$\operatorname{supp}(f_{X,Y})=\{(x,y)\in\mathbb{R}^2||x|+|y|\leq1\}$ then $X,Y$ are not independent

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Let $Z=(X,Y)$ be a absolutely coninuous random variable such that $$ \\\operatorname{supp}(f_Z)=\{(x,y)\in\mathbb{R}^2||x|+|y|\leq1\} \ $$ Show that $X,Y$ are not independent.


I don't have a good idea how to continue the proof. I need to find $(t,s)\in\mathbb{R}^2$ that setasfies $$ \\ \int_{-1}^{s}\int_{-1}^{t}f_Z(x,y)dx dy=\int_{-1}^{1}f_Z(t,y)dy\cdot\int_{-1}^{1}f_Z(x,s)dx \ $$

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Hint: Can $X$ and $Y$ simultaneously be close to $1$? Can you show that each individually has a positive probability of being close to $1$?


Hint 2: I mean the following. Note that since the set $\{x > \frac{1}{2}, y > \frac{1}{2}\}$ intersects the support of $f_Z$ trivially, we must have that $P[X > \frac{1}{2}, Y > \frac{1}{2}] = 0\,.$ To show that they're not independent, you'd just now need to show that $P[X > \frac{1}{2}] > 0$ and $P[Y > \frac{1}{2}] > 0$, since we'd then have $$P\left[X > \frac{1}{2}, Y > \frac{1}{2}\right] \neq P\left[X > \frac{1}{2}\right] \cdot P\left[Y > \frac{1}{2}\right]\,. $$