Context
Section $16.2$ of Kerson Huang's Statistical Mechanics ($2$nd edition) deals with a derivation of two-point correlation function $\Gamma({\bf r})$, defined in terms of an order parameter density $m({\bf r})$ as $$\Gamma({\bf r})\equiv \big\langle m({\bf r})m(0)\big\rangle-\big\langle m({\bf r})\big\rangle\big\langle m(0)\big\rangle\tag{1}$$ where $\langle..\rangle$ denotes the ensemble average. To be explicit, for example, $$\langle m({\bf r})\rangle\equiv \sum\limits_{m({\bf r})}m({\bf r})e^{-\beta\mathcal{H}}$$ where $\mathcal{H}$ is the Hamiltonian of the system. All this is fine but I am stuck with something that is usually a pretty trivial step!
He uses the Fourier transform and inverse transform convention $$m({\bf r})=\int \frac{d^3k}{(2\pi)^3} e^{+i{\bf k}\cdot{\bf r}}\tilde{m}({\bf k}),~~ \tilde{m}({\bf k})=\int d^3x e^{-i{\bf k}\cdot{\bf r}}m({\bf r}).\tag{2}$$ With this, he makes the problematic claim that (see just above Eq. $16.11$), $$\boxed{\big\langle\tilde{m}({\bf k})\tilde{m}({\bf p})\big\rangle=(2\pi)^3\delta^{(3)}({\bf k}+{\bf p})|\tilde{m}({\bf k})|^2.}\tag{3}$$ To derive Eq.(3), one would usually proceed $$\big\langle \tilde{m}({\bf k})\tilde{m}({\bf p})\big\rangle=\big\langle\int d^3x \int d^3x^\prime e^{-i{\bf k}\cdot{\bf r}}e^{-i{\bf p}\cdot{\bf r}^\prime} m({\bf r})m({\bf r}^\prime)\big\rangle\tag{4}$$ if the two momenta were equal. But here, I don't see any way standard way of reducing it to the expression $(3)$. So the question is, how does he get Eq.$(3)$?