Trouble with indicator functions

Question

We want to convolve random variables $X$ and $Y$ which are both uniformly distributed on $[0,1]$. We have the densities $f_1(x)=1_{[0,1]}(x)$ and $f_2(y)= 1_{[0,1]}(y)$, where $1$ is the indicator function.

Now using the formula, we get:$$\begin{aligned} f(u)=\left(f_{1} * f_{2}\right)(u) &=\int \limits_{-\infty}^{\infty} f_{1}(u-v) f_{2}(v) d v \\ &=\int \limits_{0}^{1} \mathbb{1}_{[0,1]}(u-v) d v \\ &\overset{(*)}=\int \limits_{0}^{1} \mathbb{1}_{[0,1]}(u-v) \mathbb{1}_{[0,1]}(u) d v+\int \limits_{0}^{1} \mathbb{1}_{[0,1]}(u-v) \mathbb{1}_{[1,2]}(u) d v \\ &=\int \limits_{0}^{u} \mathbb{1}_{[0,1]}(u) d v+\int \limits_{u-1}^{1} \mathbb{1}_{[1,2]}(u) d v=u \mathbb{1}_{[0,1]}(u)+(2-u) \mathbb{1}_{[1,2]}(u) \end{aligned}$$

What's happening in $(*)$?

Answer 1

Since $X$ and $Y$ take values in $[0,1]$ their sum $X+Y$ takes values in $[0,2]$. So we have to find $f(u)$ only for $0 \leq u\leq 2$. Note that $1_{[0,1]} (u)+1_{(1,2]}(u)=1$ in this case. So $\mathbb{1}_{[0,1]}(u-v)=\mathbb{1}_{[0,1]}(u-v)1_{[0,1]} (u)+\mathbb{1}_{[0,1]}(u-v) 1_{(1,2]}(u)$ (Of course, $\{1\}$ has measuer $0$).

Next step: $0\leq u-v \leq 1$ is same as $ u-1\leq v \leq u$. However, $u-1$ may be negative. So we have to consider the cases $0 \leq u \leq 1$ and $1 <u \leq 2$ separately. In the first case we have to integrate w.r.t $v$ from $0$ to $1$ and in the second case from $u-1$ to $1$. I hope this gives enough hints for you to complete the argument.

Answer 2

For $v\in[0,1]$ and $u-v\in[0,1]$ you have $u\in[0,2]$ so inside the integral $\int_0^1\mathbb{1}_{[0,1]}(u-v)dv$ you may multiply by $\mathbb{1}_{[0,2]}(u)$. You can write this indicator as the sum $\mathbb{1}_{[0,1]}(u)+\mathbb{1}_{[1,2]}(u)$ for Lebesgue-a.a. $u$ (only $u=1$ is a problem).