$X,Y$ - uniform random variables, but here's a trick. While $f_Y(y)=1$ on $[0,1]$, $f_X(x)=5$ if $0<x<0.1, 0.9<x<1$ and $f_Y(y)=0$ otherwise.
I know how find $PDF_{X+Y}$ for continuous $f_X$, but what should I do here? Do I have to write two convolution integrals for $0<x<0.1, 0.9<x<1$? I assume that if $x\in(0.2,0.8)$ the integral will be $0$.
Hint: $\displaystyle f_{X+Y}(z) = \int_{-\infty}^{\infty} f_Y(y)f_X(z-y)dy = \int_{0}^{1} f_X(z-y)dy = \int_{z-1}^{z} f_X(x)dx.$