# intuition of decomposition of $\mathbb{R}$ into disjoint union of first category and null set

Let the length of an interval $I$ be $|I|.$

A subset $B\subset \mathbb{R}$ is called a null set if for any $\varepsilon>0,$ there exists a sequence of open intervals $(I_n)_{n\in\mathbb{N}}$ such that $B\subseteq\bigcup_{n\in\mathbb{N}}I_n$ and $\sum_{n\in\mathbb{N}}|I_n|<\varepsilon.$

A subset $A\subseteq\mathbb{R}$ is of first category if $A=\bigcup_{n\in\mathbb{N}}A_n$ where $A_n$ is nowhere dense set.

Oxtoby stated the following theorem:

Theorem: The real line $\mathbb{R}$ can be decomposed into two disjoint union of first category set $A$ and null set $B$.

Idea of Proof: Let $\mathbb{Q}=\{p_n:n\in\mathbb{N}\}.$ Fix $i,j\in\mathbb{N}.$ Let $I_{i,j}$ be intervals containing $p_n$ of length $1/2^{i+j}.$ Then $A=\bigcap_{j\in\mathbb{N}}\bigcup_{i\in\mathbb{N}}I_{i,j}$ is a null set while $B = \bigcup_{j\in\mathbb{N}}\bigcap_{i\in\mathbb{N}}I_{i,j}^c$ is of first category.

Question: What is an intuition behind the construction of $A$ and $B?$ When I tried to prove the theorem on my own, I would not know that $A$ and $B$ are constructed as above.

Any help is appreciated.

EDIT: According to Dave's comment, the decomposition is due to Lebesgue outer measure being $G_{\delta}.$ However, I think that there should be a more elementary way to construct the decomposition.

In particular, given that $\mathbb{R}$ can be decomposed into disjoint union of null set and set of first category, how do we construct them solely from their definitions?

• Possibly one way to understand this is from the fact that Lebesgue outer measure is a $G_{\delta}$ regular outer measure and the (topological) separability of the reals. See Proof #1 at the top of p. 13 here – Dave L. Renfro Oct 29 '17 at 11:54
• What does $p_n$ stand for? Is this an enumeration of the rationals? – Gerry Myerson Oct 31 '17 at 3:12
• @GerryMyerson: Yes, you are right. $p_n$'s are enumeration of $\mathbb{Q}.$ – Idonknow Oct 31 '17 at 5:23

This construction is actually quite natural. The idea is that you want to construct a set $A$ which is large in the sense of category but small in the sense of measure. As a start, you might try constructing a dense open set $A_\epsilon$ with small measure (say, measure at most something like $\epsilon$). How can you do this? Well, to make $A_\epsilon$ dense, you just need it to contain the rationals. Since the rationals are countable, you can get an open set containing them with small measure: just take an interval of length $\epsilon/2^n$ around $p_n$, and the total measure will be at most $\sum_n \epsilon/2^n=2\epsilon$.

Now this $A_\epsilon$ is not quite a null set. To get a null set, you can take an intersection of such sets $A_\epsilon$ as $\epsilon$ goes to $0$. That's exactly what your $A$ is: $\bigcup_{i\in\mathbb{N}}I_{i,j}$ is just $A_\epsilon$ for $\epsilon=1/2^j$, and $A$ is their intersection. Since it's a countable intersection, its complement $B$ still has first category.

• This is precisely what I am looking for. Thanks. – Idonknow Oct 31 '17 at 5:34
• By the way, is a set of first category small? It is a countable union of 'small sets'. – Idonknow Oct 31 '17 at 5:41
• Well, "first category" is one notion of "smallness". But as this example shows, a set of first category can be "large" in other ways such as measure. – Eric Wofsey Oct 31 '17 at 5:53
• Why do we start with constructing a dense open set? – Idonknow Oct 31 '17 at 10:16
• Because sets that are "large" in the sense of category (i.e., their complement is first category) are built out of dense open sets (specifically, they are sets which contain a countable intersection of dense open sets). – Eric Wofsey Oct 31 '17 at 17:05

Here's another construction, from Gelbaum & Olmsted, Counterexamples in Analysis, Chapter 8, Example 19. It presumes that you know about Cantor sets of positive measure.

For $n=1,2,\dots$, let $A_n$ be a Cantor set in $[0,1]$ of measure $(n-1)/n$, and let $A=A_1\cup A_2\cup\cdots$. Since each $A_n$ is nowhere dense, $A$ is of first category. But also for each $n$, the measure of $A$ is at least that of $A_n$, which is $(n-1)/n$, so the measure of $A$ is 1, so the measure of its complement in $[0,1]$ is zero.

Now copy what we've done in $[0,1]$ to each interval $[m,m+1]$ for integer $m$, and take the union of the first category sets thus generated.