The Utility of Translation Invariance in Axiomatic Integration

Imposing this forces the integral of the rationals' characteristic function to be zero

This post reports an interesting result related to the question of whether an axiomatic framework is sufficient to fully determine the value of integrals. Here we work with the axioms of a linear integral proposed in a previous post:

Normalization: The function $1$ is integrable on any interval $I$ , and returns the length of the interval: $\int_I 1 = |I|$
Nonnegativity: If $f$ is integrable on $I$ and $f\geq 0$ then $\int_I f\geq 0$ .
Subdivision: If $[a,b]$ is an interval and $c\in[a,b]$ , then $f$ is integrable on $[a,b]$ if and only if its integrable on $[a,c]$ and $[c,b]$ . Furthermore, in this case $\int_{[a,b]}f = \int_{[a,c]}f +\int_{[c,b]}f$
Linearity: If $f,g$ are integrable on $I$ then so is $cf+kg$ for any $c,k\in\mathbb{R}$ and $\int_{I}(cf+kg)=c\int_I f+k\int_I g$

The question in the background here is: for all integrals which do satisfy these axioms, if they can integrate $\chi_{\mathbb{Q}}$ do they agree on its value? This note does not solve this problem, but does provide an additional condition one can assume that makes the result true: translation invariance.

Translation Invariance

An integral is said to be translation invariant if it has the following property: for any $f(x)$ which is integrable on $[a,b]$ , then $f(x-c)$ is integrable on $[a+c,b+c]$ and $\int_{[a,b]}f(x)=\int_{[a+c,b+c]}f(x-c)$

The main theorem of this post is as follows:

Integrating the Characteristic of the Rationals

Let $\int^1$ and $\int^2$ be any two translation invariant integrals, which both agree that $\chi_{\QQ}$ is integrable on some interval $[a,b]$ . Then

$\int^1_{[a,b]}\chi_{\mathbb{Q}}=\int^2_{[a,b]}\chi_\QQ$

This is the first time we’ve been able to prove such a result for a function with a set of discontinuities having positive measure. Unfortunately the proof techique is rather tailor-made to this situation, and its not clear the idea generalizes.

Steps of the Proof

First, we can make use of the fact that we know a translation invariant integral for which $\chi_{\QQ}$ is integrable: the Lebesgue integral. Since the rationals have measure zero, this function differs from the zero function on a set of measure zero, and so its Lebesgue integral agrees with that of the zero function:

$\int_{[a,b]}^{\mathrm{Leb.}}\chi_{\mathbb{Q}}=\int_{[a,b]}^{\mathrm{Leb.}}0=0$

Thus, to show any two integrals agree on $\chi_{\QQ}$ is the same as showing that any translation invariant integral must evaluate to zero.

In the two lemmas below, we reduce the general case to studying just the integral of $\chi_\QQ$ on the unit interval. The following observation proves useful in several of the arguments: since the rational numbers are closed under addition, translating the characteristic function of $\QQ$ by an element of $\QQ$ leaves the function unchanged:

$r\in\QQ\implies \chi_\QQ(x)=\chi_\QQ(x-r)\hspace{1cm}\forall x\in\mathbb{R}$

Reduction to the Unit Interval, I

If $\chi_{\QQ}$ is integrable on $[a,b]$ if and only if it is integrable on $[0,1]$ .

First, let $\int$ be a linear translation invariant integral, and $[a,b]$ and interval for which it can integrate $\chi_\QQ$ . Choose some rational $r$ such that $0\in (a+r,b+r)$ and note by translation invariance that $\chi_\QQ(x-r)$ is now integrable on this interval. But this is simply $\chi_\QQ$ itself, as $r$ is rational. Choosing $N$ such that $1/N < b+r$ gives an interval $[0,1/N]\subset [a+r,b+r]$ and subdivision implies that $\chi_\QQ$ is integrable on $[0,1/N]$ .

