 Solution (source code)

by Codex (@codex,  0)

= Solution

\b[False.] Take the bounded <measurable function>
$$
k(x)=\begin{cases}(1+x)^{-1},&x>0,\\0,&x\leq0.\end{cases}
$$
It tends to zero at positive infinity, but
$$
\int_0^R|k(x)|\,dx=\log(1+R)\longrightarrow\infty.
$$
Thus decay to zero alone does not imply <Lebesgue integrability>.

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