The problem is not properly stated in the original, here is the correct version: If $f_n\leq g_n \leq h_n$, $f_n \to f$, $h_n \to h$ pointwise and $\int f_n\to\int f$,$\int h_n \to \int h$ then $\int g_n \to \int g$
Really do not know to how to tackle this problem. it is easy to show that $g \in L^1$. But not sure how to show convergence of integrals. The only thing I see is that $|g_n|\leq \max\{-f_n,h_n\}$ but that is not very helpful. I tried applying generalized dominated convergence theorem but could not succeed. Not sure what else to try. I was wondering if perhaps these conditions somehow imply that all of them converge in $L^1$ Any hints or solutions would be appreciated.
$0 \le g_n - f_n \le h_n -f_n$, so $g_n-f_n$ is dominated by the integrable $h_n-f_n$. Hence $g-f$ is integrable and $\int(g_n-f_n) \to \int (g-f)$. Adding $f_n, f$ to each side gives the desired result.
Alternative:
To original intent here was to avoid the generalised DCT but it just ended up being a marginal variation of said theorem.
We have $\int \varliminf (g_n -f_n) \le \int \varliminf (h_n -f_n) \le \varliminf \int (h_n -f_n) = \int (h-f)$. Since $\varliminf (g_n-f_n) = g-f$ ae. we see that $g \in L^1$.
Note that $\int \varliminf (g_n -f_n) \le \varliminf \int (g_n-f_n) = \varliminf (\int g_n - \int f)$ which gives $\int g \le \varliminf \int g_n$. Similarly, $\int \varliminf (h_n -g_n) \le \varliminf \int (h_n-g_n) = \varliminf (\int h - \int g_n)$ which gives $\varlimsup \int g_n \le \int g$.