In this brief we want to prove a passage of the original GANs Paper by Ian Goodfellow (check this one if you want a full overview of the paper). Specifically, we want to prove that the following equation is satisfied:
\[\int_{\mathbf{z}}p_{z}(\mathbf{z})\log(1-\mathcal{D}(\mathcal{G}(\mathbf{z})))d\mathbf{z} = \int_{\mathbf{x}}p_{g}(\mathbf{x}) \log(1-\mathcal{D}(\mathbf{x})) d\mathbf{x}\]For a continuous random variable $\mathbf{z}$, let $\mathbf{x} = \mathcal{G}(\mathbf{z})$ and $\mathbf{z}=\mathcal{G}^{-1}(\mathbf{x})$ suppose that $\mathcal{G}$ is differentiable and that its inverse $\mathcal{G}^{-1}$ is monotonic. By the formula for inverse functions and differentiation we have that
\[\frac{d\mathbf{z}}{d \mathbf{x}}\cdot \frac{d \mathbf{x}}{d \mathbf{z}} = 1\] \[\frac{d\mathbf{z}}{d \mathbf{x}} = \frac{1}{\frac{d \mathbf{x}}{d \mathbf{z}}}\\\] \[\frac {d\mathbf{z}} {d\mathbf{x}} = \frac {1} {\frac{d\mathcal{G}(\mathcal{G}^{-1}(\mathbf{x}))}{d(\mathcal{G}^{-1}(\mathbf{x}))}}\] \[\frac {d\mathbf{z}} {d\mathbf{x}} = \frac {1} {\mathcal{G}^{'}(\mathcal{G}^{-1}(\mathbf{x}))}\] \[d\mathbf{z} = \frac {1} {\mathcal{G}^{'}(\mathcal{G}^{-1}(\mathbf{x}))}d\mathbf{x}\]so that by a change of variables we can rewrite everything in function of $\mathbf{x}$,
\[\int_{-\infty}^{\infty} p_{z}(\mathbf{z}) \log \left( 1-\mathcal{D}(\mathcal{G}(\mathbf{z})) \right) d\mathbf{z} =\] \[\int_{-\infty}^{\infty} p_{z}(\mathbf{z}) \log \left( 1-\mathcal{D}(\mathbf{x}) \right) \frac{1}{\mathcal{G}^{'}(\mathcal{G}^{-1}(\mathbf{x}))}d\mathbf{x}=\] \[\int_{-\infty}^{\infty} \color{orange}{p_{z}(\mathbf{z})} \log \left( 1-\mathcal{D}(\mathbf{x}) \right) \color{orange}{ \frac{1}{\mathcal{G}^{'}(\mathcal{G}^{-1}(\mathbf{x}))}}d\mathbf{x}=\] \[\int_{-\infty}^{\infty} \color{orange}{p_{z}(\mathbf{z})} \color{orange}{ \frac{1}{\mathcal{G}^{'}(\mathcal{G}^{-1}(\mathbf{x}))}} \log \left( 1-\mathcal{D}(\mathbf{x}) \right) d\mathbf{x}=\]Now, notice that the cumulative distribution function $P_\mathbf{X}(\mathbf{x}):\mathbb{R}^n\to[0,1]$ is $P_\mathbf{X}(\mathbf{x})=Pr(\mathbf{X}<\mathbf{x})=Pr(X_1\leq x_1,X_2\leq x_2,\dots,X_n\leq x_n)$, we observe that:
\[P_\mathbf{X}(\mathbf{x})=\] \[Pr(\mathbf{X}<\mathbf{x})=\] \[Pr(\mathcal{G}(\mathbf{Z})\leq\mathbf{x})=\] \[Pr(\mathbf{Z}\leq\mathcal{G}^{-1}(\mathbf{x}))=\] \[P_{\mathbf{Z}}(\mathcal{G}^{-1}(\mathbf{x}))\]From here
\[P_\mathbf{X}(\mathbf{x})=P_{\mathbf{Z}}(\mathcal{G}^{-1}(\mathbf{x}))\\ P_\mathbf{X}(\mathbf{x})=P_{\mathbf{Z}}(\mathbf{z})\]We take the derivative w.r.t $\mathbf{x}$:
\[\frac{dP_\mathbf{X}(\mathbf{x})}{d\mathbf{x}} =\frac{P_{\mathbf{Z}}(\mathbf{z})}{d\mathbf{x}}\] \[\frac{dP_\mathbf{X}(\mathbf{x})}{d\mathbf{x}} =\frac{P_{\mathbf{Z}}(\mathbf{z})}{d\mathbf{z}}\frac{d\mathbf{z}}{d\mathbf{x}}\] \[p_{X}(\mathbf{x}) = p_{\mathbf{z}}(\mathbf{z})\frac{1}{\mathcal{G}^{'}(\mathcal{G}^{-1}(\mathbf{x}))}\]For clarity we rename $p_{X}(\mathbf{x})$ as $p_{g}(\mathbf{x})$ since it represents the distribution learned from our generator $\mathcal{G}$, we have
\[\color{orange}{p_{g}(\mathbf{x}) = p_{\mathbf{z}}(\mathbf{z})\frac{1}{\mathcal{G}^{'}(\mathcal{G}^{-1}(\mathbf{x}))}}\]