= Solution
\b[False in both cases.] Without the zero-at-zero condition, $g=0$ and $f\equiv1$ already give an integrable input and a non-integrable composition on $\mathbb R$.
Even imposing $f(0)=0$ does not control how fast $f(t)$ approaches zero. Take $g(x)=(1+x^2)^{-1}$ and the <continuous function> $f(t)=\sqrt{|t|}$. The input has integral $\pi$, but
$$
(f\circ g)(x)=\frac1{\sqrt{1+x^2}}
$$
has a divergent integral, because its tails are comparable to $1/|x|$. Thus <continuity> at zero is weaker than the linear growth estimate that would preserve <integrability>.
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