Given random variables $D: \Omega \to \{-1,1\}$, $X: \Omega \to \{1,\dots, n\}$ and $Y: \Omega \to \{1,\dots, n\}$, such that
$$X \sim Y$$
and
$$P[D| X] = P[D|Y]$$
may one conclude that
$$(D,X) \sim (D,Y)?$$
One may be tempted to use the definition of conditional probability and argue that
$$P[D=i, X=k ] = \int_{\{X=k\}}1_{\{D=i\}} dP = \int_{\{Y=k\}}1_{\{D=i\}} dP = P[D=i, Y=k ] ,$$
but since $P[X= k ] = P[Y= k ] $ does not necessarily imply that $\{X= k \} = \{Y=k \}$, I don't necessarily think this might be true...
Is it still possible to show that $(D,X) \sim (D,Y)$?
$P\left[D\mid X\right]=P\left[D\mid Y\right]$ tells us that: $$P\left[D=i,X=k\right]=P\left[D=i\mid X=k\right]P\left[X=k\right]=P\left[D=i\mid Y=k\right]P\left[X=k\right]$$
On base of $X\sim Y$ we can proceed with: $$\cdots=P\left[D=i\mid Y=k\right]P\left[Y=k\right]=P\left[D=i,Y=k\right]$$