Does $P(Z > 1 | X=x, Y=y) < y$ $\forall x,y$, implies $P(Z>1 | Y=y) < y$? where $X, Y, Z$ are continuous RVs and they are NOT independent of each other.
My intuition is that since the RHS of the inequality $P(Z > 1 | X=x, Y=y) < y$ $\forall x,y$ is "y" and doesn't involve "x", so I can remove $X=x$ in the conditional part on the LHS.
But I'm not sure if this is a rigorous argument.
On the other hand, Z and Y depend on X (it could be a simple correlation, or Z and Y maybe a function of X), so is it OK to ignore $X=x$?
And will the answer be different if I don't use realization values $x,y$, 
i.e.
does $P(Z > 1 | X, Y) < Y$,  implies $P(Z>1 | Y) < Y$?
Thanks for any inputs you have.

Seems like it's true, provided $X,Y,Z$ have a joint density function.  Indeed, first note
\begin{align*}
P(Z > 1 \mid X = x, Y = y) & = \int_1^\infty \frac{f_{X,Y,Z}(x,y,z)}{f_{X,Y}(x,y)} \, dz < y \\
& \iff \int_1^\infty f_{X,Y,Z}(x,y,z)\, dz < y \cdot f_{X,Y}(x,y).
\end{align*}
Then,
\begin{align*}
P(Z > 1 \mid Y = y) & = \int_1^\infty \frac{f_{Y,Z}(y,z)}{f_Y(y)} \, dz \\
& = \int_1^\infty \left(\frac{\int_{-\infty}^\infty f_{X,Y,Z}(x,y,z)\, dx}{\int_{-\infty}^\infty f_{X,Y}(x,y)\, dx}\right) \, dz \\
& = \frac{1}{\int_{-\infty}^\infty f_{X,Y}(x,y)\, dx} \int_{-\infty}^\infty \int_1^\infty f_{X,Y,Z}(x,y,z)\, dz \, dx \\
& < \frac{1}{\int_{-\infty}^\infty f_{X,Y}(x,y)\, dx} \int_{-\infty}^\infty y \cdot f_{X,Y}(x,y) \, dx \\
& = y.
\end{align*}
The first equality is the definition of the conditional probability density function, the second is the definition of marginal density functions from joint densities, the third is Tonelli's theorem (valid since $f_{X,Y,Z}(\cdot,y,\cdot) \geq 0$ for all $y$ and the Lebesgue measure is sigma-finite), and the fourth equality is by assumption.
Update Just to clarify, Tonelli's theorem just provides sufficient conditions to interchange the order of integration.  A similar result is Fubini's theorem.  Anyways, depending on your background/class this might be for, I've seen most people simply exchange the order without much thought.  I'm just justifying it for myself.
