Axiomatic Integration: Translation Invariance

First, let $\int$ be a linear translation invariant integral, and $[a,b]$ an 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.

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 concatenating 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

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

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

But this last line is now ready to apply subdivision in reverse: the two pieces are the integrals of the single function $\chi_\QQ(x-\eta)$ over $[0,\eta]$ and $[\eta,1]$, so that function is integrable on all of $[0,1]$ and the sum is its integral there. Thus we’ve proven the claim:

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

Since $\chi_\QQ\geq 0$ we know by nonnegativity that $\int_{[0,1]}\chi_\QQ$ is nonnegative. We wish to prove it’s 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 $\{\xi_1,\ldots, \xi_{N-1}\}$ whose pairwise differences are also irrational (for specificity, take the square roots of the first $N-1$ primes). This is exactly the condition making the $N$ sets below pairwise disjoint: $\QQ,\,\QQ+\xi_1,\, \QQ+\xi_2,\,\ldots,\, \QQ+\xi_{N-1}$

Let $C$ be the set of points of $[0,1]$ in none of them; then we can decompose the unit interval as $[0,1]= \big(\QQ\sqcup (\QQ+\xi_1)\sqcup \cdots \sqcup (\QQ+\xi_{N-1})\big)\cap[0,1]\;\sqcup\; C$

This decomposition descends to a decomposition of the constant function $1$ on $[0,1]$ 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 $\chi_{\QQ+\xi_i}(x)=\chi_\QQ(x-\xi_i)$, so we can rewrite this as

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

Each of the translated terms is integrable on $[0,1]$ by the previous lemma. Solving for $\chi_C$, we see it’s 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 is an irrational translation of $\chi_\QQ$ by construction, and so they all 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 $a=0$.

By normalization $1$ is integrable on $[0,1]$. Translating by any integer $k$, translation invariance gives that $1$ is integrable on $[k,k+1]$ for every $k\in\ZZ$.

Now apply subdivision in reverse: $1$ integrable on $[k,k+1]$ and on $[k+1,k+2]$ implies it is integrable on $[k,k+2]$. Inducting on the number of steps, $1$ is integrable on $[m,n]$ for any pair of integers $m<n$.

Finally let $[a,b]$ be an arbitrary interval, and choose integers $m<a$ and $n>b$. Since $[a,b]\subset[m,n]$, two applications of subdivision (splitting off $[m,a]$ and then $[b,n]$) give that $1$ is integrable on $[a,b]$.

Let $I=[a,b]$. If $a=b$ this is the degenerate case, and $\int_{\{a\}}1=0=|I|$ by the integral-over-a-point argument of the first post, which used only subdivision.

So assume $a<b$ and set $x=b-a>0$. Translation invariance applied to the constant function $1$ on the interval $[0,x]$, translating by $c=a$, gives $\int_{[0,x]}1=\int_{[a,a+x]}1=\int_{[a,b]}1$ and by hypothesis the left side is $x=b-a=|I|$.

First take $x=1/n$. Subdivide the unit interval at the points $0,\tfrac1n,\tfrac2n,\ldots,\tfrac{n-1}{n},1$. Applying the subdivision axiom $n-1$ times, $\int_{[0,1]}1=\sum_{k=0}^{n-1}\int_{[k/n,\,(k+1)/n]}1$ Every interval in that sum has length $1/n$, so by translation invariance every term equals $\int_{[0,1/n]}1$. The left hand side is $1$ by normalization, so $1 = n\int_{[0,1/n]}1\implies \int_{[0,1/n]}1=\frac1n$

Now take $x=m/n$ for positive integers $m,n$. Subdivide $[0,m/n]$ at the points $k/n$ for $k=0,\ldots,m$. As before each of the $m$ pieces has length $1/n$, so translation invariance and the case above give each one the value $1/n$: $\int_{[0,m/n]}1=\sum_{k=0}^{m-1}\int_{[k/n,\,(k+1)/n]}1=m\cdot\frac1n=\frac{m}{n}$

For monotonicity, let $0\leq x< y$. Both integrals exist by the integrability lemma, and subdivision at $x$ gives $F(y)=\int_{[0,y]}1=\int_{[0,x]}1+\int_{[x,y]}1=F(x)+\int_{[x,y]}1$ The constant function $1$ is nonnegative, so the nonnegativity axiom forces $\int_{[x,y]}1\geq 0$, and hence $F(y)\geq F(x)$.

Now fix any $x>0$ and let $p,q$ be rationals with $0<p<x<q$. Monotonicity gives $F(p)\leq F(x)\leq F(q)$, and the rational case evaluates the outer two: $p\leq F(x)\leq q$ This holds for every such pair of rationals. Taking the supremum over rationals $p<x$ gives $x\leq F(x)$, and taking the infimum over rationals $q>x$ gives $F(x)\leq x$. Density of $\QQ$ in $\RR$ makes both of these sharp, so $F(x)=x$.