My textbook about probability goes slightly too quick over this, I am not following an example given in the textbook.
The theory of the combined probably $Z = X + Y$ is:
$$f_Z(z) = \int_{-\infty}^\infty f_X(z-y)f_Y(y) \, \mathrm{d}y$$
This seems quite straightforward and I understand it. Howver directly after it an example is given, for two exponential distributions $$\mathrm{Exp}(\lambda) \sim f(x)= \lambda e^{-\lambda x}$$ For $ x \geq 0$ otherwise $f(x) = 0$.
Now given $T_1 = T_2 = \mathrm{Exp(\lambda)}$ the following statements are made:
$$\begin{align} f_Z(z) = f_{T_1 + T_2}(z) &= \int_{-\infty}^\infty f_{T_1}(z-y)f_{T_2}(y) \, \mathrm{d}y \\ &= \int_{0}^z \lambda e^{-\lambda(z-y)} \cdot \lambda e^{-\lambda y} \, \mathrm{d}y \end{align}$$
Now I don't understand above line, or more specific I don't understand where the upper limit $z$ comes from. The lower limit is a consequence of the exponential distribution definition. But what about the upper limit? Is it always the "running variable" for all distributions, or is it unique to the exponential distribution - and why?
As you noted, the lower limit is $0$ because the density function of the exponential distribution is only defined for $x \geq 0$. In fact this is the same reason why the upper limit is $z$. Notice that one term in the integrand is $f_{T_1}(z-y)$, and since $f_{T_1}$ is the density function of an exponential random variable, $f_{T_1}(z-y)$ is only defined for $z-y \geq 0$, in other words $y \leq z$.