A proof for the Final Value theorem using Dominated convergence theorem

$\newcommand{\dd}{\mathrm{d}}\newcommand{\R}{\mathbb{R}}\newcommand{\C}{\mathbb{C}}\newcommand{\lap}{\mathscr{L}}$I went through the proof again, and I think that you're right. The variables are intertwined and the proof is not rigorous. Here's the correct statement:

Final value theorem. Suppose that the following conditions are satisfied:

1. $f$ is continuously differentiable and both $f$ and $f'$ have a Laplace transform
2. $f'$ is absolutely integrable, that is, $\int_0^\infty|f'(\tau)|\dd\tau$ is finite,
3. $\lim_{t\to\infty}f(t)$ exists and is finite,

Then,

\begin{equation} \lim_{t\to \infty}f(t) = \lim_{s\to 0^+}sF(s). \end{equation}

Proof. We know that the Laplace transform of the derivative is

\begin{equation} sF(s) - f(0^+) {}={} \lap\{f'(t)\}(s) {}={} \int_{0^+}^\infty f'(\tau) e^{-s\tau}\dd\tau. \end{equation} Therefore, \begin{align} \lim_{s\to 0^+} sF(s) {}={}& f(0^+) + \lim_{s{}\to{}0^+} \int_{0^+}^{\infty}f'(\tau)e^{-s\tau}\dd\tau \notag \end{align}

We want to use the dominated convergence theorem ; Define $\phi_s(\tau) = f'(\tau)e^{-s\tau}$. We have $|\phi_s(\tau)|\leq f'(\tau)$ which is assumed to be absolutely integrable (Assumption 2). Therefore,

\begin{align} \lim_{s\to 0^+} sF(s) {}={}&f(0^+) + \int_{0^+}^{\infty} \lim_{s{}\to{}0^+} f'(\tau)e^{-s\tau}\dd\tau \notag \\ {}={}&f(0^+) + \int_{0^+}^{\infty} f'(\tau)\dd\tau \notag \\ {}={}& {f(0^+)} + \lim_{t\to\infty}f(t) - {f(0^+)}, \\ {}={}& \lim_{t\to\infty} f(t). \end{align} This completes the proof. $\Box$

Comment on the Wikipedia article (8 May, 2019): it is stated that with a change of variables, $\xi = st$, (in fact, the same symbol is used for $t$ and $\xi$), the integral becomes

$$ \int_0^\infty \lim_{s\to 0^+}f(\xi/s) e^{-\xi}\dd\xi. $$

As the OP noted, there is some confusion in that proof; in fact, $\xi$ and $s$ are not independent variables (by definition), so the limit $\lim_{s\to 0^+}f(\xi/s)$ is not equal to $\lim_{t\to\infty}f(t)$.

Update: Following @LaurentLessard's comment, here is a different proof that requires $f$ to be bounded

Final value theorem #2. Let $f$ be a function in the time domain with Laplace transform $F(s)=(\mathcal{L} f)(s)$. Suppose that the following conditions are satisfied:

1. $f$ is bounded,
2. $\lim_{t\to\infty}f(t)$ exists and is finite,

then, $$ \lim_{t\to \infty}f(t) = \lim_{s\to 0^+}sF(s). $$

Proof. We observe that \begin{align} sF(s) = \int_0^{\infty} f\left(\frac{y}{s}\right)e^{-y}\dd{}y. \end{align} Define the function $f_s(y) = f\left(\frac{y}{s}\right)e^{-y}$. Since $f$ is assumed to be bounded, there is an $M>0$ such that $|f(t)| \leq M$ for all $t\geq 0$, therefore, $|f_s(y)| \leq Me^{-y}$. As a result, we can apply the dominated convergence theorem: \begin{align} \lim_{s\to0^+}sF(s) {}={} & \lim_{s\to0^+}\int_0^{\infty} f\left(\frac{y}{s}\right)e^{-y}\dd{}y \notag \\ \overset{\text{dct}}{=}{} & \int_0^{\infty} \lim_{s\to0^+}f\left(\frac{y}{s}\right)e^{-y}\dd{}y {}={} \lim_{t\to\infty}f(t). \end{align} This completes the proof. $\Box$