Iteratively applying translation invariance by $1/N$ (and the fact that $\chi_{\QQ}(x)=\chi_{\QQ}(x-k/N)$ for any $k$ ) we see that $\chi_\QQ$ is integrable on each of $[k/N,(k+1)/N]$ . From here, iteratively applying subdivision (in reverse) yields the integrability of $\chi_\QQ$ on $[0,1]$ .

Second, we assume that $\chi_\QQ$ is integrable on $[0,1]$ , and let $[a,b]$ be an arbitrary interval. Choose two integers $m,n$ with $m < a$ and $n>b$ . By translation invariance, for any $k\in\ZZ$ we have $\chi_{\QQ}(x-k)$ is integrable on $[k,k+1]$ . and since $k\in\ZZ$ this means $\chi_\QQ$ is itself integrable on this interval. Repeatedly applying the subdivision axiom for the finitely many adjacent intervals $[m,m+1]\cup[m+1,m+2]\cup\cdots\cup [n-2,n-1]\cup[n-1,n]$ shows that $\chi_\QQ$ is integrable on $[m,n]$ . But as $[a,b]\subset [m,n]$ by construction, applying subdivision once more yields that $\chi_\QQ$ is integrable on $[a,b]$ as claimed.

Next we show that knowing the value of $\int_{[0,1]}\chi_\QQ$ suffices.

Reduction to Unit Interval II

If $\int_{[0,1]}\chi_{\QQ}=0$ then $\int_{[a,b]}\chi_{\QQ}=0$ for any interval $[a,b]$ .

Assume that $\int_{[0,1]}\chi_{\QQ}=0$ , and let $[a,b]$ be any interval. Following the procedure in the above proof, we can produce an interval $[m,n]$ containing $[a,b]$ from concateinating finitely many translated copies of $[0,1]$ . Using the second part of the translation invariance property, we know the integral of $\chi_\QQ$ on each of these translated intervals is the same as its value on $[0,1]$ : zero. Summing them all up shows $\chi_\QQ$ integrates to $0$ on $[m,n]$ . Then recalling $[a,b]\subset[m,n]$ and applying subdivision gives $0=\int_{[m,n]}\chi_\QQ =\int_{[m,a]}\chi_{\QQ}+\int_{[a,b]}\chi_\QQ+\int_{[b,n]}\chi_\QQ$

Finally, using the axiom on nonnegativity, since $\chi_\QQ\geq 0$ for all $x$ , we know that all three terms on the right hand side are nonnegative. But they sum to zero! So, in fact all are zero, including the middle one

$\int_{[a,b]}\chi_{\QQ}=0$

To simplify the work involved in calculating the integral over the unit interval, we first prove a lemma about irrational translations of $\chi_\QQ$ : even though such a translation does affect the values of the resulting function, it does not affect the integral

Irrational Translations

Let $\zeta$ be an irrational number. Then translation of $\chi_\QQ$ by $\zeta$ does not affect its integral over the unit interval: $\int_{[0,1]}\chi_\QQ(x)=\int_{[0,1]}\chi_{\QQ}(x-\zeta)$

Let $\zeta$ be irrational, and write $\zeta = n+\xi$ for $\xi\in (0,1)$ . Then $\chi_\QQ(x-\zeta)=\chi_\QQ(x-\xi-n)=\chi_\QQ(x-\xi)$ So it suffices to consider the case of an irrational in $(0,1)$ . For such a $\chi$ note that by translation invariance we know $\int_{[0,1]}\chi_\QQ(x)=\int_{[\xi,1+\xi]}\chi_\QQ(x-\xi)$ This second integral can be subdivided at $1\in[\xi,1+\xi]$ , $\int_{[\xi,1+\xi]}\chi_\QQ(x-\xi)=\int_{[\xi,1]}\chi_\QQ(x-\xi)+\int_{[1,1+\xi]}\chi_\QQ(x-\xi)$ We can then apply translation invariance by $-1$ to the second integral in this sum: $\int_{[1,1+\xi]}\chi_\QQ(x-\xi)=\int_{[0,\xi]}\chi_\QQ(x-\xi +1)$ But since $1$ is rational we have $\chi_\QQ(x-\xi +1)=\chi_\QQ(x-\xi)$ Substituting this back in to the original equality,

\begin{align} \int_{[0,1]}\chi_\QQ(x)&=\int_{[\xi,1+\xi]}\chi_\QQ(x-\xi)\\ &=\int_{[\xi,1]}\chi_\QQ(x-\xi)+\int_{[1,1+\xi]}\chi_\QQ(x-\xi)\\ &= \int_{[\xi,1]}\chi_\QQ(x-\xi)+\int_{[0,\xi]}\chi_\QQ(x-\xi) \end{align}

