Inequality involving symmetric difference and probability measure Suppose we are working in a probability space and we have events $A_j,A_j'$ such that $\mathbb{P}(A_j \Delta A_j') < \epsilon 2^{-j}$ for each $j=1,2,...$, then apparently due to the finiteness of $\mathbb{P}$, we can say there is some natural number $J$ for which $\mathbb{P}(A \Delta \bigcup_{1 \leq j \leq J} A_j') \leq \epsilon$ where $A = \bigcup_{j=1}^{\infty} A_j$. I do not see how to show this inequality, I have noticed the following but do not think they are especially useful for this:
$\mathbb{P}(A \Delta \bigcup_{j=1}^{\infty}) \leq \sum_{j=1}^{\infty} \mathbb{P}(A_j \Delta A_j') = \epsilon$, and due to finiteness of $\mathbb{P}$ we have for any events $C,D$, $|\mathbb{P}(C) - \mathbb{P}(D)| \leq \mathbb{P}(C \Delta D)$. I have tried to no avail to put these ideas together to formulate some sort of proof.

Using a slight modification of QuantumSpace's idea: $(A \Delta \bigcup_{j=1}^{\infty} A_j') \Delta (A \Delta \bigcup_{j=1}^m A_j')) = \bigcup_{j=m+1}^{\infty} A_j'$ so $\mathbb{P}((A \Delta \bigcup_{j=1}^{\infty} A_j') \Delta (A \Delta \bigcup_{j=1}^m A_j'))) \rightarrow 0$ as $m \rightarrow \infty$, and thus using $|\mathbb{P}(C) - \mathbb{P}(D)| \leq \mathbb{P}(C \Delta D)$ we get $\lim_{m \rightarrow \infty} \mathbb{P}(A \Delta \bigcup_{j=1}^m A_j') = \mathbb{P}(A \Delta \bigcup_{j=1}^{\infty} A_j')$ which gives the desired result.
