3 Introduction
Nearly three months later, I am finally getting around to writing up some more posts in this series. In that time I have continued figuring out some neat things about these periodic functions I have been calling “pathological”, and have also gotten busy with an unrelated project in 3D topology that I’ll have to write about soon.
In the last post I said I would give some concrete examples of incommensurable funcitons with periodic sum. I will still do so, but first let me actually reframe things in terms of the periodic decomposition problem I referenced in the first post of the series. Honestly I should have done this from the beginning. I will be borrowing extensively from Revesz for this initial exposition.
4 The Periodic Decomposition Problem.
The periodic decomposition problem begins with the observation that if a function \(f: \mathbb{R}\to \mathbb{R}\) can be written as \[f = f_1+\ldots +f_n, \tag{4.1}\] where each \(f_i\) is periodic and has a period \(p_i\), then we must have \[\Delta_{p_1}\ldots \Delta_{p_n} f = 0. \tag{4.2}\] Thus, Equation 4.2 is a necessary condition for \(f\) to be decomposable in this manner. The decomposition problem then consists of determining when the implication can be reversed. This is a nice generalization of the fact that a function is periodic with period \(p\) if and only if \(\Delta_p f = 0\)
Notice that the identity \(Id(x) = x\) always satisfies Equation 4.2 no matter what \(p_i\) we chose. But if the \(p_i\) are all commensurable, the intersection \(\bigcap \mathbb{Z}p_i\) contains some nonzero number \(p\). This means it cannot be the case that \(Id(x)\) has a decomposition (1), for in that case the sum on the left hand side of (1) would be periodic in \(p\), and \(Id\) has no periods other than \(0\). However, For a system of \(n\) pairwise incommensurable real numbers \(p_1,\ldots, p_n\), we do have the following theorem assuming the axiom of choice which I will refer to as the Periodic Decomposition Theorem (PDT).
Theorem: Let \(p_1,\ldots, p_n\), be pairwise incommensurable. Then, for any function \(f:\mathbb{R}\to\mathbb{R}\), there exists a decomposition (1) for \(f\) if and only if Equation 4.2 holds.
I’ll sketch the proof here of the PDT for special case of \(n=2\), and briefly describe the general case. Suppose we have a function \(h:\mathbb{R}\to\mathbb{R}\), a pair \(p, q\) of incommensurable numbers, and \(\Delta_p\Delta_q h = 0\). Let \(G = \mathbb{Z}p+\mathbb{Z}q\), and consider the quotient \(\mathbb{R}/G\). For each coset in this quotient, choose a single representative (this is where we invoke AC). On a given coset with \(y\) as its representative, any \(x\) in the can be written uniquely as \(x = y + mp + nq\) for some integers \(m, n\). So, on that coset, define \(f(x) := h(y+ nq)\) and \(g(x) := h(y+mp)-h(y)\). Now do this on every coset in order to define \(f\) and \(g\) on all of \(\mathbb R\).
Then \(\Delta_p\Delta_q h = 0\) gives that \(h = f + g\); this follows from the fact that \[\Delta_p\Delta_q h(x) = h(x+p+q) - h(x+p) - h(x+q) + h(x).\] Setting \(x = y + (m-1)p + (n-1)q\), this gives \(h(y + mp + nq) = h(y + mp) + h(y + nq) - h(y)\).
The proof is basically the same for arbitrary \(n\). The key is to expand out \(\Delta_{p_1}\ldots \Delta_{p_n} f\) in terms of translates of \(f\) by sums of the \(p_i\). Writing \(T_p f(x) := f(x+p)\), we have \(\Delta_p f = (T_p - Id)f\), so \[\Delta_{p_1}\ldots \Delta_{p_n} f = (T_{p_1} - Id)\ldots (T_{p_n} - Id)f.\] Expanding out the right-hand side, you notice that other than the \(T_{p_1 + \ldots + p_n}f\) term, all others are periodic in some \(p_i\) (e.g. \(T_{p_1 + p_2}f\) is periodic in \(p_3, \ldots, p_n\)). So group together terms that share periods in order to get a decomposition of the form Equation 4.1.
