Past exam of the integration course of the University of Oxford / 2023 / 3 / b / ii / Solution

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 givesThus $g^2$ is a Lebesgue integrable function. The squared functions satisfy $|f_n{|}^2\to 0$ pointwise and $|f_n{|}^2\le |g{|}^2$ almost everywhere. The dominated convergence theorem therefore yieldsThis uses the Lp inclusion on a finite measure space to turn the assumed fourth-power bound into the second-power domination actually needed.