 Solution

For each $n$, let $N_n$ be the null set where $f_n\ne g_n$. Their union $N$ is null. At any $x\notin N$, the two sequences agree term by term, so their pointwise limits are equal: $f(x)=g(x)$. Therefore $f\sim g$. Taking the countable union first is essential: it produces one exceptional set outside which all equalities hold simultaneously.