I would like to check if I understood well the proof of "$\frac{d^3k}{(2\pi)^3}\frac{1}{2 k_0}$ is Lorentz invariant".
My question is different from the others linked to this topic because I want to check the proof in details with all the steps.
I have an arbitrary function $f(k^0,\vec{k})$.
I can proove that : (*)
$$ \int \frac{d^3k}{(2\pi)^3}\frac{1}{2 \omega(\vec{k})} f(\omega(\vec{k}),\vec{k})=(2 \pi) \int \frac{d^4k}{(2 \pi)^4} \delta(k^2-m^2) \Theta(k^0) f(k^0,\vec{k}) $$
Indeed :
$$ \delta(k^2-m^2) \Theta(k^0)=\frac{\delta(k^0-\omega(\vec{k}))}{2\omega(\vec{k})} $$
And we thus have :
$$ (2 \pi) \int \frac{d^4k}{(2 \pi)^4} \delta(k^2-m^2) \Theta(k^0) f(k^0,\vec{k})=(2 \pi) \int \frac{d^4k}{(2 \pi)^4} \frac{\delta(k^0-\omega(\vec{k}))}{2\omega(\vec{k})} f(k^0,\vec{k}) \\=\int \frac{d^3k}{(2 \pi)^3} \frac{1}{2\omega(\vec{k})} f(\omega(\vec{k}),\vec{k})$$
We also can proove (I will not do this here) that $$ \frac{d^4k}{(2 \pi)^4} \delta(k^2-m^2) \Theta(k^0) $$ is Lorentz invariant.
Thus, if we do a Lorentz transform : $k=\Lambda k'$, we have, with $g(k'^0,\vec{k'})=f(\Lambda(k'^0, \vec{k'}))$, we have according to (*) :
$$ (2 \pi) \int \frac{d^4k'}{(2 \pi)^4} \delta(k'^2-m^2) \Theta(k'^0)g(k'^0,\vec{k'}) = \int \frac{d^3k'}{(2\pi)^3}\frac{1}{2 \omega(\vec{k'})} g(\omega(\vec{k'}), \vec{k'})$$
Thus, if we have $\vec{k}=\vec{\Lambda k'}$, we have for any function $f$:
$$ \int \frac{d^3k}{(2\pi)^3}\frac{1}{2 \omega(\vec{k})} f(\omega(\vec{k}),\vec{k})=\int \frac{d^3k'}{(2\pi)^3}\frac{1}{2 \omega(\vec{k'})} f(\Lambda(\omega(\vec{k'}), \vec{k'}))$$
I just used them l.h.s of (*) with replacing the r.h.s with the line calculated above.
And it is finally equal to : $$ = \int \frac{d^3k'}{(2\pi)^3}\frac{1}{2 \omega(\vec{k'})} f(\omega(\vec{\Lambda k'}), \vec{\Lambda k'})))$$
Because $f(\Lambda(\omega(\vec{k'}), \vec{k'}))=f(\omega(\vec{\Lambda k'}), \vec{\Lambda k'})$
Which proove that $\frac{d^3k}{(2\pi)^3}\frac{1}{2 k_0}$ is Lorentz invariant.
