homework and exercises - Proving the Lorentz invariance of the Lorentz invariant phase space element

I have been looking around for a satisfactory answer to prove that

$$\frac{d^3\vec{p}}{2E_{\vec{p}}}$$

where $E_{\vec{p}}=+\sqrt{(|\vec{p}|c)^2+(mc^2)^2}$, is Lorentz invariant. The standard answer seems to be that the above measure is equal to

$$d^4p \, \delta (p^2-m^2)\theta(p_0)$$

where $p^2=p_0^2-|\vec{p}|^2$, and $\theta(x)$ is the step function. I understand these two are equivalent, but I don't understand why the second has to be Lorentz invariant, in particular why the Dirac delta has to be Lorentz invariant. I have found a document (section 2.1) that proves that $\delta^{(4)}(p-p')$ is Lorentz invariant, but I can't find a way to extend their method successfully here. In fact, all I can seem to get tells me the above isn't Lorentz invariant, and that in fact it should transform to

$$\frac{d^3\vec{p}'}{2\gamma E_{\vec{p}'}}$$

which makes sense from $d^4p$ being Lorentz invaraint and $dp_0$ transforming proportional to $\gamma$, but it isn't what everyone else says. What's the problem here?

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An alternative way to "derive" the factor of $1/\gamma$:

$$\delta (p^2-m^2)\theta(p_0)=\frac{1}{E_{\vec p}}\delta(p_0-E_{\vec{p}})$$

Now, as $dp_0$ transforms to $\gamma dp_0$, the $\delta(p_0-E_{\vec{p}})$ should transform to $\gamma^{-1}\delta(p_0-E_{\vec{p}})$, as can be shown with an analogous argument to the one shown in the document. The $E_{\vec{p}}$ also transforms to $\gamma(E_{\vec{p}}-v p_x)$ for an $x$-boost, but this doesn't seem to solve it.

Answer

I've come across an answer in Peskin's Introduction to Quantum Field theory where he looks at how to make the 3-momentum delta function invariant that might satisfy you. Look at a boost in the $p_3$ direction so that $p'_3= \gamma(p_3+\beta E),E'=\gamma(E+\beta p_3).$

$$\delta^3(p-q)= \delta^3(p'-q')\frac{dp'_3}{dp_3}$$ $$= \delta^3(p'-q')\gamma(1+\beta\frac{dE}{dp_3})= \delta^3(p'-q')\frac{\gamma}{E}(E + \beta E\frac{dE}{dp_3})$$ $$= \delta^3(p'-q')\frac{\gamma}{E}(E+\beta p_3)= \delta^3(p'-q')\frac{E'}{E}.$$

So for a 3-momentum delta function, the quantity $E\delta^3(p-q)$ is Lorentz invariant. Combined with the fact that $\delta^4(p-q)$ is invariant, we conclude $\frac{1}{E}\delta(p_0-q_0)$ is invariant.