Probability of symmetric difference inequality I think I am missing something easy here, but my book notes that
$$P\left\{\bigcup_{j=1}^\infty A_j \mathbin\triangle \bigcup_{j=1}^\infty B_j\right\} \leq \sum_{j=1}^\infty P\{A_j \mathbin\triangle B_j\}$$
where $A\mathbin\triangle B$ is the symmetric difference of the two sets $A,B$. It seems some what intuitive as the sets may overlap and the right hand side will be greater in that case, no? I have tried to write the symmetric difference of the unions as disjoint subsets of the union of symmetric differences, and then use additivity and $P(A) \leq P(B)$ for $A\subset B$ (despite this fact being proven after the original note, not that I think it depends on this) but did not succeed.
Any suggestions or hints would be helpful, thank you. 
 A: Suppose $\omega\in A_i$ and $\omega\not\in\bigcup_{j=1}^\infty B_j$.  Then $\omega$ is a member of none of the sets $B_j$, $j=1,2,3,\ldots\,$.  Therefore $\omega\in A_i\mathbin\triangle B_i$.  And so $\omega\in\bigcup_{j=1}^\infty A_j\mathbin\triangle B_j$. And the same thing applies if $\omega\in B_j$ and $\omega\not\in\bigcup_{j=1}^\infty A_j$.  Thus we have
$$
\bigcup_{j=1}^\infty A_j \mathbin\triangle \bigcup_{j=1}^\infty B_j \subseteq \bigcup_{j=1}^\infty A_j \mathbin\triangle B_j.
$$
Consequently
$$
P\left\{\bigcup_{j=1}^\infty A_j \mathbin\triangle \bigcup_{j=1}^\infty B_j\right\} \leq P\left\{ \bigcup_{j=1}^\infty A_j\mathbin\triangle B_j \right\}.
$$
Lastly, we have
$$
P\left\{ \bigcup_{j=1}^\infty A_j\mathbin\triangle B_j \right\} \leq \sum_{j=1}^\infty P\{A_j \mathbin\triangle B_j\}.
$$
A: The LHS is points in some $A_i$ and no $B_j$, points in the RHS are in some $A_i$ and not $B_i$ specifically, so you're not cutting away all the points in $A_i\setminus B_j$ for $j\ne i$.
A: Well, let's look at the base case.
$$\begin{align}(A_1\cup A_2)\triangle(B_1\cup B_2)
 & = ((A_1\cup A_2)\cap B_1^c \cap B_2^c)\cup(A_1^c\cap A_2^c\cap(B_1\cup B_2))
\\[1ex] & = (A_1\cap B_1^c\cap B_2^c)\cup(A_1^c\cap A_2^c\cap B_2)\cup (A_2\cap B_1^c\cap B_2^c)\cup (A_1^c\cap A_2^c\cap B_1)
\\[1ex] & \subseteq (A_1\cap B_1^c)\cup(A_1^c\cap B_1)\cup(A_2\cap B_2^c)\cup(A_2^c\cap B_2)
\\[2ex]\therefore (A_1\cup A_2)\triangle(B_1\cup B_2) & \subseteq (A_1\triangle B_1)\cup(A_2\triangle B_2)
\end{align}$$
Now let look at the inductive step.
$$\begin{align} \left(\bigcup_{j=1}^n A_j\;\cup A_{n+1}\right)\triangle\left(\bigcup_{j=1}^n B_j\;\cup B_{n+1}\right) & \subseteq \left(\bigcup_{j=1}^n A_j \triangle \bigcup_{j=1}^n B_j\right)\cup(A_{k+1}\triangle B_{k+1})
\end{align}$$
And we can clearly see where that is going.   Thus we argue that because it is true for two, and inductive step also holds, therefore it shall be true for more, ad infinitum.
Then we apply measure theory to this result and reach the requisite conclusion.
$\Box$