But this last line is now ready to apply subdivision in reverse: its just the integral of $\chi_\QQ(x-\xi)$ over all of $[0,1]$ ! Thus we’ve proven the claim:

$\int_{[0,1]}\chi_\QQ(x)=\int_{[0,1]}\chi_\QQ(x-\xi)$

Now we are ready to put this to work.

The Unit Interval Case

Let $\int$ be any translation invariant integral (satisfying the original axioms + linearity), and $\chi_\QQ$ the characteristic function of the rationals. Then if $\chi_\QQ$ is integrable on $[0,1]$ , $\int_{[0,1]}\chi_\QQ=0$

Since $\chi_\QQ\geq 0$ we know by nonnegativity that $\int_{[0,1]}\chi_\QQ$ is nonnegative. We wish to prove its precisely zero, so assume for the sake of contradiction that the integral is some positive value $\int_{[0,1]}\chi_\QQ=a$

Let $N$ be some positive integer such that $Na>1$ , and choose a collection of $N-1$ irrational numbers $\{\zeta_1,\ldots, \zeta_{N-1}\}$ which are linearly independent over $\QQ$ (there are uncountably many such to choose from, but for specificity, take the square roots of the first $N-1$ primes). Because each of these are $\QQ$ linearly independent, we see the $N$ sets below are pairwise disjoint $\QQ,\,\QQ+\xi_1,\, \QQ+\xi_2,\,\ldots,\, \QQ+\xi_{N-1}$

Let $C$ be the complement of their union; then we can decompose the unit interval as $[0,1]= \QQ\sqcup (\QQ+\xi_1)\sqcup (\QQ+\xi_2)\sqcup \cdots \sqcup (\QQ+\xi_{N-1})\sqcup C$

This decomposition descends to a decomposition of $1$ into a sum of characteristic functions

$1=\chi_\QQ+\chi_{\QQ+\xi_1}+\chi_{\QQ+\xi_2}+\cdots+\chi_{\QQ+\xi_{N-1}}+\chi_{C}$

For each $i$ , we see $\chi_{\QQ+\xi_1}(x)=\chi_\QQ(x-\xi_i)$ so we can rewrite this as

$1=\chi_\QQ(x)+\chi_{\QQ}(x-\xi_1)+\chi_{\QQ}(x-\xi_2)+\cdots+\chi_{\QQ}(x-\xi_{N-1})+\chi_{C}(x)$

Solving this for $\chi_C$ , we see that this is a linear combination of functions we know to be integrable, and so is integrable by the linearity axiom. Using the second part of that axiom, we expand to get

$\int_{[0,1]}1=\int_{[0,1]}\chi_\QQ(x)+\int_{[0,1]}\chi_\QQ(x-\xi_1)+\cdots +\int_{[0,1]}\chi_\QQ(x-\xi_{N-1})+\int_{[0,1]}\chi_C$

Now by our lemma, each of the first $N$ integrals here are irrational translations of $\chi_\QQ$ by construction, and so they integrate to the same value $a$ . Applying the normalization axiom to the left hand side, this gives

$1=Na +\int_{[0,1]}\chi_C$

But since $\chi_C$ is a characteristic function it only takes the values $0,1$ and so the nonnegativity axiom implies its integral is $\geq 0$ . But this is a contradiction! We specifically chose $N$ so that $Na>1$ . Thus our assumption that $a$ is positive must be wrong, and the only option is that $a=0$ .