1. Let $g : [0,1]^m \rightarrow \mathbb{R}^+$ be a bounded measurable nondecreasing function.
Let $i \in \{1,..,m\}$ and $u$ such that $\mathbb{P}(\hat{p_i} \le u) >0$.
We note $\hat{p} = (\hat{p_1}, .., \hat{p_m})$ the random vector of the p-values. The function $g$ being measurable and positive, $g(\hat{p})$ is a positive random variable.
For all positive random variable $X$ and for all density $\mathbb{P}$, we have
$$\mathbb{E}(X) = \int_0^\infty \mathbb{P}(X \geq t)dt.$$
We now apply this property for $X = g(\hat{p})$ and $\mathbb{P} = \mathbb{P}(.|\hat{p_i}\le u)$, then we have
$$\mathbb{E}(g(\hat{p}) | \hat{p_i} \le u) = \int_0^\infty \mathbb{P}(g(\hat{p}) \geq t | \hat{p_i} \le u) dt.$$
2. Let $g : [0,1]^m \rightarrow \mathbb{R}^+$ be a bounded measurable nondecreasing function.
Let $t \geq 0$. Lets show that $\hat{p} \mapsto \textbf{1}_{g^{-1}([t,+\infty[)}(\hat{p})$ is nondecreasing.
Let $\hat{p}, \hat{q}$ such that $\hat{p} \le \hat{q}$ (ie : $\hat{p_i} \le \hat{q_i}$ for all $i$).
If $\textbf{1}_{g^{-1}([t,+\infty[)}(\hat{p}) = 0$, then $\textbf{1}_{g^{-1}([t,+\infty[)}(\hat{p}) \le \textbf{1}_{g^{-1}([t,+\infty[)}(\hat{q})$.
Now if $\textbf{1}_{g^{-1}([t,+\infty[)}(\hat{p}) = 1$, then $\hat{p} \in g^{-1}([t,+\infty[)$ and then $g(\hat{p})\geq t$. The function $g$ being nondecreasing, we have $g(\hat{q}) \geq g(\hat{p}) \geq t$, so $\hat{q} \in g^{-1}([t,+\infty[)$, then $\textbf{1}_{g^{-1}([t,+\infty[)}(\hat{q}) = 1$. We have again $\textbf{1}_{g^{-1}([t,+\infty[)}(\hat{p}) \le \textbf{1}_{g^{-1}([t,+\infty[)}(\hat{q})$, prooving that $\hat{p} \mapsto \textbf{1}_{g^{-1}([t,+\infty[)}(\hat{p})$ is nondecreasing.
Show that $\hat{p}$ fullfills the WPRDS property.
Let $g : [0,1]^m \rightarrow \mathbb{R}^+$ be a bounded measurable nondecreasing function and take $i \in \{1,..,m\}$. By hypothesis, we have that $\hat{p}$ fulfills WPRDS for any nondecreasing indicator function.
But we know that $\textbf{1}_{g^{-1}([t,+\infty[)}$ is nondecreasing. So we have, for all $u \le v$,
$$\mathbb{E}(\textbf{1}_{g^{-1}([t,+\infty[)}(\hat{p}) | \hat{p_i} \le u) \le \mathbb{E}(\textbf{1}_{g^{-1}([t,+\infty[)}(\hat{p}) | \hat{p_i} \le v).$$
But $\mathbb{E}(\textbf{1}_{g^{-1}([t,+\infty[)}(\hat{p}) | \hat{p_i} \le u) = \mathbb{P}(g(\hat{p}) \geq t | \hat{p_i} \le u)$. So we have
$$\mathbb{P}(g(\hat{p}) \geq t | \hat{p_i} \le u) \le \mathbb{P}(g(\hat{p}) \geq t | \hat{p_i} \le v),$$
that give us, integrating on all the space,
$$\mathbb{E}(g(\hat{p}) | \hat{p_i} \le u) \le \mathbb{E}(g(\hat{p}) | \hat{p_i} \le v).$$