Let $P_R$ be a probability on $R$, let $\mathcal{Y}$ be the $\sigma$-algebra generated by the intervals that contain the positive reals $(x\geq 0)$, and let $\mathcal{Z}$ be the $\sigma$-algebra generated by the intervals that contain the negative reals $(x<0)$. Suppose $P_{R/\mathcal{Z}}(z)(A)=P_R[A|\mathcal{Z}](z)$ is a probability on $R$ (in fact it is) for each $z$. We can therefore condition that probability by $\mathcal{Y}$ and thus define $P_{R/\mathcal{Y}/\mathcal{Z}}(y,z)(A)=P_R[A|\mathcal{Y}|\mathcal{Z}](y,z)$. What would $P_{R/\mathcal{Y}/\mathcal{Z}}(y,z)(A)$ be?
And $E[X|\mathcal{Y}|\mathcal{Z}](y,z)\quad (X(x)=x)$?
I would bet on: \begin{align*} &\hspace{-6cm}P_{R/\mathcal{Y}/\mathcal{Z}}(y,z)(A)=I_{R^{^-}}(z)I_{A^{^-}}(y)+I_{R^{^+}}(y)I_{A^{^+}}(z)\\[10pt] &\hspace{-6cm}E[X|\mathcal{Y}|\mathcal{Z}](y,z)=\begin{cases} y& \text{ if }\ y,z< 0 \\ z& \text{ if }\ y,z\geq 0\\ indifferent&\ otherwise \end{cases} \end{align*}