$Y=X+Z$ and $X \perp Z$ and $Y$, $X$, $Z$ continuous. Is $X \perp Z|Y$? Let $Y=X+Z$. Assume that $Y$, $X$, $Z$ are continuously distributed. Let $X \perp Z$. Prove / disprove that $X \perp Z | Y$.
I've seen a bunch of these types of problems on math.SE (and almost all disprove by counterexample) but in all problems I've seen the counterexamples use discrete distributions. I suspect there are still counterexamples when the variables are continuous but I couldn't find a counterexample and just wanted to make sure that there is nothing special about discrete distributions.
Here's a counterexample if the variables are discrete: Let $X \in \{0,1\}$ and $Z \in \{0,1\}$ independently. Then conditional on $Y=2$, we know that $X=Z=1$.
In the continuous case I was trying to convince myself that the statement is false by letting $Z$ take on large values and $X$ small values then if $Y$ takes on large values then $Z$ must also be large but then couldn't quite make the step that this would imply that $X$ and $Z$ are dependent.
Quick note on notation:  $X \perp Z$ means $X$ is independent of $Z$ and $X \perp Z|Y$ means $X$ is independent of $Z$ conditional on $Y$.
Thanks for help.

Other than in degenerate examples, intuitively, it seems obvious that $X,Z$ are dependent given $Y$ because given both $Z$ and $Y$, the value of $X$ is determined exactly. This applies for both discrete and continuous random variables.
In the continuous case, for any distributions of $X,Z:\;\;$ $f_{X,Z\mid Y}(x,z\mid y) = 0$ whenever $x+z\neq y$ but in general $f_{X\mid Y}(x\mid y)\;f_{Z\mid Y}(z\mid y) \neq 0$. This is sufficient to show that $X,Z$ are dependent given $Y$.
