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Let $X$ and $Y$ be two dependent random vectors in $\mathbb{R}^d$, such that $X\neq Y$ with probability 1, whose joint probability measure has density $\mu(x,y)$ with respect to the Lebesgue measure. For measurable sets $A$ and $B$, does the inequality $$ \mathbb{P}(X-Y \in A, Y\in B) \leq \sup _{t\in B}\mathbb{P}(X \in A+t) $$ hold true? Herein, operations are meant elementwise and $A+t:=\{x+t:x \in A\}$.

I was thinking to go through something like:

$$ \mathbb{P}(X-Y \in A, Y\in B)= \int_{B}\int_{A+y}\mu(x,y)dxdy\\ \leq \sup_{t\in B}\int_{B}\int_{A+t}\mu(x,y)dxdy\\ =\sup_{t\in B}\int_{A+t}\int_B\mu(x,y)dydx\\ \leq \sup _{t\in B}\int_{A+t}\mu(x)dx\\ =\sup _{t\in B}\mathbb{P}(X \in A+t) $$

where $\mu(x)$ denote the marginal density of $X$, but I have doubts about the second and third lines, I'm not sure they're correct. I pass from the third to the fourth line by using the fact that $\mu(x,y)$ is nonnegative and $\int_B\mu(x,y)dy \leq \int_{\mathbb{R}^d}\mu(x,y)dy=\mu(x)$.

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