Let $ X\sim \exp(\lambda) $ and let $ (U|X=k) \sim U[x-1,x+1] $ for $ x\ge 0 $. Why $ P(X=k) \ne 0 $?

Question

Iv'e encountered the following definition:

$X\sim \exp(\lambda)$ and let $(U|X=x) \sim \mathcal U[x-1,x+1]$ for $x\geq 0$.

Now I know that since $X$ is continuous, $P(X=x)=0$ for any $x$. So how can we even discuss $(U|X=x)$?

Answer 1

$E(X|Y)$ can be written as $f(Y)$ for some measurable function $f: \mathbb R \to \mathbb R$ and $E(X|Y=x)$ is the notation used for the function $f(x)$.