Let $f:\mathbb{R}\to\mathbb{R}$ be given (possibly with some conditions to be added later?).
Prove the following statement:
$$ \lim_{x\to\infty}\frac{1}{x}\int_0^xdy\int_0^y\,dz\,f(z)=\lim_{s\to0}\hat{f}(s)$$ where $\hat{f}\equiv\int_0^\infty e^{-\cdot x}f(x)dx$ is the Laplace transform of $f$.
I tried to use the final value theorem, which states that if the LHS side of the following equality exists, then the equality is true: $$ \lim_{x\to\infty}g(x)=\lim_{s\to0}s\hat{g}(s) \,.$$
Applying this to $g(x):=\frac{1}{x}\int_0^xdy\int_0^y\,dz\,f(z)$ does not get me very far however.
It's certainly true if $\int_0^\infty|f(t)|\,dt<\infty$. This seems too easy to be interesting; hard to say whether it could be of any use to the OP, but since he asked:
In that case Dominated Convergence shows that $$\lim_{s\to0}\hat f(s)=\int_0^\infty f(t)\,dt: =I.$$If we say $$A(y)=\int_0^y f(z)\,dz$$then $A$ is continuous on $[0,\infty)$ and $\lim_{y\to\infty}A(y)=I$, which makes it easy to show that $$\lim_{x\to\infty}\frac 1x\int_0^x A(y)\,dy=I.$$