What is a counterexample to the converse of this corollary related to the Dominated Convergence Theorem?

Question

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.

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Re an answer:

$\int_{[-1,1]} |f_n-f|\mathrm d\lambda=2$?

$$\int_{[-1,1]} |f_n-f|\mathrm d\lambda$$

$$ = \int_{[-1,1]} |f_n|\mathrm d\lambda$$

$$ = \int_{[-1,1]} |n1_{(0,1/n)} - n1_{(-1/n,0)}|\mathrm d\lambda$$

$$ \color{red}{=?} \int_{[-1,1]} |n1_{(0,1/n)}| + |n1_{(-1/n,0)}|\mathrm d\lambda$$

$$ = \int_{[-1,1]} n1_{(0,1/n)} \mathrm d\lambda + \int_{[-1,1]} n1_{(-1/n,0)}\mathrm d\lambda$$

$$ = 1 + 1 = 2$$

Answer 1

Yes, the Scheffe's lemma does not hold if $f \notin \mathscr{L}^1 (S, \Sigma, \mu)$.

Counterexemple: Let $S=(0,1]$, $\Sigma$ be the Borel $\sigma$-algebra of $(0,1]$ and $\mu$ be the Lebesgue measure. Let $f$ be defined as $f(x)={1\over x}$. For each $n\in \mathbb{N}$, let $f_n =\chi_{[{1 \over n},1]} f$.

Then, for all $n\in \mathbb{N}$, $f_n$ is nonnegative and $f_n \in \mathscr{L}^1 (S, \Sigma, \mu)$, $\lim_{n \to \infty} f_n(s) = f(s) \forall s \in S$ and $\lim_{n \to \infty} \int_S f_n d\mu = +\infty = \int_S f d\mu$, BUT, for all $n\in\mathbb{N}$, $\int_S |f_n - f| d\mu =+\infty$.

Answer 2

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, $\int_{[-1,1]} |f_n-f|\mathrm d\lambda=2$.