The bounded convergence theorem says that if $\{f_n\}$ is such that $f_n$ are measurable, bounded by $M$ for all $n$, supported on a set $E$ of finite measure and $f_n(x)\to f(x)$ a.e. as $n\to\infty $ then $f$ is measurable, supported on $E$ for a.e. x ans $$\int|f_n-f|\longrightarrow 0$$
Is the fact that $m(E)<\infty $ important ? For exemple, if $f_n(x)=e^{-nx^2}$ then $f_n$ are bounded, measurable, supported on all $\mathbb R$ (and thus on a set of measure $\infty $) and we have that $$\lim_{n\to\infty }\int_{-\infty }^\infty f_n(x)\mathrm dx=0=\int_{-\infty }^\infty \lim_{n\to\infty }f_n(x)\mathrm dx,$$
so may be $m(E)<\infty $ is not important... What do you think ?
If $E$ is not of finite measure, $f_n, f$ might not be integrable and so the conclusion does not make sense. (For example, $f = \chi_{[-\infty,n]}$ converges pointwisely to $f = 1$).
Even if $f_n, f$ are integrable, there are still counterexample: $f_n = \chi_{[n,n+1]}$ converges to the zero function, but $$\int |f_n - f| = 1$$ for all $n\in \mathbb N$.