Given the function $F(t)=2-2e^{-t}$ defining a measure on $(\mathbb{R}_+,\mathfrak{B}(\mathbb{R}_+))$ and I want to calculate the convolution of this function with itself. I tried to do that by using the derivative of F that is given by $f(t)=2e^{-t}$ respectively using Radon Nikodym (as far as I am concerned we should have $F \ll \lambda$ where $\lambda$ is the Lebesgue measure).
My calculation is then: $$F^{*2}(r)=F*F(r) =\int\limits_{0}^{\infty} 1_{\{s\leq r\}} \, \mathrm{d}F(s) \, \mathrm{d} F(t)=\int\limits_{0}^{\infty} F(r-t) \, \mathrm{d} F(t) = \int\limits_{0}^{\infty} 2-2e^{-(r-t)}\,\mathrm{d}F(t) \\ = \int\limits_{0}^{\infty} 2e^{-t}(2-2e^{-(r-t)})\,\mathrm{d}t=\int\limits_0^{\infty} 4e^{-t}\,\mathrm{d}t - 4e^{-r}\int\limits_{0}^{\infty} e^t e^{-t} \,\mathrm{d} t = \int\limits_0^{\infty} 4e^{-t}\,\mathrm{d}t - 4e^{-r}\int\limits_{0}^{\infty} 1 \,\mathrm{d} t. $$
My observation here is that the function "$1$" is not integrable (at least not on the interval $(0,\infty)$), so that I cannot continue. What am I missing or where am I doing something I am not supposed to do?