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The answer is yes. Indeed, let $u_n:=f_n-H$ and $v:=g-H$, where \begin{equation} H(x):=\int_0^x(x-t)h(t)\,dt \end{equation} for $x\in[0,1]$. Then $u_n\to v$ pointwise and $H''=h$ almost everywhere, so that $u_n''\le0$ almost everywhere. So, $u_n$ is concave and hence $v$ is concave. So, $u_n\to v$ uniformly and hence $f_n\to g$ uniformly, which implies $\lim \int_0^1 f_n=\int_0^1 g$.