This is regarding Lorentz invariant phase space volume,
I am doing in following way,
$$I=\int dk^0 \delta(k^2-m_0^2) \Theta(k_0) = \int dk^0 \delta(k_0^2-E_k^2) \Theta(k_0)$$ where $$E_k = \sqrt{\mathbf{k^2}+m_0^2}$$
From mass shell (Einstien relation) relation ..
$$k_0^2 = E_k$$ which means that $\delta$ function is singular, if it is for mass shell particle. hence we can integrate around small uncertainty around $k_0$ , $\Delta k_0$,
which means that, \begin{equation} I = \int_{\Delta k_0} dk_0 \frac{1}{2k_0} (\delta(k_0 - E_k) + \delta(k_0+E_k))\Theta(k_0) = \frac{1}{2E_k} \end{equation} hence $$\int\frac{d^3k}{2E} = \int d^3k \int_{\Delta k_0} dk_0 \delta(k^2-m_0^2)\Theta(k_0)$$ since $d^3k$ integration is all over three dimensional momentum space and $$k_0 = E_k = \sqrt{\mathbf{k}^2+m_0^2}$$ integration over small uncertainty of $k_0$ wiil also converted into all over k_0 space (or 4-momentum space)becaus of its dependence on $\mathbf{k}$ which runs over all three dimensional momentum space.
hence, $$\int \frac{d^3k}{2E} = \int d^4k \delta(k^2-m_0^2)\Theta(k_0)$$. I found here un the help page that
integral I can also be written as(instead of putting the delta integration see eq 1)
$$I = dk_0 \delta(k_0 -E_k)\frac{1}{2k_0}\Theta(k_0)=\frac{1}{2E_k}$$ $$\int \frac{d^3k}{2E} = \int d^4k\delta(k_0^2-m_0^2)\Theta(k_0)$$
Please help me if my understanding is correct
