Problem with an Earlier Question This question has been already asked on this site before here - 
Proving Fatou type lemma
I found that the solution provided is quite sophisticated.
There was a suggestion in the comments section to first prove it for intervals 
and then for a countable union of intervals. I was also thinking along the same lines. I have shown that the required inequality is actually an equality
if $U$ is a connected open subset of $\mathbb{R}$ or if $U$ is a finite union of disjoint connected open subsets of $\mathbb{R}$. But I am unable to generalize it for countable unions. I thought of using Fatou's Lemma, but to no avail. So can someone please provide me a proof using basic measure theory,i.e using convergence theorems, regularity properties, etc.? Thanks for any help.
 A: I guess you can proof it by using standard measure theory:
First define:
$$\mu_n(U) := \int_U f_n(x)dx$$
And $$\mu(U) := \lim_{n\to\infty} \mu_n(U)$$
Then $\mu$ is a $\sigma$-finite measure on $\mathcal{B}(\Bbb R)$
On the other hand you have with $$\psi(U) :=\int_U f(x)dx$$ a second $\sigma$-finite measure on $\mathcal{B}(\Bbb R)$.
Now consider $$\mathcal{E} = \left\{ (-\infty,y] \;|\; y \in \Bbb R\right\}$$ which is an intersection-stable generator of $\mathcal{B}(\Bbb R)$
But by your given properties it holds $$\mu|_\mathcal{E} = \psi|_\mathcal{E}$$
And so $\mu \equiv \psi$ on $\mathcal{B}(\Bbb R)$, hence especially for open sets $U \in \mathcal{B}(\Bbb R)$
A: You may use Borel regularity of the measure $d\mu = f\, dx$.
Given an open $U$ and $\epsilon>0$ there is a compact set $K\subset U$ with $\mu(U\setminus K)<\epsilon$. By compactness there are finitely many open disjoint intervals $I_1,...,I_k$ with $K\subset S=I_1\cup...\cup I_k\subset U$. Now, with $\mu_n = f_n \, dx$ we have $\mu_n(S) \rightarrow \mu(S)$ so $$\liminf_n \;\mu_n(U)\;\geq \;\lim_n \mu_n(S) = \mu(S) \geq \mu(U)-\epsilon.$$
