Special case of Fatou's lemma Hi everyone: Suppose that $D$ is a domain of $\mathbb{R}^{n}$  $(n \geq1)$, $Y$ a locally compact topological space, and $\mu$ a measure on $Y$. Let $f(x,y): D\times Y\rightarrow[0,+\infty)$ be measurable.
My question is: How can we use Fatou's lemma (and under which condition(s)) to get
$$\int_{Y}\liminf_{\zeta\rightarrow x}f(\zeta,y)d\mu(y)\leq\liminf_{\zeta\rightarrow x}\int_{Y}f(\zeta,y)d\mu(y)?$$
Is there a smart trick to transform the above limit infimum into a sequencial limit infimum? Thanks for considering.
 A: As long as the integrand on the left hand side is measurable, you can indeed reduce your question to the sequencial case:
There exists a sequence $(\zeta_n)_n$ in $\Bbb{R}^n$ with $\zeta_n \rightarrow x$ and
$$\liminf_{\zeta \rightarrow x} \int_Y f(\zeta, y)\,d\mu(y) = \lim_n \int_Y f(\zeta_n, y)\, d\mu(y).$$
This is true, because we are only talking about the liminf of a single(!) number here.
Fatou now yields
$$\int_Y \liminf_n f(\zeta_n, y) d\mu(y) \leq \liminf_n \int_Y f(\zeta_n, y) \, d\mu(y) = \liminf_{\zeta \rightarrow x} \int_Y f(\zeta, y) \, d\mu(y).$$
Noting $\liminf_{\zeta \rightarrow x} f(\zeta, y) \leq \liminf_n f(\zeta_n, y)$ for all $y \in Y$ completes the proof.
I am quite sure that in general the integrand on the left hand side of your equation does not have to be measurable. I will edit the post if I come up with an explicit counterexample.
EDIT: Let me elaborate a bit. If you have a sequence $(z_n)_n$ of real numbers, then there is always a subsequence $(x_{n_k})_k$ with $\lim_k x_{n_k} = \liminf_n z_n$ (show this!). This is what I meant with "the $\liminf$ of a single(!) number". Perhaps I should have said "of a single sequence".
Now consider the case of "not a single sequence", by which I mean e.g. a sequence of functions (like the integrand above). As an easy example, consider
$$
f_{n}:\left\{ 0,1\right\} \rightarrow\mathbb{R},x\mapsto\begin{cases}
\left(-1\right)^{n}, & x=0\\
\left(-1\right)^{n+1}, & x=1.
\end{cases}
$$
Here, $\liminf f_n \equiv -1$, but if you choose any subsequence $(f_{n_k})_k$, you always have $f_{n_k}(0) = - f_{n_k}(1)$, which shows that $(f_{n_k})_k$ can not converge pointwise to $-1$.
The same holds for the "continuous" liminf (i.e. $\liminf_{\zeta \rightarrow x}$), but here you have to replace the subsequence with an actual sequence $\zeta_n \rightarrow x$ (show this!). This is what I have done above for the liminf
$$
\liminf_{\zeta \rightarrow x} \int_Y f(\zeta, y) \, d\mu(y).
$$
Note that we are again in the case of a single "sequence" because the expression depends only on $\zeta$, not on $y$ (or any other variable).