5
This theorem in fact holds in any model of ZF in which there exists a Vitali Set, meaning a set containing exactly one element from each equivalence class in \(\mathbb{R}/\mathbb{Q}\). This is the quintessential example of a nonmeasurable set. AC of course implies the existence of such a set, but the other way around is not true (this is not obvious; see for example Blaas). Moreover, there are models of ZF where all subsets of the real line are measurable, meaning a Vitali set cannot exist in such a model. In other words, Like AC, the assertion that Vitali sets exist is independent of ZF, but it is strictly weaker than the full Axiom of Choice itself.
Lets begin with an elementary construction that requires no Axiom of Choice. The pointwise behavior of this elementary example will be highly irregular, though the almost-everywhere behavior will be highly boring: the functions in this example will be almost everywhere constant.
To find examples that are not equivalent in measure to a constant function, we will need to exit the world of measurable functions entirely, and thus begin our foray into some fascinating set-theoretic territory.
Before we begin, a quick reminder of notation from previous posts: \(P_f\) denotes the period module of \(f\), which is always a \(\mathbb{Z}\)-submodule of \(\mathbb{R}\). Two periodic functions \(f, g\) are incommensurable when \(P_f \cap P_g = \{0\}\). We write \(\mathbb{Z}_{p,q} := p\mathbb{Z} + q\mathbb{Z}\) for the \(\mathbb{Z}\)-span of \(\{p, q\}\), and similarly \(\mathbb{Z}_{p,q,r}\) for three generators. Throughout, \(p, q\) (and sometimes \(r\)) will be real numbers chosen so that \(p/q \notin \mathbb{Q}\) (and \(p, q, r\) are \(\mathbb{Q}\)-linearly independent when three generators are involved).
6 The Foundational Example: \(E_1\)
We begin with the simplest possible example answering Question 1 from Part I: a pair of incommensurable periodic functions whose sum is periodic.
Construction \(E_1\). Fix \(p, q \in \mathbb{R}\) with \(p/q \notin \mathbb{Q}\), and let \(G := \mathbb{Z}_{p,q} = p\mathbb{Z} + q\mathbb{Z}\). Define \(f, g: \mathbb{R} \to \mathbb{R}\) by \[ f(mp + nq) := n, \qquad g(mp + nq) := -m, \qquad \text{for } mp+nq \in G, \] and \(f(x) = g(x) := 0\) for \(x \notin G\).
Note that since \(p/q \notin \mathbb{Q}\), every element of \(G\) has a unique representation as \(mp + nq\) with \(m, n \in \mathbb{Z}\), so \(f\) and \(g\) are well-defined.
Let us verify the basic properties of this construction.
Periodicity of \(f\) and \(g\). Take any \(x = mp + nq \in G\). Then \(f(x + p) = f((m+1)p + nq) = n = f(x)\), so \(p \in P_f\). For \(x \notin G\), we have \(x + p \notin G\) as well (since \(p \in G\)), and so \(f(x+p) = 0 = f(x)\). Thus \(p \in P_f\). By the same reasoning \(q \in P_g\).
No other periods of \(f\) in \(G\). Suppose \(t = m'p + n'q \in G\) is a period of \(f\). Then for \(x = 0\): \(f(t) = n' = f(0) = 0\), so \(n' = 0\). Hence \(P_f \cap G \subseteq p\mathbb{Z}\). Combined with the previous paragraph, \(P_f \cap G = p\mathbb{Z}\).
No periods of \(f\) outside \(G\). Suppose \(t \notin G\) is a period of \(f\). Then \(t + G \neq G\), so \(t \notin G\) means in particular that \(0 + t = t \notin G\), hence \(f(t) = 0 = f(0)\). So far consistent. But take \(x = q \in G\): \(f(q) = 1\), while \(q + t \notin G\) (because \(t \notin G\) and \(q \in G\) means their sum is in the coset \(t + G \neq G\)), so \(f(q + t) = 0 \neq 1\). So \(t \notin G\) is not a period of \(f\).
