 Solution

False in both cases. Without the zero-at-zero condition, $g=0$ and $f\equiv 1$ 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 $\pi$, buthas 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.