What is the meaning of propagator in the context of lattice theory? Say in $1+1D$ free fermion theory, it is easy to calculate the propagator in the (effective) field theory to be 
$$\langle \psi^\dagger(z)\psi(z')\rangle = \frac{1}{2\pi}\frac{1}{z-z'}$$
(in the notation of https://arxiv.org/abs/cond-mat/9908262). What is the meaning of this back in the lattice theory? Especially, there is a singularity when $z\rightarrow z'$ which is not present in the lattice theory. How to explain this discrepancy?
 A: I would prefer to answer your general question in a different context, because the example you mention has a difficulty which is special to the chosen QFT and not to the question of the meaning of singularities in the two-point functions. In particular, there is the fermion doubling problem, namely that it is impossible to put a single chiral non-interacting fermion on the lattice.
Therefore allow me to discuss your question in the simplest case: a scalar field in $(0+1)$ dimensions. Our variable is a field $\phi : \mathbb{Z} \rightarrow \mathbb{R},\ x \mapsto \phi(x).$ For simplicity i will set the lattice constant to $1$. This just means i am measuring distances in units of the lattice constant.
Now consider the Green's function
$$ G(x-y) = \langle \phi(x) \phi(y) \rangle \tag{1} $$
This solves the difference equation
$$ 2G(x) - G(x+1) - G(x-1) = \delta_{x,0} \ . $$
This may be solved by a Fourier transform; write
$$ G(x) = \int_{-\pi}^\pi \frac{ d p}{2\pi} \hat{G}(p) e^{ i p x} \ . $$
Inserting in equation $(1)$ gives
$$ \hat{G}(p) = \frac{1}{2}\frac{1}{1-\cos(p)}  \ .$$
This is smooth except at $0$. To further analyze the singularity at $p=0$ rewrite:
$$2 - 2 \cos(p) = p^2 - \frac{1}{12} p^4 + \mathcal{O}(p^6) = p^2\left(1 -\frac{1}{12} p^2 + \mathcal{O}(p^4)\right) $$
so that
$$ \hat{G}(p) = \frac{1}{p^2}\frac{1}{1 - \frac{1}{12} p^2 + \mathcal{O}(p^4)} = \frac{1}{p^2}\left(1 + \frac{1}{12} p^2 + \mathcal{O}(p^4)\right) = \frac{1}{p^2} + \frac{1}{12} + \mathcal{O}(p^2) \ . \tag{2}$$
Hence
$$ \hat{G}(p) - \frac{1}{p^2} $$
Is smooth in $[-\pi,\pi]$. After these preparatory comments, we write
$$ G(x) = G_c(x) + G_1(x) + G_2(x) $$
with 
$$ G_c(x) = \int_{-\infty}^{\infty} \frac{dp}{2\pi} \frac{1}{p^2} e^{i p x}  $$
The continuums Green's function as will be explained shortly, then there are two error terms:
$$ G_1(x) = \int_{-\pi}^\pi \frac{ d p}{2\pi} \left[ \hat{G}(p) - \frac{1}{p^2} \right]e^{ i p x} \ . $$
As was shown in equation $(2)$, the integrand is smooth. By the Riemann-Lebesgue Lemma we may conclude immediately that $G_1(x)$ decays exponentially with increasing $x$!
The other error term is
$$G_2(x) = - \int_{\pi}^\infty \frac{d p}{\pi} \frac{\cos(p x)}{p^2} \ .$$
$G_2$, again by Riemann-Lebesgue, decays exponentially with increasing $x$!
On the other hand, $G_c(x)$ satisfies the continuum equation
$$-G_c''(x) = \int_{-\infty}^{\infty} \frac{dp}{2\pi}  e^{i p x} = \delta(x) $$
and hence $G(x) = - \frac{|x|}{2}$. Hence if we consider large distances, the exponentially decaying parts $G_1,G_2$ will become more and more irrelevant, while the linearly rising continuum part $G_c$ will become more and more important!
On the other hand, for small distances, there is no reason to neglect the terms $G_1,G_2$.
A: Because they are propagators they contain much detailed dynamic information, and because they are expectation values in some statistical ensemble they contain all statistical mechanical information.
