eqnarray}
P((A\setminus B)\cup(B\setminus A))<\epsilon.
\end{eqnarray} 
I am stuck on how to start with the problem. I am aware of the "principle of appropriate sets", but I am not sure how to invoke it. I would appreciate any initial ideas. 
EDIT: I have made a small attempt at solving the problem by assuming that any $B\in \mathcal{F}$ can be expressed as a countable union of an increasing sequence of sets, i.e.,
\begin{eqnarray}
B=\bigcup\limits_{n=1}^{\infty}A_{n},~A_{n}\subseteq A_{n+1}, A_{n}\in\mathcal{A}.
\end{eqnarray}
This assumption is made upon knowing that the smallest monotonic class generated by $\mathcal{A}$ is the same set as $\mathcal{F}$. 
Can anyone please tell me whether or not the above procedure is correct, and if the assumption made is justified?
 A: Dynkin's $\pi-\lambda$ theorem will do the trick. (In fact, I would suggest try doing that without reading below, as it is an extremely useful theorem in measure theory to get familiar with)
Let $\Lambda$ be the set of subsets $B$ of $\mathcal{F}$ such that the property required holds. It is clear that $\mathcal{A}\subset\Lambda$, and it is also trivial that $\mathcal{A}$ is a $\pi$-system. To apply the $\pi-\lambda$ theorem, it remains to show that $\Lambda$ is a $\lambda$-system.
The fact that $\Omega\in\Lambda$ follows from the fact that $\Omega\in\mathcal{A}$, because of the definition of an algebra. If $B_1,B_2\in\Lambda$, $B_2\subset B_1$, then the fact $(B_1\setminus B_2)\in\Lambda$ follows from the identity
$$
(B_1\setminus B_2)\setminus(A_1\setminus A_2)\subset (B_1\setminus A_2)\cup(A_2\setminus B_2).
$$
Finally, if $B_1\subset B_2\subset\cdots$ are all elements of $\Lambda$, then taking $A_1,A_2,\ldots$ such that
$$
P((B_i\setminus A_i)\cup(A_i\setminus B_i))<\varepsilon/(2^i),
$$
one has
$$
P((\cup_{i=1}^{\infty}B_i\setminus \cup_{i=1}^{\infty}A_i)\cup(\cup_{i=1}^{\infty}A_i\setminus \cup_{i=1}^{\infty}B_i)<\varepsilon.
$$
By the countable additivity of the measure $P$, there is a finite index $N$ so that
$$
P((\cup_{i=1}^{\infty}A_i\setminus \cup_{i=1}^{N}A_i)<\varepsilon
$$
and therefore
$$
P((\cup_{i=1}^{\infty}B_i\setminus \cup_{i=1}^{N}A_i)\cup(\cup_{i=1}^{N}A_i\setminus \cup_{i=1}^{\infty}B_i)<2\varepsilon.
$$
Since $\cup_{i=1}^{N}A_i\in\mathcal{A}$, this shows that $\cup_{i=1}^{\infty}B_i\in\Lambda$, hence $\Lambda$ is a $\lambda$-system and therefore $\sigma(\mathcal{A})\subset\Lambda$.
