Two-point Le Cam lower bound for squared-error risk

Put $c=(t_0+t_1)\mathrm{/}2$ and define the randomized test $\varphi (x)=K(x,[c,\mathrm{\infty }))$. It is measurable and lies in $[0,1]$ by [F2]. Under parameter $0$, every action in $[c,\mathrm{\infty })$ has squared error at least ${\mathrm{\Delta }}^2\mathrm{/}4$; under parameter $1$, every action in $(-\mathrm{\infty },c)$ has squared error at least ${\mathrm{\Delta }}^2\mathrm{/}4$. Hence $R_0(K)+R_1(K)\ge \frac{{\mathrm{\Delta }}^2}{4}(\int \varphi d{P}_0+\int (1-\varphi )d{P}_1)\mathrm{.}$

For each integer $N\ge 1$, let ${\varphi }_N=N^{-1}{\sum }_{r=1}^N{\mathbf{1}}_{\{\varphi \ge r\mathrm{/}N\}}$. Then $0\le \varphi -{\varphi }_N\le 1\mathrm{/}N$. By [F4], each event term satisfies $P_0(\varphi \ge r\mathrm{/}N)-P_1(\varphi \ge r\mathrm{/}N)\ge -\mathrm{\parallel }{P}_0-P_1{\mathrm{\parallel }}_{\mathrm{T}\mathrm{V}}$, so $\int {\varphi }_{N}d{P}_0-\int {\varphi }_{N}d{P}_1\ge -\mathrm{\parallel }{P}_0-P_1{\mathrm{\parallel }}_{\mathrm{T}\mathrm{V}}$. The two integral errors from replacing $\varphi$ by ${\varphi }_N$ have total absolute value at most $2\mathrm{/}N$; letting $N\to \mathrm{\infty }$ gives $\int \varphi d{P}_0+\int (1-\varphi )d{P}_1\ge 1-\mathrm{\parallel }{P}_0-P_1{\mathrm{\parallel }}_{\mathrm{T}\mathrm{V}}\mathrm{.}$ Combining with step 1.1 and using $\max (R_0,R_1)\ge (R_0+R_1)\mathrm{/}2$ proves the risk bound. Taking the infimum over $K$ and applying [F5] proves the minimax assertion.