Let $f_n:\Omega\to[0,\infty)$ be a sequence of functions such that $f_n\to f$ pointwise almost everywhere in $\Omega\subset\mathbb{R}^N$ is bounded. For $k\geq 1$, define $T_k(f_n(x))=\min\{k,f_n(x)\}$ for $x\in\Omega$. Suppose $f_n\to f$ in $L^p(\Omega)$ and $\nabla T_k(f_n)\to \nabla T_k(f)$ strongly in $L^p(\Omega)$, $p>1$ as $n\to\infty$ for every fixed $k\geq 1$. Then upto a subsequence, $\nabla f_n\to \nabla f$ pointwise almost everywhere in $\Omega$.
To prove the above result, let us fix $x_0\in\Omega$. Then it is enough to prove $\nabla f_n(x_0)\to\nabla f(x_0)$ as $n\to\infty$.
Note that $f_n(x_0)\to f(x_0)$ as $n\to\infty$, so $f_n(x_0)\leq k$ for some $k\geq 1$. Therefore, $\nabla f_n(x_0)=\nabla T_k(f_n(x_0))$ which upto a subsequence converges to $\nabla T_k(f(x_0))=\nabla f(x_0)$ by the given condition $\nabla T_k(f_n)\to\nabla T_k(f)$ in $L^p(\Omega)$. Note that $\nabla T_k(f(x_0))=\nabla f(x_0)$, since $f_n(x_0)\leq k$ implies $f(x_0)\leq k$.
So for $x_0\in\Omega$, we obtain $\nabla f_n(x_0)\to\nabla f(x_0)$ upto a subsequence as $n\to\infty$. If we take another point, say $x_1\in\Omega$, then we obtain $\nabla f_n(x_1)\to\nabla f(x_1)$ upto a another subsequence as $n\to\infty$. But this subsequence may be different from the above which we found for $x_0$ and similarly we can find for all other points in $\Omega$. But this does not prove the result.
Can someone please help from here.
Thanks.