I am solving the following exercise.
I have followed the below solution for the first part, and it seems okay. Except for why (S2) and (S3) implies that $g_n\to f$ almost everywhere. We should not get $g_n\to f$ almost everywhere just because $f_n\to f$ and $g_n\in\text{conv}\{f_1,f_2,\dots,f_n\}$. Right? What am I missing here?
I am a bit unsure about how things change in the second part. I do not know where to start either. Any help is much appreciated. (The book in question is Brezis.)
Let $x\in \Omega$. Let $\varepsilon\gt 0$ be fixed. Let $n\geqslant 1$: since $g_n(x)\in\operatorname{conv}\left\{f_k(x),k\geqslant n \right\}$, there exists an integer $m=m(x,n)\geqslant n$ and numbers $\lambda_k=\lambda_k(x,n)\in \left[0,1\right]$ such that $\sum_{k=n}^m\lambda_k=1$ and $$ \left\lvert g_n(x)-\sum_{k=n}^m \lambda_kf_k(x)\right\rvert\lt \varepsilon. $$ Since $\sum_{k=n}^m\lambda_kf(x)=f(x)$, it follows that \begin{align} \left\lvert g_n(x)-f(x) \right\rvert &=\left\lvert g_n(x)- \sum_{k=n}^m \lambda_kf_k(x)+\sum_{k=n}^m \lambda_kf_k(x) -\lambda_kf(x) \right\rvert \\ &\leqslant\left\lvert g_n(x)- \sum_{k=n}^m \lambda_kf_k(x) \right\rvert+\left\lvert \sum_{k=n}^m \lambda_kf_k(x) -\lambda_kf(x) \right\rvert\\ &\leqslant\varepsilon+ \sum_{k=n}^m \lambda_k\left\lvert f_k(x) - f(x) \right\rvert\\ &\leqslant \varepsilon+\sup_{k\geqslant n}\left\lvert f_k(x) - f(x) \right\rvert \sum_{k=n}^m \lambda_k. \end{align} We got that for all $x\in\Omega$ and all positive $\varepsilon$ and all integer $n$, $$\left\lvert g_n(x)-f(x) \right\rvert\leqslant\varepsilon+\sup_{k\geqslant n}\left\lvert f_k(x) - f(x) \right\rvert $$ hence letting $\varepsilon\to 0$ gives $$\left\lvert g_n(x)-f(x) \right\rvert\leqslant \sup_{k\geqslant n}\left\lvert f_k(x) - f(x) \right\rvert, $$ which is sufficient to derive the almost sure convergence of $(g_n)$ to $f$.