Prove a collection is a $\sigma$-algebra I have to prove that a collection of sets is a $\sigma$-algebra. I'm stuck with the axiom of closure under countable unions.
The collection is
$$
\mathcal{A}=\{A\in\mathcal{B}:m(A\Delta T^{-1}A)=0\}
$$
(The framework I'm working in is a measure space $(X,\mathcal{B},m)$, with $T:X\to X$ measurable function)
I know that these are just set theoretical operations, but I'm a bit rusty.
 A: So here goes the answer hopefully.
We will have to take care of two problems. One uses the countable additivity of the measure. The other depends on how symmetric difference behaves with respect to union.
Let's deal with the second problem first. We need to look at the following two sets $\bigcup_{i\in\omega}A_i\triangle T^{-1}(\bigcup_{i\in\omega} A_i)$ and $\bigcup_{i\in\omega} (A_i\triangle T^{-1}(A_i))$. Since we will want to bound the measure of the first one it would really be nice if we could get it to be a subset of the second. As it turns this works fine, let's show it.
First notice that $T^{-1}(\bigcup_{i\in\omega} A_i)=\bigcup_{i\in\omega}T^{-1}(A_i)$. I will leave this for the reader to prove since it is quite easily done by the same argument I will use for the next part. 
Next we need to look at $\bigcup_{i\in\omega}A_i\triangle T^{-1}(\bigcup_{i\in\omega} A_i)= \bigcup_{i\in\omega}A_i\triangle \bigcup_{i\in\omega}T^{-1}(A_i)$ . We write down what it means for an element $x$ to be in this set. 
Since we are looking at a symmetric difference an element $x$ is in it, if it is in the set on the left and not in the set on the right or if it is in the set on the right but not the set on the left. That means $x\in \bigcup_{i\in\omega}A_i\triangle \bigcup_{i\in\omega}T^{-1}(A_i)$ if $\exists i\;(x\in A_i)\wedge \forall i\;(x\not\in T^{-1}(A_i))\;$ or if $\forall i\;(x\not\in A_i)\wedge \exists i\;(x\in T^{-1}(A_i))$ . 
Now since in both cases we have an $\exists$ and a $\forall$ we can only gain new elements by changing the $\forall$ to an up front $\exists$. In math terms $\{x;\exists i\;(x\in A_i)\wedge \forall i\;(x\not\in T^{-1}(A_i))\} \subseteq\{x;\exists i (x\in A_i)\wedge (x\not\in T^{-1}(A_i))\}$ . 
Similarly $\{x;\forall i\;(x\not\in A_i)\wedge \exists i\;(x\in T^{-1}(A_i))\} \subseteq\{x;\exists i (x\not\in A_i)\wedge (x\in T^{-1}(A_i))\}$ .
But since our set $\bigcup_{i\in\omega}A_i\triangle \bigcup_{i\in\omega}T^{-1}(A_i)=\{x;\exists i\;(x\in A_i)\wedge \forall i\;(x\not\in T^{-1}(A_i))\}\cup \{x;\forall i\;(x\not\in A_i)\wedge \exists i\;(x\in T^{-1}(A_i))\}$ 
We get  $\bigcup_{i\in\omega}A_i\triangle \bigcup_{i\in\omega}T^{-1}(A_i)\subseteq \{x;\exists i (x\in A_i)\wedge (x\not\in T^{-1}(A_i))\}\cup \{x;\exists i (x\not\in A_i)\wedge (x\in T^{-1}(A_i))\}$ .
Now notice that $\{x;\exists i (x\in A_i)\wedge (x\not\in T^{-1}(A_i))\}\cup \{x;\exists i (x\not\in A_i)\wedge (x\in T^{-1}(A_i))\} = \bigcup_{i\in\omega} (A_i\triangle T^{-1}(A_i))$ . This follows because $\{x;\exists i (x\in A_i)\wedge (x\not\in T^{-1}(A_i))\}=\bigcup_{i\in\omega}\{x;(x\in A_i)\wedge(x\not\in T^{-1}(A_i)\}$  and similarly for the second set.
Altogether from the above we get the important inclusion $\bigcup_{i\in\omega}A_i\triangle T^{-1}(\bigcup_{i\in\omega} A_i)\subseteq \bigcup_{i\in\omega} (A_i\triangle T^{-1}(A_i))$ . 
The result now pretty much writes itself.
We can see $\text{m}(\bigcup_{i\in\omega}A_i\triangle T^{-1}(\bigcup_{i\in\omega} A_i))\leq \text{m}(\bigcup_{i\in\omega} (A_i\triangle T^{-1}(A_i)))\leq \sum_{i\in\omega}\text{m}(A_i\triangle T^{-1}(A_i))=0$ .
Where the first inclusion follows from monotonicity of measure and the second follows from countable subadditivity.
Ps.: If anyone sees typos please feel free to fix. This ended up being much longer then I expected.
