Let $F(s):[0,T]\rightarrow[0,T]$ be the Laplace transform of a probability distribution $f(t)$ on the real line with finite support. This means that $$F(s)=\int_0^{L}e^{-st}f(t)dt$$ for all $s$, while $$\int_0^{L}f(t)dt=1$$ and $f(t)\geq0$ for all $t\in[0,L]$. The numbers $T$ and $L$ are positive real constants. I found that the inverse problem: obtain $f(t)$ from $F(s)$, is ill-posed. Specifically, we can make a sequence of Laplace transforms $F_n(s)$ converge uniformly to some limiting $F(s)$, while the sequence of PDFs $f_n(t)$ does not converge pointwise to $f(t)$. I'll give an example at the end of the question.

However, it seems to be different for the cumulative distribution function. Let the CDF be defined by
$$C(t):= \int_0^t f(t')dt'.$$
If the Laplace transforms $F_n(s)$ of $f_n(t)$ converge uniformly to the Laplace transform $F(s)$ of $f(t)$, does it follow that the CDFs $C_n(t)$ converge pointwise to $C(t)$?



  [1]: https://mathoverflow.net/questions/265832/upper-bounds-on-the-inverse-laplace-transform