Measure of difference of countable unions

Question

Let $(X,\mathcal{M},\mu)$ a measurable space, $\mu$ positive measure.

For all $n\in \mathbb{N}$, let $A_n,B_n\in \mathcal{M}$ s.t. $A_n\subseteq B_n$ and $\mu(B_n\setminus A_n)=0$.

I want to deduce $$\mu\left(\bigcup_{n\in\mathbb{N}}B_n \setminus \bigcup_{n\in\mathbb{N}}A_n \right)=0$$

I have that: $$\bigcup_{n\in\mathbb{N}}B_n \setminus \bigcup_{n\in\mathbb{N}}A_n =\bigcup_{n\in\mathbb{N}}B_n \cap\bigcap_{n\in\mathbb{N}}(X\setminus A_n)\overset{(\ast)}{=}\bigcup_{n\in\mathbb{N}}(B_n\setminus A_n)$$

and then I can conclude using the sigma sub additivity of $\mu$. But $(\ast)$ is correct? Is there a simpler way to see it? I think that $(\ast)$ is wrong but I don't know how to obtain $\mu(B_n\setminus A_n)$ somewhere.

Answer 1

The equality is not true in general (example below), but you don't need the equality. It is enough for you to show that $$ \bigcup_{n\in\mathbb{N}}B_n \setminus \bigcup_{n\in\mathbb{N}}A_n \subset\bigcup_{n\in\mathbb{N}}(B_n\setminus A_n), $$ which is fairly easy.

For an example of strict containment, consider for instance $B_1=[0,2]$, $B_2=[1,2]$, $A_1=(0,2)$, $A_2=(1,2)$. Then $$ (B_1\cup B_2)\setminus (A_1\cup A_2)=\{0,2\}, $$ while $$ (B_1\setminus A_1)\cup(B_2\setminus A_2)=\{0,1,2\}. $$