This question may be simple, though I'm not managing to find an answer. Let $X$ and $Y$ be two dependent random vectors in in $\mathbb{R}^d$, with joint probability density $\mu(x,y)$ (with respect to the Lebesgue measure). For any subset $A \subset \mathbb{R}^d$ and vector $t \in \mathbb{R}^d$, define 
$$
A+t=\{x+t=(x_1+t_1, \ldots, x_d+t_d): x \in A\}.
$$
Is it true that 
$$
P(X-Y \in A, Y \in B) \leq \sup_{t \in B}P(X \in A+t)
$$
where $A$ and $B$ are measurable proper subset of $\mathbb{R}^d$? The inequality is trivially true if $X$ and $Y$ are independent:
$$
P(X-Y \in A, Y \in B) =\int_B \left[\int_{A+y} \mu(x|y)dx\right] \mu(y)dy\\
=\int_B \left[\int_{A+y} \mu(x)dx\right] \mu(y)dy\\
=\int_B P(X\in A+y)\mu(y)dy\\
\leq \sup_{y \in B}P(X\in A+y)
$$
where $\mu(x|y)$, $\mu(x)$ and $\mu(y)$ are the conditional density of $X$ given $Y=y$, the marginal density of $X$ and the marginal density of $Y$, respectively. What about the case where $X$ and $Y$ are dependent (i.e. $\mu(x|y)\neq \mu(x)$)?

**EDIT** Here is my attempt:

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

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 
$$
\int_B\mu(x,y)dy \leq \int_{\mathbb{R}^d}\mu(x,y)dy=\mu(x).
$$