For a counter-example to reverse Fatou lemma without the domination hypothesis, take $f_n:=\chi_{(n,n+1)}$, with $X$ the real line, Borel $\sigma$-algebra and Lebesgue measure. We have $\limsup_{n\to +\infty}f_n(x)=0$ for all $x$ but $\int f_nd\mu=1$.
The question has been edited. The updated question now asks to prove:
Suppose $\{f_n\}_{n \in \mathbb{N}}, f \in \mathscr{L}^1 (S, \Sigma, \mu)$ and $\lim_{n \to \infty} f_n(s) = f(s)$ a.e. in $S$. Then $\lim_{n \to \infty} \int_S |f_n - f| d\mu = 0$ iff $\lim_{n \to \infty} \int_S |f_n| d\mu = \int_S |f| d\mu$
Proof: (=>) It is trivial, since, from Minkowski's inequality, we have
$$\left | \int_S|f_n|d\mu-\int_S|f|d\mu \right |\leqslant \int_S|\,|f_n| -|f|\,| d\mu\leqslant \int_S|f_n -f| d\mu$$
(<=) Note that $|f_n -f|\leqslant |f_n| +|f|$. So, for each $n$, the function $|f_n| +|f| - |f_n -f|$ is non-negative and using Fatou's Lemma, we have
\begin{align}
2 \int_S|f|d\mu
&=\int_S \lim\inf(|f_n| +|f| - |f_n -f|)d\mu \leqslant \lim\inf \int_S (|f_n| +|f| - |f_n -f|)d\mu = \\
&=\lim\inf \left (\int_S|f_n|d\mu +\int_S|f|d\mu - \int_S|f_n -f|d\mu \right) = \\
&= \left(\lim\inf\int_S|f_n|d\mu\right) +\int_S|f|d\mu - \left(\lim\sup\int_S|f_n -f|d\mu\right) = \\
&=2\int_S|f|d\mu - \left(\lim\sup\int_S|f_n -f|d\mu\right)
\end{align}
So we have
$$2 \int_S|f|d\mu \leqslant 2\int_S|f|d\mu - \left(\lim\sup\int_S|f_n -f|d\mu\right) $$
Since $f \in \mathscr{L}^1 (S, \Sigma, \mu)$ , we know that $\int_S|f|d\mu<+\infty$, and so we get
$$\lim\sup\int_S|f_n -f|d\mu \leqslant 0$$
So we can conclude that
$$\lim\int_S|f_n -f|d\mu = 0$$
Remark: There is another way to prove the (<=) part, which uses the Dominated Convergence Theorem (instead of Fatou's Lemma). However such way (for the question as currently stated) is a little bit "trickier" than the one presented above using Fatou's Lema. Here it is:
(<=) Consider $|f_n| \wedge |f|$ defined by $(| f_n | \wedge |f|)(x)=\min\{|f_n(x)|,|f(x)|\}$, for each $x \in \Omega$. Consider also
$$ \sigma(f_n,f)(x) = \left \{\begin{aligned} &= -1 &\textrm{ if } f_n(x)f(x)<0 \\
&= 0 &\textrm{ if } f_n(x)f(x)=0 \\ &= 1 &\textrm{ if } f_n(x)f(x)>0 \end{aligned}\right.$$ for each $x \in \Omega$.
Since $\{f_n\}$ converges to $f $ a.e.,
we have that $\{|f_n| \wedge |f|\}$ converges to $|f|$ a.e., and $\{\sigma(f_n,f)\}$ converges to $\chi_{[f\neq 0]}$ a.e.. So, $\{\sigma(f_n,f)(|f_n| \wedge |f|)\}$ converges to $|f|$ a.e.. But we know that, for all $n$, $\vert \sigma(f_n,f)(|f_n| \wedge |f|) \vert = |f_n| \wedge |f| \leqslant |f| $ and $\int_{\Omega } |f| d\mu< \infty $. So we can apply Lebesgue Dominated Convergence Theorem and we have that
$$\lim_{n \to \infty}\int_{\Omega } \sigma(f_n,f)(|f_n| \wedge |f|) d\mu = \int_{\Omega } |f| d\mu$$
To conclude the proof, note that
$$\vert f_n-f\vert = |f_n|+|f|-2\sigma(f_n,f)(|f_n| \wedge |f|)$$
So
$$\int_{\Omega } \vert f_n-f\vert d\mu = \int_{\Omega } |f_n| d\mu +\int_{\Omega } |f| d\mu -2\int_{\Omega } \sigma(f_n,f)(|f_n| \wedge |f|) d\mu $$
And so, since $\lim_{n \to \infty}\int_{\Omega } |f_n| d\mu = \int_{\Omega } |f| d\mu$ and $\int_{\Omega } |f| d\mu<+\infty$, we have
$$ \lim_{n \to \infty}\int_{\Omega } \vert f_n-f\vert d\mu =0$$
Best Answer
You cannot use an specific example to show that a general statement is true. For this you need a "proof".
But you can use an specific example to show that a general statement is not true.
For instance, the example $f(x)=|x|$ shows that the statement "All continuous functions are differentiable" is not true. But the fact that $105=3\cdot5\cdot7$ does not prove that any $n\in\mathbb{N}$ can be decomposed into the product of primes.