Rao-Blackwell improvement for a nonrandomized estimator

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Context: please consider a parametric statistical model $(\mathcal{Y},\{P_\theta:\theta\in\Theta\})$ and suppose that we are estimating $g(\theta)$. Associated with this is the set of decisions $\mathcal{D}=\{g(\theta):\Theta\}$ and a loss function $L$. Then it is proved in Monfort and Gourieroux (1996) that given any sufficient statistic $S$ and any randomized decision rule $m$, there exists a randomized decision rule $m^*$ that is a function of $S$ and that has the same risk as $m$.

The proof defines $m^*$ as follows: $$ \forall D\subset\mathcal{D}:\quad m_y^*(D)=E_\theta[m_Y(D)|S(Y)=S(y)]\\ =\int_\mathcal{Y}m_{\bar{y}}(D)dP_\theta^{Y|S(Y)=S(y)}(\bar{y})=\int_\mathcal{Y}\left(\int_{d\in D}dm_{\bar{y}}(d)\right)dP_\theta^{Y|S(Y)=S(y)}(\bar{y}).\tag{i} $$ My question: what is $m^*$ when $m$ is a nonrandomized estimator $\delta:y\mapsto\delta(y)\in\mathcal{D}$? The text claims that $m^*$ is also a nonrandomized estimator, say $\delta^*$, satisfying $\delta^*(Y)=E_\theta[\delta(Y)|S(Y)]$. Since a nonrandomized estimator can be thought of as a degenerate randomized estimator, I tried to use the definition (i) to derive $m^*$ but I'm not getting far.

Remarks on notation and progress: in the text, a randomized decision rule $m$ assigns each observation $y$ to a distribution $m_y$ on the set of decisions $\mathcal{D}$. In definition (i) above, $\bar{y}$ is just a dummy, written in that manner to distinguish it from $Y$ and $y$. Moreover $$ P_\theta^{Y|S(Y)=S(y)} $$ denotes the conditional distribution of $Y$ given that $S(Y)=S(y)$. Because $S$ is sufficient, $P_\theta^{Y|S(Y)=S(y)}$ actually doesn't depend on $\theta$ so the expectation in (i) doesn't depend on $\theta$ either.

So far, I can only note that when $m$ is degenerate, $m_\bar{y}(D)=1$ if $\delta(\bar{y})\in D$ and $0$ otherwise. Thus, $$ \int_\mathcal{Y}m_{\bar{y}}(D)dP_\theta^{Y|S(Y)=S(y)}(\bar{y})=\int_{\bar{y}\in\mathcal{Y}:\delta(\bar{y})\in D}dP_\theta^{Y|S(Y)=S(y)}(\bar{y})=P_\theta^{Y|S(Y)=S(y)}[\delta^{-1}(D)] $$ but then I'm stuck. Thanks for your consideration!