QUESTION
Find p.d.f of $Y=X_1+X_2$, where $X_1$and $X_2$ are two independent random variables $X_i$~Uniform(0,1), $i=1,2$
I read few question about convolution of two uniform in here. But I have different case using indicator function. I don't think my solution is strictly correct. What is the possible solution with indicator function?
MY SOLUTION
For $0\lt y-x_2\lt 1$ and $0\lt x_2\lt 1$
$$f_1(y-x_2)f_2(x_2)=1$$
$$f_Y(y)=\frac{d}{dy}F_Y(y)=\int_{-\infty}^{\infty}f_1(y-x_2)f_2(x_2)=\int_{-\infty}^{\infty}I(0\lt y-x_2\lt 1)I(0\lt x_2\lt1)dx$$
$$=\int_{-\infty}^{\infty}I(0\lt x_2\lt y\lt x_2+1\lt 2)dx=\int_{-\infty}^{\infty}I(0\lt x_2\lt y)I(y-1\lt x_2\lt 1)dx$$
Thus
$f_Y(y)=\int_0^ydx$, if $0\lt y\lt 1$ and $=\int_{y-1}^1dx$, if $1\lt y\le 2$.
$\huge\color{green}{\checkmark}$That is correct.
$$\begin{align} f_Y(y) &= \int_\Bbb R f_X(y-x)f_X(x)\,\mathrm d x\\[1ex]&= \int_\Bbb R \mathbf 1_{0\leqslant y-x\leqslant 1}\mathbf 1_{0\leqslant x\leqslant 1}\,\mathrm d x\\[1ex]&=\int_{\max\{0,y-1\}}^{\min\{1,y\}}\mathbf 1_{0\leqslant y\leqslant 2}\,\mathrm d x\\[1ex]&= (\min\{1,y\}-\max\{0,y-1\})\,\mathbf 1_{0\leqslant y\leqslant 2}\\[1ex]&= y\,\mathbf 1_{0\leqslant y<1}+(2-y)\,\mathbf 1_{1\leqslant y\leqslant 2} \end{align}$$
This is an example of a Triangular Distribution.
To check your answer, you could have verified that the integral over the support does equal $1$. $$\int_0^1 y\,\mathrm d y+\int_1^2(2-y)\,\mathrm d y = 1$$