A proof for the Final Value theorem using Dominated convergence theorem I'm going over the proof for the Final Value theorem using the Dominated Convergence theorem on Wikipedia, I don't understand how from the equation
$$sF\left(s\right)=\int_{0}^{\infty}f\left(\frac{t}{s}\right)e^{-t}dt$$
and the fact that $\left|f\left(\frac{t}{s}\right)e^{-t}\right|$ is dominated by $Me^{-t}$ they get to the limit
$$\lim_{s\searrow0}sF\left(s\right)=\int_0^{\infty}{\alpha}e^{-t}dt=\alpha$$
 A: $\newcommand{\dd}{\mathrm{d}}\newcommand{\R}{\mathbb{R}}\newcommand{\C}{\mathbb{C}}\newcommand{\lap}{\mathscr{L}}$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:

*

*$f$ is continuously differentiable and both $f$ and $f'$ have a Laplace transform

*$f'$ is absolutely integrable, that is, $\int_0^\infty|f'(\tau)|\dd\tau$
is finite,

*$\lim_{t\to\infty}f(t)$ 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:

*

*$f$ is bounded,

*$\lim_{t\to\infty}f(t)$ 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. $\Box$
