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 1** 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 
$$
\int\_B\mu(x,y)dy \leq \int\_{\mathbb{R}^d}\mu(x,y)dy=\mu(x).
$$


**EDIT 2** Are there conditions (different from independence) under which the inequaliy holds true? In the somewhat patological case $X=Y$ with probability one, as highlighted in the answer below, the inequality may not be true. But what if $X \neq Y$ with probability 1? Are there conditions under which, in such an istance, the inequality is satisfied?



Is this probability inequality true?
This is false:


Let the probability distribution be $P(X = 0, Y = 0) = P(X = 1, Y = 1) = \frac{1}{2}$ and let $A = \{ 0\}$, $B = \{ 0, 1\}$. Then the left-hand side of your inequality is $1$ while the right-hand side is $\frac{1}{2}$.


If you want your densities to be continuous just convolve with some smooth highly concentrated function (both $X, Y$ and $A, B$).


Actually, if $X=Y$ with probability one and $A = \{0\}$, $B = \mathbb{R}$ then your inequality is false as long as distribution of $X$ is non-constant. For example, if $X$ is normal then left-hand side is $1$ while right-hand side is zero.


