# sum of sigma finite measure

I want to demonstrate:

Finite sum of sigma-finite measures is a sigma finite measure.

I try this: Suppose that are m sigma measures in a space ($\mathcal{X}$,$\mathcal{A}$), denoted by $\mu_k$ with $k \in {1,...,m}$. Then exist $A_n^k$ for each k,and $\mathcal{X}=\bigcup\limits_{n \in \mathbb{N}}A_n^k$.

And i want to prove that $A_n=\bigcup \limits_{k=1}^{m} A_n^k$ is the decomposition of the sum measure $\mu=\sum \limits_{k=1}^{m}\mu_k$, but I have problems with the finite measure of each $A_n$.

My definition of $A_n$ is wrong?

Is an easier solution?

Thanks

Let us do the case $m=2$ (the general one follows by induction).
The problem is that we may have $\mu_1(A_n^2)=+\infty$.
However, if we define $B_{n_1,n_2}:=A_{n_1}^1\cap A_{n_2}^2$, then the family $\left(B_{n_1,n_2}\right)_{(n_1,n_2)\in\mathbb N^2}$ is countable and $$(\mu_1+\mu_2)\left(B_{n_1,n_2}\right) \leqslant \mu_1\left(A_{n_1}^1\right)+\mu_2\left(A_{n_2}^2\right),$$ which is finite.