Convolution of two identically distributed uniform random variables

I'm not sure why $f_{X + Y}(a) = \int_{0}^{1} f_{X}(a - y) \mathop{dy}$. Is that because $f_{Y}(a) = 1$ on the interval from $0$ to $1$? But what about when $1 < a < 2$? Wouldn't the integral just become $0$ in this case then?

No, it is because $f_Y(y)=1$ when $y\in(0;1)$ else $f_Y(y)=0$, and because $y$ is the bound variable of integration.$$\begin{align}f_{X+Y}(a)&= \int_{-\infty}^\infty f_{X,Y}(a-y,y)~\mathsf d y\\[1ex] &=\int_{-\infty}^\infty f_{X+Y}(a-y)~f_Y(y)~\mathsf dy \\[1ex] &= \int_0^1 f_{X+Y}(a-y)~\mathsf dy\end{align}$$

Now $f_X(a-y)=1$ when $0\lt a-y\lt 1$ else $f_X(a-y)=0$ and we know $(0;2)$ is the support for $f_{X+Y}$

$$\begin{align}f_{X+Y}(a) &= \int_0^1~\mathbf 1_{0\lt a\lt 2~,~ a-1\lt y\lt a}~\mathsf dy\\[1ex] &= \mathbf 1_{0\lt a\lt 2}\int_{\max(0,a-1)}^{\min(1,a)}~\mathsf dy \\[2ex] &=\mathbf 1_{0\lt a\lt 1}\int_0^a~\mathsf d y+\mathbf 1_{1\leq a\lt 2}\int_{a-1}^1~\mathsf d y\end{align}$$

Okay, we have the supports, $0<y<1$ for $f_Y(y)$ and $a-1<y<a$ for $f_X(a-y)$, then since the support of $y$ for their product is the intersection of both, and it is thusly $\max(0,a-1)< y< \min(1,a)$ . This is relative to $a$, which itself is supported over $0<a<2$.

Those min/max functions are a little annoying, so we shall partition the support for $f_{X+Y}(a)$ into $0<a<1$ union $1\leqslant a< 1$.   Thusly in the first part, the bounds for the integraton are $0$ to $a$ while for the second part they are $a-1$ to $1$.

PS: The notation is an indicator function, having value of 1 when the indice is true, and 0 otherwise. $$\mathbf 1_{E}=\begin{cases}1&:& E\\0&:&\text{otherwise}\end{cases}$$