Showing a measure is $\sigma$-finite Let ${(X_n, \mathcal A_n, \mu_n)_{n \in \mathbb N}}$ be measure spaces. Let $X_n$ be pairwise disjoint.
Define $(X, \mathcal A, \mu)$ by$X=\bigcup_{i=1}^{\infty} X_n$
$\mathcal A=\{E \subset X: E \cap X_n \in \mathcal A_n \; \forall n \in \mathbb N\}$ and $\mu(E)=\sum_n \mu_n(E \cap X_n)$
I want to show that if all $\mu_n$ are $\sigma$-finite than $\mu$ is $\sigma$-finite aswell.
My attempt: 
I wanted to take the sequence of the unions of the sequences that make $\mu_n$ sigma finite. So the first element is the union of all the first parts of the each sequence and so on.
Showing that the union is of all these equals $X$ is easy, but I fail at showing that each element has finite measure and that each element is contained in $A$ and hoped that someone could help me here. 
 A: As the comments indicate, you need to accept that countable unions of countable sequences are still countable. See here for example.
Now you just need to show that the measure $\mu$ is $\sigma$-finite, i.e. you need to show that $X$ is a countable union of sets all of which have finite measure. As you said, do this using the sequences which make each $\mu_n$ $\sigma$-finite.
Let $\{U_{i,k}: k \in \mathbb{N}\}$ be a countable sequence of sets in $\mathcal{A}_i$ which satisfy


*

*$\mu_i(U_{i,j}) < \infty$ for all $j \in \mathbb{N}$

*$\bigcup_j U_{i,j} = X_i$


We will consider collection $\{U_{i,j}, i \in \mathbb{N}, j \in \mathbb{N}\}$. All we need to show is that this collection 1) is countable, 2) unions up to $X$, and 3) all elements have finite measure under $\mu$.
Using the fact above, 1 is immediate. 2 is also obvious. 3 is just a tiny bit of work. Pick an arbitrary $U_{i,j}$ from that collection. This set is in $\mathcal{A}$ because


*

*$U_{i,j} \subset X$

*$U_{i,j} \cap X_i = U_{i,j}$ which is in $\mathcal{A}_i$, and

*$U_{i,j} \cap X_k$ for $i \neq k = \emptyset$ since the $X_k$'s are pairwise disjoint, and $\emptyset \in \mathcal{A}_k$ for any $k$.


We already know that $\mu_i(U_{i,j}) < \infty$ and obviously $\mu_j(\emptyset) = 0$ for any $j \neq i$ so that $\mu(U_{i,j}) < \infty$ also.
