Based on Williams' Probability w/ Martingales:
Let $(S, \Sigma, \mu)$ be a measure space.
Dominated Convergence Theorem: Suppose $\{f_n\}_{n \in \mathbb{N}}$, $f$ are $\Sigma$-measurable $\forall n \in \mathbb{N}$
s.t. $\lim_{n \to \infty} f_n(s) = f(s) \forall s \in S$ or a.e. in S
and $\exists g \in \mathscr{L}^1 (S, \Sigma, \mu)$ s.t. $|f_n(s)| \le
g(s) \forall s \in S$ (I guess: or a.e. in S). Then $\lim_{n \to
\infty} \int_S |f_n – f| d\mu = 0$.
The last sentence implies
$$\lim_{n \to \infty} \int_S f_n d\mu = \int_S f d\mu$$
It is apparently false to say that
$$\lim_{n \to \infty} \int_S f_n d\mu = \int_S f d\mu \to \lim_{n \to \infty} \int_S |f_n – f| d\mu = 0$$
Hence, we have things like
Scheffé's Lemma Part (i): Suppose nonnegative $\{f_n\}_{n \in \mathbb{N}}, f \in \mathscr{L}^1 (S, \Sigma, \mu)$ and $\lim_{n \to \infty} f_n(s)
= f(s) \forall s \in S$ or a.e. in S. Then $\lim_{n \to \infty} \int_S |f_n – f| d\mu = 0$ iff $\lim_{n \to \infty} \int_S f_n d\mu = \int_S
f d\mu$
So what is a counterexample saying that $\lim_{n \to \infty} \int_S f_n d\mu = \int_S f d\mu$ implies $\lim_{n \to \infty} \int_S |f_n – f| d\mu = 0$?
Based on Scheffé's Lemma, I'm guessing a counterexample might have something to do with at least one of the functions not being integrable or something.
Re an answer:
$\int_{[-1,1]} |f_n-f|\mathrm d\lambda=2$?
$$\int_{[-1,1]} |f_n-f|\mathrm d\lambda$$
$$ = \int_{[-1,1]} |f_n|\mathrm d\lambda$$
$$ = \int_{[-1,1]} |n1_{(0,1/n)} – n1_{(-1/n,0)}|\mathrm d\lambda$$
$$ \color{red}{=?} \int_{[-1,1]} |n1_{(0,1/n)}| + |n1_{(-1/n,0)}|\mathrm d\lambda$$
$$ = \int_{[-1,1]} n1_{(0,1/n)} \mathrm d\lambda + \int_{[-1,1]} n1_{(-1/n,0)}\mathrm d\lambda$$
$$ = 1 + 1 = 2$$
Best Answer
A potential counter-example with pointwise convergence cannot be a sequence which is dominated by an integrable function.
For example, we can consider the interval $[-1,1]$ endowed with the Borel $\sigma$-algebra and the Lebesgue measure and the functions $f_n:=-n\mathbf 1_{(-1/n,0)}+n\mathbf 1_{(0,1/n)}$, $n\geqslant 1$. Then $f_n(s)\to f(s):=0$ for each $s\in [-1,1]$ and $\int_{[-1,1]} f_n\mathrm d\lambda=0=\int_{[-1,1]} f\mathrm d\lambda$. However, $\int_{[-1,1]} |f_n-f|\mathrm d\lambda=2$.