# Up to a null set, is the countable intersection of open sets $O_1,O_2,\dots\subset\mathbb R$ equal to a countable union of disjoint closed intervals?

Suppose $$\{O_i\}_{i=1}^\infty$$ is any sequence of open sets in the reals $$\mathbb R$$. Is is true that their intersection $$\bigcap_{i=1}^\infty O_i$$ is, up to a null set, equal to some at-most countable union of disjoint closed intervals?

Modulo edge cases like $$O_i=\emptyset$$ and $$O_i=\mathbb R$$, I think the answer is yes. In trying to prove this, I looked at the sets $$A_n = \bigcap_{i=1}^n O_i$$ and drew them for a random collection $$\{O_i\}_{i=1}^\infty$$. It seems like as $$n$$ increases, the sets $$A_n$$ form "islands", so to speak, of decreasing open intervals. To be precise, there seems to be a collection of decreasing sequences of open sets $$(B_i^1)_{i\in\mathbb N}, (B_i^2)_{i\in\mathbb N},(B_i^3)_{i\in\mathbb N},\dots$$ such that $$A_n=\bigcup_{k=1}^\omega B^k_n$$, where the collection is at-most countable and the union is disjoint for each $$n$$.

If this holds, the result follows immediately because each decreasing sequence $$(B_i^k)_{i\in\mathbb N}$$ must converge to the empty set, a singleton, or some interval $$(a,b)$$, $$(a,b]$$, $$[a,b)$$, $$[a,b]$$; we can take the null set to be the boundary points of these limiting intervals. So, does this actually hold?

Any help is greatly appreciated.

• I presume you meant “closed interval” as in non-degenerate ones? Otherwise every set is a union of disjoint singletons. Commented May 4 at 5:33
• If we are talking about non-degenerate intervals, then this is not true. For example, a fat Cantor set is a $G_\delta$ set, but it almost contains no nonempty open set, so it almost contains no non-degenerate closed interval. Commented May 4 at 5:37
• @DavidGao yes, and I also wanted the union to be at-most countable. I'll edit my post to clarify this. Commented May 4 at 5:38
• Well, in any case my comment did provide a counterexample. I’ll write a full answer. Commented May 4 at 5:42
• True, the set of irrational numbers is also $G_\delta$. But it is up to null set equal to $\mathbb{R}$, which I’d consider to be a closed interval. Commented May 4 at 5:51

Let $$A \subset \mathbb{R}$$ be a fat Cantor set. In particular, $$m(A) > 0$$, $$A$$ is closed and nowhere dense. Now, any closed subset of $$\mathbb{R}$$ is a countable intersection of open sets:
$$A = \bigcap_n \{x: d(x, A) < \frac{1}{n}\}$$
But if $$O \subset \mathbb{R}$$ is an open set that is almost contained in $$A$$, then $$O = \varnothing$$. Indeed, $$O \setminus A$$ must be of measure zero. But it is an open set, so $$O \setminus A = \varnothing$$, i.e., $$O \subset A$$. But $$A$$ has no interior, so $$O = \varnothing$$. In particular, $$A$$ contains no non-degenerate closed interval, even up to null sets. Since $$m(A) > 0$$, this means $$A$$ cannot be a countable union of closed intervals, even up to null sets.