Show $ρ( A_i \bigcup B_j ) = ρ( A_i - B_j ) + ρ(B_j - A_i ) + ρ( A_i \bigcap B_j ) $ finite If  $\mu$ and $\nu$  are $\sigma$-finite measures on $(\mathscr \Omega , \mathscr F )$
and   $\nu$ is absolutely continuous with respect to $\mu$ : ( $\nu << \mu $ )
and let $ ρ  = \mu + \nu $
I want to prove that $ρ$ is  $\sigma$-finite and that  $\nu << ρ $ :
I tried the below:
$\text {  There exists   } \{A_{n}\}   \text { , such that  }\mu(A_n)<\infty  \text { and } \bigcup_{n=1}^\infty A_n = \Omega $ .
$\text { Also}, \text {  There exists   } \{B_{n}\}   \text { , such that  }\nu(B_n)<\infty  \text { and } \bigcup_{n=1}^\infty B_n = \Omega $ .
$\text { Then , for } \bigcup_{i=1}^\infty \bigcup_{j=1}^\infty (A_i \bigcup B_j)   = \Omega $ ,
$ρ( A_i \bigcup B_j ) = ρ( A_i - B_j ) +  ρ(B_j -  A_i )   + ρ( A_i \bigcap B_j ) $
$\text {at this point , it is enough to show that each of the above terms are finite ,              
             because that would mean } ρ( A_i \bigcup B_j )<\infty   \text {  so ,   ρ is }  \sigma-finite. $
How can we show that$\ ρ( A_i - B_j ) \text { and }  ρ(B_j -  A_i ) \text { are finite} $?
 A: I understand that $\mu$ and $\nu$ are non-negative measures.
Actually, you don't need the condition  $\nu \ll \mu $.

Let $\mu$ and $\nu$  are $\sigma$-finite measures on $(\mathscr \Omega , \mathscr F )$ and let $ \rho = \mu + \nu $. Then $\rho$ is $\sigma$-finite and $\mu \ll \rho$ and $\nu \ll \rho$.

Proof:
Since $\mu$ is a $\sigma$-finite measure, there are $\{A_{n}\}_n $ such that $\mu(A_n)<\infty$ and $\bigcup_{n=0}^\infty A_n = \Omega $.
Since $\nu$ is a $\sigma$-finite measure, there are $\{B_{n}\}_n $ such that $\nu(B_n)<\infty$ and $\bigcup_{n=0}^\infty B_n = \Omega $.
Now consider $\{A_i\cap B_j : i, j \in \mathbb{N}\}$. This is a countable collection of set in $\mathscr F $, so we can enumerate it as $\{E_n\}_n$. Then we have:
For all $n \in \mathbb{N}$, there are $i, j \in \mathbb{N}$ such that $E_n=A_i\cap B_j$ and so
$$\rho(En) = \mu(A_i\cap B_j) + \nu(A_i\cap B_j) \leqslant  \mu(A_i) + \nu(B_j) <+\infty$$
and
$$ \bigcup_{n=0}^\infty E_n  = \bigcup_{i,j=0}^\infty (A_i\cap B_j)= \bigcup_{i=0}^\infty \bigcup_{j=0}^\infty (A_i\cap B_j)= \bigcup_{i=0}^\infty \left [\bigcup_{j=0}^\infty (A_i\cap B_j) \right ]= \bigcup_{i=0}^\infty (A_i \cap \Omega)  = \Omega$$
So $\rho$ is $\sigma$-finite.
Now, for all $E \in \mathscr F$, if $\rho(E)=0$ then $\mu(E)+\nu(E)=0$ and so $\mu(E)=0$ and $\nu(E)=0$. So, $\mu \ll \rho$ and $\nu \ll \rho$.
