Symmetric Difference Inequality in a Probability Space Let (Ω, $\mathcal B$, $P$) be a probability space with $\mathcal B$ = σ($\mathcal A$), where $\mathcal A$ is an algebra of subsets
of Ω. Show that for $B$ ∈ $\mathcal B$ and all $\epsilon > 0$, there exists a set $A_\epsilon$ ∈ $\mathcal A$ such that
P($B\Delta A_\epsilon$) < $\epsilon$  and so $|P(B) − P(A_\epsilon)|$ < $\epsilon$ 
My attempt:
Let $\epsilon > 0 $ be given. Consider the countable disjoint set $ \{A_i\}_{i =1}^\infty \subseteq \mathcal A $ such that $ B \subseteq \bigcup_{i =1}^\infty A_i $ and $ P(\bigcup_{i =1}^\infty A_i) = \sum_{i =1}^\infty P( A_i) \le \frac{\epsilon}{2} + P(B) $. 
Since $ P(B) \le 1 $ then $ \sum_{i =1}^\infty P( A_i) $ converges which means we can find some $ N_\epsilon \in \mathbb N $ such that for every $ n \ge N_\epsilon, \; \sum_{i =n+1}^\infty P( A_i) < \frac{\epsilon}{2} \; $. Let $ A_\epsilon = \sum_{i =1}^n P( A_i) $.  
\begin{align} P(B \Delta A_\epsilon) & = P(B \setminus A_\epsilon) + P(A_\epsilon \setminus B) \\\\ & \le P(\bigcup_{i=n+1}^\infty A_i) + P(\left(\bigcup_{i=1}^\infty A_i \right) \setminus B) \\\\
& \le \sum_{i=n+1}^\infty P(A_i) + \sum_{i=1}^\infty P(A_i) - P(B) \\\\
& \le \epsilon  \end{align}
So suppose $ B$ and $A_\epsilon$ are disjoint. Then $|P(B)-P(A_\epsilon)| \le P(B \Delta A_\epsilon) =  P(B) + P(A_\epsilon)  < \epsilon. $ 
I cant figure out how to show this ineqality if $ B$ and $A_\epsilon$ are not disjoint. Also how do I show that such a such a set like $ \{A_i\}_{i =1}^\infty $ exists? 
 A: There is no need to consider $B$ and $A_{\epsilon}$ disjoint or not. Use the fact that
$$
C\subset D\cup (C\bigtriangleup D)
$$
The proof is as follows:
\begin{align}
D\cup (C\bigtriangleup D)&=D\cup ((C-D)\cup (D-C))
\\
&=(C\cup D)\cup (D-C)
\\
&=C\cup D\tag{$D-C\subset C\cup D$}
\\
&\supset C
\end{align}
So in this case, we have
$$
B\subset A_{\epsilon}\cup (B\bigtriangleup A_{\epsilon})
$$
And thus
$$
P(B)\leqslant P(A_{\epsilon})+P(B\bigtriangleup A_{\epsilon})\quad\text{or }\quad P(B)-P(A_{\epsilon})\leqslant P(B\bigtriangleup A_{\epsilon})
$$
Likewise
$$
A_{\epsilon}\subset B\cup (B\bigtriangleup A_{\epsilon})
$$
And
$$
P(A_{\epsilon})-P(B)\leqslant P(B\bigtriangleup A_{\epsilon})
$$
Thus
$$
|P(B)-P(A_{\epsilon})|\leqslant P(B\bigtriangleup A_{\epsilon})
$$
A: Since $B\cup (A_\epsilon\setminus B) =(B\setminus A_\epsilon)\cup A_\epsilon$, we have
$$
P(B)+P(A_\epsilon\setminus B)=P(B\setminus A_\epsilon)+P(A_\epsilon)
$$
and so
$$
P(B)-P(A_\epsilon)=P(B\setminus A_\epsilon)-P(A_\epsilon\setminus B).
$$
Therefore,
$$
|P(B)-P(A_\epsilon)|\le P(B\setminus A_\epsilon)+P(A_\epsilon\setminus B)=P(B\Delta A_\epsilon).
$$
