Let $\nu$ be a sigma-finite measure, let $\mu$ be a measure absolutely continuous with respect to $\nu$, and let $(\mu_n)$ be a sequence of measures such that $\mu_n(E)\uparrow\mu(E)$ for all measurable sets $E$. Show that $\frac{d\mu_n}{d\nu}\uparrow\frac{d\mu_n}{d\nu}$ $\nu$-almost everywhere.
I managed to prove it in the case when $\mu$ is finite, and in the case when $\mu$ is sigma-finite, but I'm not sure how to prove it in the general case. In the case when $\mu$ is finite I did it by expressing $\mu(E)-\mu_n(E)$ and $\mu_n(E)-\mu_{n+1}(E)$ in terms of integrals involving Radon-Nikodym derivatives, which is valid since you're never going to get differences of infinities. Then when $\mu$ the sigma-finite case I broke up the set $X$ into sets $X_1,X_2,X_3,...$ each of which has finite $\mu$-measure and then applied the result of the previous case. But I'm not sure what to do if $\mu$ is neither finite nor sigma-finite.
First prove that following facts :
1) If $h,g$ are non-negative extended real valued measurable and $\int_Ahd\nu≤\int_A gd\nu,\forall A$ measurable, then $h≤g$ a.e. w.r.t. $\nu$.
2)If $h,g$ are non-negative extended real valued measurable and $\int_Ahd\nu=\int_A gd\nu,\forall A$ measurable, then $h=g$ a.e. w.r.t. $\nu$.
Now let $f_n:X\rightarrow [0,\infty]$ are Radon-Nikodym derivative of $\mu_n$ w.r.t. $\nu$. Also let $f:X\rightarrow [0,\infty]$ is the Radon-Nikodym derivative of $\mu$ w.r.t. $\nu$. Then $\mu_n(E)$ increases to $\mu(E)$ for each measurable $E$ implies that $f_n≤f$ a.e. w.r.t. $\nu$. In a similar manner one can show $f_n≤f_{n+1}$ a.e. w.r.t. $\nu$. Now by monotone convergence theorem $\int_E(\lim_{n\rightarrow \infty} f_n)d\nu=\int_Efd\nu$ for each measurable $E$. Therefore applying fact 2) we get, $\lim_{n\rightarrow \infty} f_n(x)\uparrow f(x)$ for almost all $x\in X$.
Edit:-- I prove the fact 1) . So write $A_n\uparrow X$ with $\nu(A_n)<\infty$. Let $$B_n=[0≤g<h,g≤n]\implies \int_{A_n\cap B_n} h d\nu≤\int_{A_n\cap B_n} gd\nu<\infty\implies \int_X(\chi_{A_n\cap B_n}h-\chi_{A_n\cap B_n}g)d\nu=0\implies \nu(A_n\cap B_n)=0\implies \nu[0≤g<h,g<\infty]=0\implies h≤g\ a.e.\ w.r.t.\ \nu.$$