I'm trying to find the convolution $f*g$ where $f=g= \mathbb{1}_{\{ -1≤x≤1 \} }$
**Here's my attempt: $f*g = \int_{\mathbb{R} } \mathbb{1}_{\{ -1≤y≤1 \} }\mathbb{1}_{\{ -1≤x-y≤1 \} }dy$
We therefore have that $x \in [-2, 2]$, therefore
$f*g = \int_{-2}^2dy = 4$
Is that the right approach to the problem?
On
Watch out! The convolution is given by
$$\int 1_{\{-1 \le y \le 1 \}} \ 1_{\{-1 \le x-y \le 1\}} dy,$$ as you wrote. However the function (with variable $y$)
$$1_{\{-1 \le y \le 1 \}} \ 1_{\{-1 \le x-y \le 1\}}$$
is $1$ when $-1 \le y \le 1$ $\textbf{and}$ $x-1 \le y \le x+1$ and then $0$ otherwise. Therefore your convolution is $0$ when $x$ does not belong to $[-2,2]$ and is given by $x+1 - (-1) = x+2$ when $x \in [-2,0]$ and by $1-(x-1)=2-x$ when $x \in [0,2]$.
$$f*g = \int_{\mathbb{R} } \mathbb{1}_{\{ -1≤y≤1 \} }\mathbb{1}_{\{ -1≤x-y≤1 \} }dy=\int_{-1}^11_{\{-1\leq x-y\leq 1\}}dy\\=\int_{-1}^11_{\{x-1\leq y\leq x+1\}}dy$$ which is equal to the length of the interval $(x-1,x+1)\cap(-1,1)$, which you can express as various cases based on the values of $x$.