Question: Does the Bounded Convergence Theorem hold if $\operatorname{m}\left(E\right)<+\infty$ but we drop the assumption that the sequence $\{f_n\}$ is uniformly bounded on $E$ but still pointwise bounded?
I know that the answer is no, however, I cannot think of a counterexample. I though about the sequence of functions $f_n:[0,1]\to\mathbb{R}$ defined by:
$$f_n(x):=n\cdot\chi_{(0,1/n[}.$$
But this sequence is not pointwise bounded.
For reference, this is a modification of a question out of Roydens Real Analysis text. Moreover, the question out of Roydens Real Analysis text is the following:
Question: Does the Bounded Convergence Theorem hold if $\operatorname{m}\left(E\right)<+\infty$ but we drop the assumption that the sequence $\{f_n\}$ is uniformly bounded on $E$?
Obviously, the sequence of functions $\{f_n\}$ defined above suffices as a counterexample for this question.
Best Answer
Define $(f_{n})_{n \in \mathbb{N}}$ in $(0,1]$ with Lebesgue measure as follows: \begin{align*} f_{n}(x) = x^{-1} \chi_{[n^{-2},n^{-1}]}(x) \end{align*} We have $\lim_{n \to \infty} f_{n} = 0$ pointwise in $(0,1]$. Further, $f_{n}(x) \leq x^{-1}$ so we are pointwise bounded. However, $\lim_{n \to \infty} \int_{0}^{1} f_{n}(x) \, dx \neq 0$ as \begin{equation*} \int_{0}^{1} f_{n}(x) \, dx = \int_{n^{-2}}^{n^{-1}} x^{-1} \, dx = \log(n^{-1}) - \log(n^{-2}) = \log(n). \end{equation*}