This formula is presented without explanation as the CDF for the convolution of any two independent random variables in the textbook I'm reading (introduction to stochastic modeling), and doesn't seem to be a typo as I've seen other answers on here refer to this formula (but none that explain it). I don't understand how it makes sense, given that this would imply $F_Y(\xi)$ can range in value from $-\infty$ to $\infty$, but CDF can only range from 0 to 1 in valuation? Thank you for anyone who can answer
EDIT: after reading comments, I think this is the solution?
By LOTP, \begin{align}F_z(z) &= P(Z \leq z)\\ &= P(X+Y \leq z)\\ &= \int_{-\infty}^\infty P(X+Y\leq z | Y = \xi)\ \mathrm dF_Y(\xi)\\ &= \int_{-\infty}^\infty P(X\leq z-\xi)\ \mathrm dF_y(\xi)\\ &= \int_{-\infty}^\infty F_X(z-\xi)\ \mathrm dF_y(\xi)\\ &= \int_{-\infty}^\infty F_Y(z-\eta)\ \mathrm dF_x(\eta). \end{align}
I am sure this has already been answered elsewhere on this site, but I'll still attach the simple proof. By definition of the CDF for $F_{X+Y}(z)$ we have: \begin{align*} F_{X+Y}(z) &= P \{ X + Y \le z \} \\ &= \iint_{x+y \le z} f_X(x) f_Y(y) \, dx \, dy \\ &= \int_{-\infty}^{\infty} \int_{-\infty}^{z-y} f_X(x) f_Y(y) \, dx \, dy \\ &= \int_{-\infty}^{\infty} \left( \int_{-\infty}^{z-y} f_X(x) \, dx \right) f_Y(y) \, dy \\ &= \int_{-\infty}^{\infty} F_X(z-y) \, dF_Y(y) \\ \end{align*} QED.