I am trying to find the density $f_Z(u)$ where $X \in U(0,1)$, $Y \in U(0,\alpha)$ and $Z = X + Y$. I want to use the convolution formula $$f_Z(u) = \int_{-\infty}^{\infty} f_X(v) f_Y(u-v) dv. $$
I've been looking online for help, and I have found some answers/suggested solutions, but I don't understand them. For example, it states that in the case where $\alpha < 1$, we have $$\int_{v=0}^u f_X(v)f_Y(u-v)dv. $$ In that same example, they had used $V = X$ as an auxiliary variable, so that the above could also be written as $$\int_{x=0}^u f_X(x)f_Y(u-x)dx. $$
Since $X \in U(0,1)$, we should have $0 < x < 1$, and since we have an intergral $$\int (\cdot) \hspace{1mm} dx, $$ shouldn't the boundaries of the integral simply be $0$ and $1$, no matter whether $\alpha$ is smaller than $1$ or not?
I must be very confused about something regarding convolutions, so I'm hoping that someone can help clarify.
Let us recall the densities of $X$ and $Y$ first: \begin{align} f_X(x)=\mathbf{1}_{(0,1)}(x), \ \ \ \ \ f_Y(y)=\frac{1}{\alpha}\mathbf{1}_{(0,\alpha)}(y) \end{align} where $\mathbf{1}_A(\cdot)$ is the indicator function.
Let $z\in (0,1+\alpha)$. So you seem to know that $f_Z(z)$ is given by: \begin{align} f_Z(z)=\int_\mathbb{R} f_X(t)f_Y(z-t)\,dt \end{align} But both $f_X$ and $f_Y$ are zero most of the time. Where aren't they zero? We check $f_X$ and $f_Y$ one by one. We start with $f_X(t)$; it is nonzero when $0<t<1$. We have $f_Y(z-t)$ is nonzero when $0<z-t<\alpha$. That means $t<z$ and $z-\alpha<t$. We want to satisfy all these inequality at once. So that means $\max\{z-\alpha,0\}<t<\min\{1,z\}$. Hence: \begin{align} f_Z(z)=\int_{\mathbb{R}}f_X(t)f_Y(z-t)\,dt=\int^{\min\{1,z\}}_{\max\{z-\alpha,0\}}\frac{1}{\alpha}\,dt=\frac{\min\{1,z\}- \max\{z-\alpha,0\}}{\alpha} \end{align} To come back to your question, do the bounds should be from $0$ to $1$? No.