= Solution
\b[True.] Replace the dominating representative by $|g|$ if necessary, and take the countable union of the exceptional <Lebesgue null sets> so that every bound holds off a single null set. Since the interval has <Lebesgue measure> one, <Cauchy-Schwarz inequality> gives
$$
\int_0^1|g|^2\,dx\leq\left(\int_0^1|g|^4\,dx\right)^{1/2}\left(\int_0^1 1\,dx\right)^{1/2}<\infty.
$$
Thus $g^2$ is a <Lebesgue integrable function>. The squared functions satisfy $|f_n|^2\to0$ pointwise and $|f_n|^2\leq|g|^2$ <almost everywhere>. The <dominated convergence theorem> therefore yields
$$
\boxed{\|f_n\|_2^2=\int_0^1|f_n|^2\,dx\longrightarrow0.}
$$
This uses the <Lp inclusion on a finite measure space> to turn the assumed fourth-power bound into the second-power domination actually needed.
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