Why does the Law of the Unconscious Statistician work here for the pathwise estimator

Question

https://arxiv.org/abs/1906.10652

So there are these two parts "Continuous distributions have a simulation property that allows both a direct and an indirect way of drawing samples from them, making the following sampling processes equivalent: $$ \hat{\mathbf{x}} \sim p(\mathbf{x} ; \boldsymbol{\theta}) \equiv \hat{\mathbf{x}}=g(\hat{\boldsymbol{\epsilon}}, \boldsymbol{\theta}), \quad \hat{\boldsymbol{\epsilon}} \sim p(\boldsymbol{\epsilon}) $$ and states that an alternative way to generate samples $\hat{\mathbf{x}}$ from the distribution $p(\mathbf{x} ; \boldsymbol{\theta})$ is to sample first from a simpler base distribution $p(\epsilon)$, which is independent of the parameters $\theta$, and to then transform this variate through a deterministic path $g(\boldsymbol{\epsilon} ; \boldsymbol{\theta}) ;$ we can refer to this procedure as either a sampling path or sampling process. For invertible paths, this transformation is described by the rule for the change of variables for probability $$ p(\mathbf{x} ; \boldsymbol{\theta})=p(\boldsymbol{\epsilon})\left|\nabla_{\boldsymbol{\epsilon}} g(\boldsymbol{\epsilon} ; \boldsymbol{\theta})\right|^{-1} $$" "Equipped with the pathwise simulation property of continuous distributions and LOTUS, we can derive an alternative estimator for the sensitivity analysis problem (2) that exploits this additional knowledge of continuous distributions. Assume that we have a distribution $p(\mathbf{x} ; \boldsymbol{\theta})$ with known differentiable sampling path $g(\epsilon ; \boldsymbol{\theta})$ and base distribution $p(\boldsymbol{\epsilon})$. The sensitivity analysis problem (2) can then be reformulated as $$ \begin{aligned} \eta &=\nabla_{\theta} \mathbb{E}_{p(\mathbf{x} ; \theta)}[f(\mathbf{x})]=\nabla_{\boldsymbol{\theta}} \int p(\mathbf{x} ; \boldsymbol{\theta}) f(\mathbf{x}) d \mathbf{x} \\ &=\nabla_{\boldsymbol{\theta}} \int p(\boldsymbol{\epsilon}) f(g(\boldsymbol{\epsilon} ; \boldsymbol{\theta})) d \boldsymbol{\epsilon} \\ &=\mathbb{E}_{p(\boldsymbol{\epsilon})}\left[\nabla_{\boldsymbol{\theta}} f(g(\boldsymbol{\epsilon} ; \boldsymbol{\theta}))\right] . \\ \overline{\boldsymbol{\eta}}_{N} &=\frac{1}{N} \sum_{n=1}^{N} \nabla_{\boldsymbol{\theta}} f\left(g\left(\hat{\boldsymbol{\epsilon}}^{(n)} ; \boldsymbol{\theta}\right)\right) ; \quad \hat{\boldsymbol{\epsilon}}^{(n)} \sim p(\boldsymbol{\epsilon}) \end{aligned} $$ In Equation (29a) we first expand the definition of the expectation. Then, using the law of the unconscious statistician, and knowledge of the sampling path $g$ and the base distribution for $p(\mathbf{x} ; \theta)$, we reparameterise this integral $(29 \mathrm{~b})$ as one over the variable $\epsilon$. The parameters $\theta$ have now been pushed into the function making the expectation free of the parameters. This allows us to, without concern, interchange the derivative and the integral $(29 \mathrm{c})$, resulting in the pathwise gradient estimator (29d)."

Particularly, I am looking at this equation and not exactly understanding the equality between these two. \begin{aligned} \nabla_{\boldsymbol{\theta}} \int p(\mathbf{x} ; \boldsymbol{\theta}) f(\mathbf{x}) d \mathbf{x} &=\nabla_{\boldsymbol{\theta}} \int p(\boldsymbol{\epsilon}) f(g(\boldsymbol{\epsilon} ; \boldsymbol{\theta})) d \boldsymbol{\epsilon}\end{aligned}