Let $\mu_n$ be a sequence of positive measures on a measurable space $(X,M)$ converging to a measure $\mu$,i.e. $\lim_{n\rightarrow \infty} \mu_n(A)=\mu(A)$ for any measurable set $A$. I want to check if $L^1(\mu)=\bigcap_n L^1(\mu_n)$.
1) First suppose that $\mu_n$ is an increasing sequence of measures. I think in this case the relation above does not hold. Here is my counterexample. Let $\mu_n:= n\gamma$ where $\gamma$ is the counting measure on $\mathbb R$. Let $\phi:= \chi_{\{0\}}$. Then $\int \phi d\mu_n=\mu_n(\{0\})=n$ for all $n$ so that $\phi\in \bigcap_n (L^1(\mu_n))$, but $\int \phi d\mu=\infty$ so that $\phi\notin L^1(\mu)$. Is this correct? Is there any better counterexample?
2) Now suppose that $\mu_n$ is decreasing. I think for this case as well the relation above is wrong. Again take $u_n$ as in the case above. We have $0=\int d\mu_n$ for each $n$ but $\infty= \int d\mu$. Hence $\chi_{\mathbb R}$ is a counterexample. Is this correct? What about any better counterexample?
The answer is clearly no in general.
To see this, pick an arbitrary measure $\nu$ such that $L^1(\nu) \subsetneq L^1(\mu)$. Now replace $\mu_1$ by $\nu$.