= Solution
For a <Lebesgue measurable function> $f:\mathbb R\to\mathbb R$, the <preimage> $f^{-1}(U)$ of every open <set> $U$ is a <Lebesgue measurable set>. For a <Borel measurable function>, each such preimage must instead be a <Borel set>. Equivalently, both definitions can test all Borel target sets: the collection
$$
\mathcal C_f=\{B\subseteq\mathbb R:f^{-1}(B)\text{ belongs to the source sigma-algebra}\}
$$
is a <sigma-algebra>, since preimages preserve complements and countable unions. Containing all open sets therefore makes it contain the <Borel sigma-algebra>.
Now let $U$ be open. Since $h$ is a <Borel measurable function>, $h^{-1}(U)$ is Borel. Since $g$ is a <Lebesgue measurable function>, $g^{-1}(h^{-1}(U))$ is a Lebesgue measurable set. But
$$
(h\circ g)^{-1}(U)=g^{-1}(h^{-1}(U)).
$$
Thus \b[$h\circ g$ is a <Lebesgue measurable function>]. The intermediate sigma-algebra is Borel; this is why <Borel postcomposition preserves Lebesgue measurability>.
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