This question is somewhat related to this answer by @copper.hat about a similar question.
Let $X,Y$ be two independent, uniformly distributed random variables over $[0,1]$, i.e $X,Y\sim \mathcal U[0,1]$.
We want to evaluate the density function of the sum of $Z=X+Y$. I'm already familiar with the interval decomposition solution. I went for the convolution here. Denote $\mu$ as the Lebesgue measure on $\mathbb R$.
We have
\begin{align*} f_Z(z)&=\int_{-\infty}^{+\infty}f_X(z-y)f_Y(y)dy\\ &=\int_{0}^1 f_X(z-y)dy\hspace{1cm}\text{ substitute }t=z-y\\ &=\int_{z-1}^{z}\textbf{1}_{[0,1]}(t)dt\\ &=\mu\big([z-1,z]\cap[0,1]\big) \end{align*} According to what I understood, we can use the property $$\mu(A\cup B)=\mu(A)+\mu(B)-\mu(A\cap B)$$ Since both $\mu([z-1,z])=\mu([0,1])=1$, we're left with evaluating $\mu([z-1,z]\cup[0,1])$.
Notice that if $0\leq z\leq 1$, then $\mu([z-1,z]\cup[0,1])=|1-(z-1)|=|z|$, and if $1\leq z \leq 2$, we have $\mu([z-1,z]\cup[0,1])=|z-0|=|z|$.
So finally, we get that $$\mu([z-1,z]\cap[0,1])=\mu([z-1,z])+\mu([0,1])-\mu([z-1,z]\cup[0,1])=(2-|z|)$$
I don't know whether I'm correct. I must end up with a density function that resembles the one from the interval decomposition solution. It's missing the indicator function, plus it should verify that the overall integral equals 1
Any help or guidance is highly appreciated.
Turns out there was a mistake in writing the union :
The problem with my attempt is that I didn't notice that if for example $z=0$, then $\mu([z-1,z]\cup[0,1])=\mu([-1,0]\cup[0,1])=2\neq |z|=0$.
Finally
$$\mu([z-1,z]\cap[0,1])=(2-(2-|z|))\textbf{1}_{[0,1]}(z)+(2-|z|)\textbf{1}_{[1,2]}(z)$$ $$=(|z|)\textbf{1}_{[0,1]}(z)+(2-|z|)\textbf{1}_{[1,2]}(z)=z\textbf{1}_{[0,1]}(z)+(2-z)\textbf{1}_{[1,2]}(z)$$
which is identical to the interval decomposition solution.