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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 $u_n$ is concave for each $n$ (see the Detail below). Hence, by https://math.stackexchange.com/questions/193346/sequence-of-convex-functions, $u_n\to v$ uniformly and hence $f_n\to g$ uniformly, which implies $\lim \int_0^1 f_n=\int_0^1 g$.

Detail: By Taylor's theorem with the integral form of the remainder and the definition of $H$, for $x\in[0,1]$,
\begin{equation} u_n(x)=f_n(x)-H(x)=f_n(0)+f'_n(0)x+\int_0^1(x-t)_+[f''_n(t)-h(t)]\,dt. \end{equation} Now the concavity of $u_n$ follows because $(x-t)_+$ is convex in $x$ and $f''_n<h$.

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 $u_n$ is concave for each $n$ (see the Detail below). So, by Lemma 1 below, $u_n\to v$ in $L^1[0,1]$ and hence $\lim \int_0^1 f_n=\int_0^1 g$.

Detail: By Taylor's theorem with the integral form of the remainder and the definition of $H$, for $x\in[0,1]$,
\begin{equation} u_n(x)=f_n(x)-H(x)=f_n(0)+f'_n(0)x+\int_0^1(x-t)_+[f''_n(t)-h(t)]\,dt. \end{equation} Now the concavity of $u_n$ follows because $(x-t)_+$ is convex in $x$ and $f''_n<h$.

Lemma 1. Suppose that $f_n$ are convex real-valued functions on $[0,1]$ converging pointwise to a real-valued function $f$. Then $f_n\to f$ in $L^1[0,1]$.

Proof. The function $f$ is real-valued and convex and hence bounded from below. So, by Corollary 3, all the functions $f_n$ are uniformly bounded from below. On the other hand, all the convex functions $f_n$ are uniformly bounded from above by $\sup_n(f_n(0)\vee f_n(1))$. So, Lemma 1 follows by dominated convergence.

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 $u_n$ is concave for each $n$ (see the Detail below). Hence, $v$ is concave, and so, $u_n$ and $v$ are continuous and piecewise-monotone. So (see e.g. https://math.stackexchange.com/questions/834126/sequence-of-monotone-functions-converging-to-a-continuous-limit-is-the-converge), $u_n\to v$ uniformly and hence $f_n\to g$ uniformly, which implies $\lim \int_0^1 f_n=\int_0^1 g$.

Detail: By Taylor's theorem with the integral form of the remainder and the definition of $H$, for $x\in[0,1]$,
\begin{equation} u_n(x)=f_n(x)-H(x)=f_n(0)+f'_n(0)x+\int_0^1(x-t)_+[f''_n(t)-h(t)]\,dt. \end{equation} Now the concavity of $u_n$ follows because $(x-t)_+$ is convex in $x$ and $f''_n<h$.

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 $u_n$ is concave for each $n$ (see the Detail below). Hence, by https://math.stackexchange.com/questions/193346/sequence-of-convex-functions, $u_n\to v$ uniformly and hence $f_n\to g$ uniformly, which implies $\lim \int_0^1 f_n=\int_0^1 g$.

Detail: By Taylor's theorem with the integral form of the remainder and the definition of $H$, for $x\in[0,1]$,
\begin{equation} u_n(x)=f_n(x)-H(x)=f_n(0)+f'_n(0)x+\int_0^1(x-t)_+[f''_n(t)-h(t)]\,dt. \end{equation} Now the concavity of $u_n$ follows because $(x-t)_+$ is convex in $x$ and $f''_n<h$.

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 $u_n$ is concave for each $n$ (see the Detail below). Hence, $v$ is concave, and so, $v$ is piecewise-monotone. So (see e.g. https://math.stackexchange.com/questions/834126/sequence-of-monotone-functions-converging-to-a-continuous-limit-is-the-converge), $u_n\to v$ uniformly and hence $f_n\to g$ uniformly, which implies $\lim \int_0^1 f_n=\int_0^1 g$.

Detail: By Taylor's theorem with the integral form of the remainder and the definition of $H$, for $x\in[0,1]$,
\begin{equation} u_n(x)=f_n(x)-H(x)=f_n(0)+f'_n(0)x+\int_0^1(x-t)_+[f''_n(t)-h(t)]\,dt. \end{equation} Now the concavity of $u_n$ follows because $(x-t)_+$ is convex in $x$ and $f''_n<h$.

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 $u_n$ is concave for each $n$ (see the Detail below). Hence, $v$ is concave, and so, $u_n$ and $v$ are continuous and piecewise-monotone. So (see e.g. https://math.stackexchange.com/questions/834126/sequence-of-monotone-functions-converging-to-a-continuous-limit-is-the-converge), $u_n\to v$ uniformly and hence $f_n\to g$ uniformly, which implies $\lim \int_0^1 f_n=\int_0^1 g$.

Detail: By Taylor's theorem with the integral form of the remainder and the definition of $H$, for $x\in[0,1]$,
\begin{equation} u_n(x)=f_n(x)-H(x)=f_n(0)+f'_n(0)x+\int_0^1(x-t)_+[f''_n(t)-h(t)]\,dt. \end{equation} Now the concavity of $u_n$ follows because $(x-t)_+$ is convex in $x$ and $f''_n<h$.