Why does the Klein-Gordon propagator $D(x)$ depend on the sign of $x^0$? In A. Zee's Quantum field theory in a nutshell p. 24, it says the Klein-Gordon propagator depends on the sign of $x^0$. Here $x=(x^0,x^1,x^2,x^3)$ with Minkowski sign convention $(+,-,-,-)$. 
$$D(x)=-i\int \frac{d^3k}{(2\pi)^32\omega_k}[e^{-i(\omega_k x^0-\vec{k}\cdot\vec{x})}\theta(x^0)+e^{i(\omega_k x^0-\vec{k}\cdot\vec{x})}\theta(-x^0)].\tag{I.3.23}$$
I don't understand why the sign of $x^0$ is Lorentz invariant as Zee claims. A typical Lorentz transformation is like
$(x{^0})'=\gamma(x^0-v x^1)$. So there must be transformations that change the sign of $x^0$ by carefully choosing the form of $x^1$.
 A: OP got a point: Zee presumably only meant orthochronous Lorentz transformations when he claimed that the sign of $x^0$ [...] do not change. 
Nevertheless, eq. (I.3.23) in Zee's book (and mentioned by OP) is directly derived from the previous equation
$$ D(x-y)~=~\int_{\mathbb{R}^4}\! \frac{d^4k}{(2\pi)^4} \frac{e^{ik\cdot(x-y)}}{k^2-m^2+i\epsilon}, \tag{I.3.22} $$
which is manifestly Lorentz invariant (which can be seen by performing the same Lorentz transformation on the $k^{\mu}$-integration variables as the $x^{\mu}$-variables), so eq. (I.3.23) better also be invariant under the full Lorentz group. In fact, eq. (I.3.23) can be rewritten as
$$  D(x)~=~-i\int_{\mathbb{R}^3}\! \frac{d^3\vec{k}}{(2\pi)^3} e^{-i(\omega_{\vec{k}}|x^0|\pm\vec{k}\cdot\vec{x})},\tag{I.3.23'} $$
thereby avoiding the Heaviside step function. Eq. (I.3.23') is clearly invariant under time reversal $T$ and parity $P$. One can check that eq. (I.3.23') is also invariant under restricted (and thereby all) Lorentz transformations. 
See also this & this related Phys.SE posts.
A: If you consider the propagator to be the following(check it for sign of metric):
$$D(x-y)=-i\int \frac{d^3k}{(2\pi)^32\omega_k}[e^{-i(\omega_k (x^0-y^0)-\vec{k}\cdot(\vec{x}-\vec{y}))}\theta(x^0-y^0)+e^{i(\omega_k (x^0-y^0)-\vec{k}\cdot(\vec{x}-\vec{y})}\theta(-(x^0-y^0)]$$
Then you are obviously right. The time-ordering of $x^0$ and $y^0$ is obviously not Lorentz invariant when the separation is space-like. It is clearly stated in many textbooks but let us provide just few references for those who are curios: see here-page 211 ,here-page 32, here-chapter4 or here-chapter10. I definitely recommend chapter 4 of Sakurai's old text. 
But now let's get back to your question. We know that sign of time coordinate doesn't change under Lorentz transformation. So if we have $x^0>0$ and $y^0>0$  in one frame it will remain so. But the sign of their difference is not invariant, indicating that it can change when separation is space-like(as we said in above paragraphs). Now suppose that you take $y$ in the above formula to be zero. What happens after a Lorentz transformation? It is simple: zero remains zero, and if $x^0$ was positive or negative it will remain so. The result is that in this special case $sign{(x^0)}$ is Lorentz invariant. 
Indeed the statement that $sign(x^0-y^0)$ is Lorentz invariant is a silly statement that you will never find in any book.
