special relativity - Lorentz-invariance of step function

I was reading about the Lorentz invariant integration measure $\int \frac{d^3k}{2E_K}$, and ways to prove that this was Lorentz invariant. Many of the proofs I have read use the step function (or Heaviside function) $\theta(k^0)$, and claim that this is obviously Lorentz-invariant (e.g Alex Nelson's notes). Other sources claim that there is some requirement that the Lorentz transformations have determinant = 1, or that $k^\mu$ is timelike. What are the actual conditions on the vector $k$ for the step function to be lorentz-invariant?

Answer

Proper, orthochronous Lorentz transformations can't change the sign of a $x^0$ component in a time-like four-vector - $dx^\mu dx_\mu$ and $\text{sign}\,x^0$ are both Lorentz invariants if $x_\mu$ is time-like. Because $\text{sign}\,x^0$ is Lorentz invariant, the step function $\theta(x^0)$ is Lorentz invariant - the condition is that $x$ is time-like.

A general Lorentz transformation that wasn't orthochronous could violate causality by making one observer see time-like events occurring in a different order.