# Is there such splitting?

Does there exist a set $A \subset [0,1]$ s. t. for every polynomial $P(x)$ of degree less than or equal to $4$ the relation between the Riemann integrals $$\int_A P(x)\,dx= \int_{[0,1]\setminus A}P(x)\,dx$$ holds?

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Here is a "Cantor-like" construction that works for any degree $n\geq 4$. Define a partition $(A_n,B_n)$ of $[1,2^{n+1}]$ by induction as follows : $(A_0,B_0)=(\lbrace 1\rbrace,\lbrace 2 \rbrace)$ and for $n\geq 2$, $A_n=B_{n-1}\cup (A_{n-1}+2^{n+1}), B_n=A_{n-1}\cup (B_{n-1}+2^{n+1})$.

Lemma. If $g$ is a polynomial of degree $\leq n$ in $x$, then $\sum_{k\in A_n}g(x+k)=\sum_{k\in B_n}g(x+k)$.

Proof of lemma : Use induction on $n$. When $n=0$, the lemma states that $g(x+1)=g(x+2)$ when $g$ is constant, and this is certainly true. Suppose that the lemma holds for $n-1$ where $n\geq 1$. Let $g$ have degree $\leq n$. Define $h$ by $h(x)=g(x+2^{n+1})-g(x)$. Then $h$ has degree $\leq n-1$, so by the induction hypothesis we have $(1)\sum_{k\in A_{n-1}}h(x+k)=\sum_{k\in B_{n-1}}h(x+k)$. Now, since $h(x+k)=g(x+2^{n+1}+k)-g(x+2^{n+1})$ the LHS in (1) above is equal to $\sum_{k\in A_{n-1}+2^{n+1}}g(x+k)-\sum_{k\in A_{n-1}}g(x+k)$, while the RHS in (1) above is equal to $\sum_{k\in B_{n-1}+2^{n+1}}g(x+k)-\sum_{k\in B_{n-1}}g(x+k)$. It is clear then that (1) shows that the lemma is true for $n$. This concludes the proof by induction.

Returning to the initial problem, if $f$ is a polynomial of degree $\leq n$, then $F(k)=\int_{\frac{k-1}{2^{n+1}}}^{\frac{k}{2^{n+1}}} f(x)dx$ is a polynomial of degree $\leq n$ in $k$. It follows then from the lemma that the following $A$ works :

$$A=\bigcup_{k\in A_n} [\frac{k-1}{2^{n+1}},\frac{k}{2^{n+1}}]$$

When $n=4$, one has

$$\begin{array}{lcl} A &=& \bigcup_{k\in \lbrace 1; 4; 6; 7; 10; 11; 13; 16; 18; 19; 21; 24; 25; 28; 30; 31 \rbrace}\big[\frac{k-1}{32},\frac{k}{32}\big] \\ [0,1] \setminus A &=& \bigcup_{k\in \lbrace 2; 3; 5; 8; 9; 12; 14; 15; 17; 20; 22; 23; 26; 27; 29; 32\rbrace}\big[\frac{k-1}{32},\frac{k}{32}\big] \end{array}$$

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Very interesting. What motivates that choice of partition, or put another way, why might lead one to something like the lemma? –  Travis Aug 27 '14 at 17:01
@ Ewan Delanoy: Can you state your answer in detail, especially Lemma? +1 at the moment. –  user64494 Aug 27 '14 at 17:02
@user64494 I'm adding some details. –  Ewan Delanoy Aug 27 '14 at 17:06
@ Ewan Delanoy : It would be interesting to compare your answer with the answer by Travis in the case $n=4$. –  user64494 Aug 27 '14 at 17:18
@Travis regular subdivisions are the most natural subdivisions. The power of $2$ in the denominator is quite natural as you divide by $2$ every time you take one further "regularity condition" into account. –  Ewan Delanoy Aug 27 '14 at 17:32

Yes.

First, since integration is linear over $\mathbb{R}$, it is enough to produce a set for which the equality holds for all elements of a basis of the vector space of polynomials $P(x)$ of degree at most four, say, for $(1, x, x^2, x^3, x^4)$. Then, the condition is a system of five equations, so to try to find a set $A$, it's natural to pick a class of choices parameterized by five variables and impose those relations.

The simplest choices of set are unions of finite intervals, which we may as well take to be nonoverlapping, so we choose the ansatz

$A := [0, a_1] \cup [a_2, a_3] \cup [a_4, a_5]$, with $0 \leq a_1 \leq \cdots \leq a_5 \leq 1$;

its compliment is

$A^c := (a_1, a_2) \cup (a_3, a_4) \cup (a_5, 1]$.

For each basis element $x^k$, $k \in \{0, 1, 2, 3, 4\}$, evaluating both sides of the integral condition gives

$\tfrac{1}{k + 1} (a_1^{k + 1} + a_3^{k + 1} - a_2^{k + 1} + a_5^{k + 1} - a_4^{k + 1}) = \tfrac{1}{k + 1} (a_2^{k + 1} - a_1^{k + 1} + a_4^{k + 1} - a_3^{k + 1} + 1 - a_5^{k + 1})$,

and rearranging gives

$a_1^{k + 1} - a_2^{k + 1} + a_3^{k + 1} - a_4^{k + 1} + a_5^{k + 1} = \tfrac{1}{2}$.

Substituting the system of five equations so produced into a CAS gives the unique (and pleasingly symmetrical) solution

$A := [0, \frac{1}{2} - \frac{1}{4}\sqrt{3}] \cup [\frac{1}{4}, \frac{1}{2}] \cup [\frac{3}{4}, \frac{1}{2} + \frac{1}{4}\sqrt{3}]$.

A priori it's conceivable that a partition of $[0, 1]$ into fewer intervals might do the trick, but (at least according to the CAS) there are no solutions for the corresponding systems.

At least for other small $N$, the same technique produces sets $A$ satisfying the integral condition for all polynomials of degree at most $N$ by partitioning $[0, 1]$ into $N + 1$ intervals.

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@ Travis: +1. Your approach is efficient for small $N$. How about an arbitrary degree $N$? I will be waiting for the answer in the general case during some time. –  user64494 Aug 27 '14 at 16:47