Let M be a $\sigma$-algebra over X and let $\mu$ and $\nu$ be $\sigma$-additive functions. Additionally let $\mu \ge\nu$ for all $E\in M$ and let $\nu$ be $\sigma$-finite. Then there exists a $\sigma$-additive function $\lambda$ such that $\mu(E)=\nu(E)+\lambda(E)$.
I want to proof this, but I dont know how to do this. This is what I already have:
Defining $\lambda=\mu-\nu$ doesnt work, because you might end up with $\infty-\infty$. However, for $\nu(E)<\infty$ this works so I only have to find $\lambda(E)$ for $\nu(E)=\infty$.
My first attempt was to simply set $\lambda(E)=\infty$ for $\nu(E)=\infty$. However, I am of the opinion that this is wrong since if $\lambda$ was $\sigma$-additive and $\bigcup_1^\infty A_i$ with $\nu(\bigcup_1^\infty A_i)=\infty$ and $\nu(A_i)<\infty$ you could compute
$$\lambda(\bigcup_1^\infty A_i)=\infty$$ and $$\lambda(\bigcup_1^\infty A_i)=\sum_1^\infty\lambda(A_i)=\sum_1^\infty\mu(A_i)-\nu(A_i)$$ and for example if $\mu(A_i)-\nu(A_i)=\frac{1}{i^2}$ this would converge and be less than infinity what would be a contradiction to the assumption that $\lambda$ is $\sigma$-additive.
So does this case happen? And if so what is the propper way to solve this problem?
Exploit this!
Let $(E_n)_n$ be a sequence of disjoint measurable sets with $X=\bigcup_{n=1}^{\infty}E_n$ and $\nu(E_n)<\infty$ for every $n\in\mathbb N$.
Prescribe $\mu_n$ on $M$ by $E\mapsto\mu(E\cap E_n)$ and $\nu_n$ by $E\mapsto\nu(E\cap E_n)$.
Then for a fixed $n\in\mathbb N$ for every $E\in M$ there is a unique $r\in[0,\infty]$ such that $\mu_n(E)=\nu_n(E)+r$ so we can prescribe $\lambda_n(E)$ in such a way that it satisfies $\mu_n(E)=\nu_n(E)+\lambda_n(E)$.
Finally prescribe $\lambda$ by $E\mapsto\sum_{n=1}^{\infty}\lambda_n(E\cap E_n)$.
It remains to be proved that $\lambda$ is $\sigma$-additive.
Give that a try yourself. I would start with proving that the $\lambda_n$ are $\sigma$-additive.