= 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> give
$$
\int_{\mathbb R}\int_{\mathbb R}|g(x-y)h(y)|\,dx\,dy
=\int_{\mathbb R}|h(y)|\,\|g\|_1\,dy
=\|g\|_1\|h\|_1<\infty.
$$
<Fubini theorem> therefore shows that the section in $y$ is integrable for <almost every> $x$. Define $(g*h)(x)$ 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>, with
$$
\boxed{\|g*h\|_1\le\|g\|_1\|h\|_1.}
$$
Its almost-everywhere defining formula is precisely the desired <convolution>. This construction explains why an everywhere-existing pointwise <integral> is not required.
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