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This answer shows that the Lebesgue sigma algebra, i.e. the sigma algebra of Lebesgue-measurable subsets of $\mathbb{R}$, is actually the Borel algebra of a topology $T$ on $\mathbb{R}$ defined as follows: $U\in T$ if $U=O-N$, where $O$ is open in the standard topology on $\mathbb{R}$ and $N$ has Lebesgue measure zero.

My question is, what are the properties of this topology $T$? Which of the separation axioms does it satisfy? Which of the countability axioms does it satisfy? And is it metrizable?

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Some observations: this topology is ccc (in $(\mathbb{R}, \mathcal{T})$, every family of pairwise disjoint non-empty open subsets is at most countable), e.g. because of the $\sigma$-finiteness of the Lebesgue measure. But it's not separable (so also not second countable), as every countable set is closed (being measure $0$, or a null set). So it's certainly not metrisable (see e.g. my answer here which implies that ccc implies separable for second metrisable spaces). It is not Lindelöf as the Cantor set is uncountable closed and discrete as a subspace.

It's Hausdorff, being finer than the Euclidean topology, but not normal or regular because you cannot separate $\sqrt{2}$ from the closed set $\mathbb{Q}$.

It's not first countable, by a diagonalisation argument: if $U_n, n \in \omega$ would be a local base at $x$, for each $n$ pick $p_n \in U_n\setminus \{x\}$ and note that no $U_n$ is contained in the open neighbourhood $\mathbb{R}\setminus \{p_n: n \in \omega\}$ of $x$. In fact any convergent sequence in this space is eventually constant, just like we can show for the coarser co-countable topology.

So it's a pretty badly behaved topology in all. I seem to recall it has some standard name, but I cannot recall it right now. I found one paper online that considers it and calls it the "essential topology" on $\mathbb{R}$.

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