Past exam of the options II course of the University of Oxford / 2005 / D / 4 / ii / Solution

Choose measurable versions of the two Lebesgue integrable functions and consider $F(x,y)=g(x-y)h(y)$. Translation and the corresponding linear change of variables preserve Lebesgue measurability; one can first use Borel versions and then disregard the resulting null-set changes. Tonelli's theorem and translation invariance giveFubini theorem therefore shows that the section in $y$ is integrable for almost every $x$. Define $(g\ast h)(x)$ by that section integral where it exists and by zero on the exceptional null set. Fubini theorem also makes this a measurable Lebesgue integrable function, withIts almost-everywhere defining formula is precisely the desired convolution. This construction explains why an everywhere-existing pointwise integral is not required.