Given the following half-maximum convention of the Heaviside function at $1$, $$ F(u) = \begin{cases} \ 1 & \text{if \(u>1\)} \\ \ 1/2 & \text{if \(u=1\)} \\ \ 0 & \text{if \(u<1\)} \\ \end{cases} $$ evaluate the Lebesgue integral: $$ \int_{[0,\infty)} e^{-\lambda u}\, \text{d}F. $$
I somehow get the thought on the usual Heaviside function without the $\frac{1}{2}$ but I just can't wrap-up the idea on how to proceed for the half-maximum version. Any thoughts?
Since $\mathrm d F = 0$ (say in the sense of distributions) on $[2,\infty)$, it is sufficient to look at the integral on $[0,2]$. Let $f(x) = e^{-\lambda x}$. By definition of the Stieltjes integral, one has to compute $$ S_X = \sum_{k=1}^n f(c_k)\left(F(x_{k+1})-F(x_k)\right) $$ for every $n\in\Bbb N$ and every partition $X=(x_k)_{k=1...n}$ with $x_1 = 0$ and $x_n = 2$, and with $c_k\in[x_k,x_{k+1}]$ and make the size of the longest subinterval $\delta_X = \sup_{0\leq k<n} |x_{k+1}-x_k|\to 0$.
If $x_k$ is never $1$, then there is $k_0$ such that $x_{k_0}<1<x_{k_0+1}$, and so $$ S_X = f(c_{k_0}) $$ with $x_{k_0}<1<x_{k_0+1}$ and so $|c_{k_0}-1|\leq \delta_X$.
If there is $k_0$ such that $x_{k_0} = 1$, then $$ S_X = \frac{1}{2}\left(f(c_{k_0})+f(c_{k_0-1})\right) $$ with $x_{k_0-1}\leq c_{k_0-1}\leq 1 \leq c_{k_0}\leq x_{k_0+1}$ and so $|c_{k_0}-1|\leq \delta_X$ and $|c_{k_0-1}-1|\leq \delta_X$.
In all the cases, when $\delta_X\to 0$, $c_{k_0}\to 1$ and $c_{k_0-1}\to 1$, and so since $f$ is continuous, one deduces that $S_X\to f(1)$. Hence $$ \int_0^\infty e^{-\lambda \,x}\,\mathrm d F(x) = e^{-\lambda}. $$ So the $1/2$ does not change anything, one still gets $\mathrm d F = \delta_1$.