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? 
 A: 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.
A: 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. 
