On the proof of The Bounded Convergence Theorem

If all that you assume is that $(f_n)_{n\in\Bbb N}$ converges pointwise to $f$, then all you know is that $\lim_{n\to\infty}f_n(x)=f(x)$ for each individual $x\in E$. But you cannot deduce from this that, given $\varepsilon>0$, you have $\bigl|f(x)-f_n(x)\bigr|$ for every sufficiently large $N$ and for all $x\in E$.

For instance, take $E=[0,1]$. For each $x\in E$ and each $n\in\Bbb N$, defined$$f_n(x)=\begin{cases}n^2x&\text{ if }x<\frac1{2n}\\-n^2\left(x-\frac1n\right)&\text{ if }x\in\left[\frac1{2n},\frac1n\right]\\0&\text{ if }x>\frac1n,\end{cases}$$Then $(f_n)_{n\in\Bbb N}$ converges pointwise to the null function, but, for each $n\in\Bbb N$, $\int_0^1f_n(x)\,\mathrm dx=\frac14$.

If the $(f_n)$ are uniformly bounded and converge pointwise to $f$, the condition $$ |f(x)-f_n(x)|<\epsilon/m(E)\qquad\forall x\in E\ \forall n\ge N\tag1$$ may never be satisfied. As a counterexample, on $E:=[0,1]$ consider $f_n$ to be a rectangle of height $1$ over the interval $(0,\frac1n)$. Then for any $\epsilon>0$ there is always a set $A$ of positive measure where $|f-f_n|>\epsilon$ on $A$. So the proof doesn't go through using (1). (The result that $\int f_n\to\int f$ is still true, though, by dominated convergence.)