The proof rests on a Fubini-type exchange of the order of integration, followed by a pointwise comparison. Define the marginal density $m(x) = \int_\Theta f(x, \theta)\pi(\theta)\, d\theta \geq 0$, which integrates the joint density of $(X, \theta)$ over $\theta$. Then \begin{align*} R_\pi(\delta) &= \int_\Theta \int_{\mathcal{X}} L(\delta(x), \theta)\, f(x, \theta)\, \pi(\theta)\, dx\, d\theta \\ &= \int_{\mathcal{X}} \int_\Theta L(\delta(x), \theta)\, \frac{f(x, \theta)\pi(\theta)}{m(x)} \, d\theta \cdot m(x)\, dx \\ &= \int_{\mathcal{X}} \mathbb{E}_\Pi[L(\delta(x), \theta) \mid x]\, m(x)\, dx. \end{align*} The key step is multiplying and dividing by $m(x)$ in the inner integral; this normalises the integrand $f(x,\theta)\pi(\theta)$ to the posterior density $f(x,\theta)\pi(\theta)/m(x)$, converting the inner integral into the posterior risk $\mathbb{E}_\Pi[L(\delta(x),\theta)|x]$. Swapping the order of integration requires $m(x) \geq 0$, which holds by definition. Let $\delta_\Pi$ minimise the posterior risk pointwise, so $\mathbb{E}_\Pi[L(\delta_\Pi(x), \theta)|x] \leq \mathbb{E}_\Pi[L(\delta(x), \theta)|x]$ for all $x$. Multiplying both sides by $m(x) \geq 0$ and integrating over $\mathcal{X}$ yields $R_\pi(\delta_\Pi) \leq R_\pi(\delta)$.