Solution
= Solution
Take the truncated dyadic approximation
$$
\boxed{g_n(x)=\min\{n,2^{-n}\lfloor2^nf(x)\rfloor\}}.
$$
It has finitely many values and measurable level sets, so is simple. Downward rounding and truncation ensure $0\le g_n\le f$. For a fixed finite $f(x)$, eventually $n>f(x)$, and then $0\le f(x)-g_n(x)<2^{-n}$. Thus $g_n(x)\to f(x)$ everywhere. If extended value infinity is permitted, define $g_n=n$ there, also giving the required limit.