Let $X$ be a metrizable space.
The Lebesgue-Hausdorff theorem states that the minimal class $\mathcal{C}$ of functions $f :X \to \mathbb{R}$ closed under pointwise limits of sequences, such that $C(X) \subset \mathcal{C}$, is exactly Borel measurable functions (denote them by $\mathcal{B}(X,\mathbb{R})$).
I wandering iff it is OK to state this theorem as this: $\overline{C(X)} = \mathcal{B}(X,\mathbb{R})$ in $\mathbb{R}^X$ with pointwise convergence topology, or equivalently the product topology.
The problem with this approach is that the product topology of $\mathbb{R}^X$ may not be a sequential topology, as property of being sequential is not preserved under any products in general. However, sets closed under pointwise sequence limits, are clearly can be used as closed sets for some topology. But every $\mathbb{R}^n$ is a sequential space as they are metrizable. So, may this topology correspond to normal pointwise convergence topology in case of $\mathbb{R}^X$, even taking in account that it is not metrizable? I failed to find any counter-example to this conjecture.
This question also can be posed like this: Are Borel measurable functions closed in pointwise topology? I have already proved that they are sequentially closed. But that's all.
Best Answer
It's quite clear that already the continuous functions are dense in $\Bbb R^X$ (for a Tychonoff space $X$) so the set of measurable functions too is dense. A set can be sequentially closed and still dense that's how non-sequential large products are... so $\overline{C(X)} = \overline{\mathcal{B}(X)} =\Bbb R^X$.