Limit of measures, two questions on limits of integrals Suppose $\mu_n$ is a sequence of measures on $(X, \mathcal{A})$ such that $\mu_n(X) = 1$ for all $n$ and $\mu_n(A)$ converges as $n \to \infty$ for each $A \in \mathcal{A}$. Call the limit $\mu(A)$. I can show that $\mu$ is a measure. I have two questions.
First question. Do we have necessarily have that $\int f\,d\mu_n \to \int f\,d\mu$ whenever $f$ is bounded and measurable?
Second question. Do we have necessarily have that$$\int f\,d\mu \le \liminf_{n \to \infty} \int f\,d\mu_n$$whenever $f$ is nonnegative and measurable?
Thanks in advance!
 A: First question is yes.
Given $|f|\leq M$ is bounded, we can approximate it uniformly using simple functions by cutting $[-M, M]$ into finitely many intervals $I_n$ with length $\epsilon$  and defining $\phi_n (x) =\sum_n \inf\{ I_n\}   \chi_{f^{-1}(I_n)}(x) $ . We would have $\|f - \phi_n\|_\infty \leq \epsilon$.
$$\lim_m\int f d\mu_m = \lim_m \lim_n \int \phi_n d\mu_m$$
since the limit in $n$ is uniform, which I mean for each $\epsilon > 0$ if we choose $n$ large enough and independent of $m$, we have 
$$\bigg|\int_X fd\mu_m - \int_X \phi_n d\mu_m \bigg|\leq \int_X |f-\phi_n| d\mu_m \leq \|f-\phi_n\|_\infty \mu_m(X) \leq \|f-\phi_n\|_\infty \leq \epsilon,$$
by Moore-Osgood theorem, we can switch the two limits and we would have 
 $$\lim_n \lim_m \int \phi_n d\mu_m = \int f d\mu.$$
The second question is also yes. 
For each simple function less than $f$
we have 
$$\int \phi d\mu_n \leq \int f d\mu_n$$
take the $\liminf$ we have 
$$\int \phi d\mu \leq \liminf_n \int f d\mu_n$$
then take the $\sup$ over all simple functions less than $f$. By definition of Lebesgue integral, we have
$$\int f d\mu \leq \liminf_n \int f d\mu_n.$$
