Theorem (Neyman-Pearson): Let $f_0 = f(\mathbf{x}; \theta_0)$ and $f_1 = f(\mathbf{x}; \theta_1)$ be the densities under $H_0$ and $H_1$. The test $\phi^*$ that rejects $H_0$ when $f_1(\mathbf{x})/f_0(\mathbf{x}) > k$ (where $k$ is chosen to give size $\alpha$) is the most powerful level-$\alpha$ test.

Step 1: Define the tests.

Let $\phi^*$ be the likelihood ratio test with rejection region $R^* = \{\mathbf{x} : f_1(\mathbf{x}) > k \cdot f_0(\mathbf{x})\}$ and size $\alpha^* = \int_{R^*} f_0\,d\mathbf{x} \leq \alpha$.

Let $\phi$ be any other test with rejection region $R$ and size $\alpha_\phi = \int_R f_0\,d\mathbf{x} \leq \alpha$.

We want to show $\beta_1(\phi^*) \geq \beta_1(\phi)$, i.e., $\int_{R^*} f_1\,d\mathbf{x} \geq \int_R f_1\,d\mathbf{x}$.

Step 2: Decompose the power difference.

$\int_{R^*} f_1 - \int_R f_1 = \int_{R^* \setminus R} f_1 - \int_{R \setminus R^*} f_1$

On $R^* \setminus R$: since $\mathbf{x} \in R^*$, we have $f_1(\mathbf{x}) > k \cdot f_0(\mathbf{x})$, so $\int_{R^* \setminus R} f_1 > k \int_{R^* \setminus R} f_0$

On $R \setminus R^*$: since $\mathbf{x} \notin R^*$, we have $f_1(\mathbf{x}) \leq k \cdot f_0(\mathbf{x})$, so $\int_{R \setminus R^*} f_1 \leq k \int_{R \setminus R^*} f_0$

Step 3: Combine the inequalities.

$\int_{R^*} f_1 - \int_R f_1 > k\left(\int_{R^* \setminus R} f_0 - \int_{R \setminus R^*} f_0\right) = k\left(\int_{R^*} f_0 - \int_R f_0\right) = k(\alpha^* - \alpha_\phi)$

Step 4: Use the size constraint.

Since $\alpha^* = \alpha$ (by construction of $k$) and $\alpha_\phi \leq \alpha$, we have $\alpha^* - \alpha_\phi \geq 0$, so: $\int_{R^*} f_1 - \int_R f_1 \geq k \cdot 0 = 0$

Therefore the power of $\phi^*$ is at least as large as the power of any other level-$\alpha$ test $\phi$: $P_{\theta_1}(\phi^* \text{ rejects}) \geq P_{\theta_1}(\phi \text{ rejects})$

This holds for every level-$\alpha$ test $\phi$, so $\phi^*$ is the most powerful level-$\alpha$ test. $\square$