The theorem for Uniqueness of Laplace transform is as below:
Suppose $f$ and $g$ are continuous functions. If $Lf(s)$ = $Lg(s)$ then $f(t)$ = $g(t)$.
I am following the proof given in the notes in the following link:
http://web.mit.edu/jorloff/www/18.03-esg/notes/laplaceuniqueness.pdf
By following the proof in the above notes, I have proved the Lemma for it, which says that
If $h(u)$ is continuous on $[0,1]$ and $\int_{0}^{1} h(u)u^ndu=0$ for $n= 1, 2, ...$ then $h(u) = 0$.
In the proof of the above theorem, I have omitted the proof from the start, it's easy and you can go through it in the link provided above. The last equation of the proof is:
$0 = Lf(s) = \int_{0}^{\infty} f(t)e^{-nt}e^{-s_0t}e^{-t}dt=\int_{0}^{1} u^n(u^{s_0}f(- \ln(u)))du$
As you can see, here the function $h(u)$ is $u^{s_0}f(-\ln(u))$. When Lemma is applied, we straight forward got $u^{s_0}f(- \ln(u)) = 0$ as mentioned in the notes. We also have to prove that this $h(u)$ is continuous on $[0,1]$. i.e; $u^{s_0}f(-\ln(u))$ is continuous on $[0,1]$ and to prove the continuity of this function, we have to prove the continuity of $\ln (u)$ as well in the interval $[0,1]$, whereas we know that $\ln(u)$ is undefined at $0$. How will we get away with this singularity problem at lower bound of interval $[0,1]$. i.e; at $0$.
Your earliest response will highly be appreciated. Thanks in advance.