Combining: \(P_f = p\mathbb{Z}\) exactly. By the same argument, \(P_g = q\mathbb{Z}\). So \(f\) is strictly periodic with fundamental period \(p\), \(g\) is strictly periodic with fundamental period \(q\), and they are incommensurable since \(P_f \cap P_g = p\mathbb{Z} \cap q\mathbb{Z} = \{0\}\).
Periodicity of the sum. Now consider \(h := f + g\). On \(G\): \(h(mp + nq) = n - m\). Take \(t = p + q\): then \(h(x + p + q) = h((m+1)p + (n+1)q) = (n+1) - (m+1) = n - m = h(x)\). ✓ Outside \(G\): \(h \equiv 0\), trivially \((p+q)\)-periodic. So \(p + q \in P_h\).
Verifying that \(P_h = (p+q)\mathbb{Z}\) exactly is straightforward and left to the reader. The key takeaway: \(h\) is strictly periodic with fundamental period \(p + q\), and so \(E_1\) provides our first example of incommensurable periodic functions whose sum is periodic. Settles Question 1 from Part I in the affirmative.
6.1 The properties of \(E_1\)
\(\mathbb{Q}\)-dependence the fundamental periods \(p, q, p+q\) are pairwise incommensurable (pairwise irrational ratios), yet they form a \(\mathbb{Z}\)-linearly dependent set, since \(p + q - (p+q) = 0\). In other words, \(P_{h}\subset P_{f}+P_g\). This has the additional consequence that \(\Delta_q^2 f = \Delta_p^2 g = 0\), and naturally \(\Delta_q^2 h = \Delta_q^2 h = 0\). This relationship amongst the period modules of \(h, f , g\) and the consequent difference operator identities are directly responsible for the next two features:
\(\mathbb{Q}\)-span inside \(\mathscr{P}\). Consider any \(\alpha, \beta \in \mathbb{Q}\). Is \(\alpha f + \beta g\) periodic? Writing \(\alpha = a/c\), \(\beta = b/c\) with common denominator \(c\), we can scale: \(\alpha f + \beta g = (1/c)(a f + b g)\). On \(G\): \((af + bg)(mp + nq) = an - bm\). Take \(t = bp + aq\): then \((af + bg)(x + bp + aq) = a(n + a) - b(m + b) = an - bm\). ✓ So \(bp + aq \in P_{af + bg}\), and by extension \(\alpha f + \beta g\) is periodic. Hence \(\operatorname{span}_\mathbb{Q}(\{f, g\}) \subseteq \mathscr{P}\). In other words, a \(\mathbb{Q}\)-linear combination of \(f\) and \(g\) translates into a \(\mathbb{Q}\)-linear combination of their fundamental periods, which is a period of the resulting function. Indeed, as the reader can verify, anytime we have \(P_{h}\subset P_{f}+P_g\) for an incommensurable decomposition \(h = f+g\), of a periodic function \(h\), we automatically get that \(\operatorname{span}_\mathbb{Q}(\{f, g\}) \subseteq \mathscr{P}\).
\(\mathbb{R}\)-span NOT inside \(\mathscr{P}\). Take \(\alpha = \sqrt{2}\): then \(\sqrt{2} f + g\) on \(G\) equals \(\sqrt{2} n - m\), and for any \(t = m'p + n'q\) to be a period we would need \(\sqrt{2} n' = m'\) for all integer choices — impossible unless \(m' = n' = 0\). So \(\sqrt{2} f + g\) has no nontrivial period, and in particular is not in \(\mathscr{P}\). In order to obtain \(operatorname{span}_\mathbb{R}(\{f, g\}) \subseteq \mathscr{P}\) for incommensurable \(f, g\), we would need there to be an uncountable family of pairwise incommensurable periods \(p_\alpha\) for each \(h_\alpha:=\alpha f + g\) and every \(\alpha\in \mathbb{R}\), which naturally could not all be generated by a single pair of fundamental periods \(p, q\). I went back and forth for about a month on whether or not this was possible, until finally last month I found a way to prove that this and much more holds under the Axiom of Choice; this will be the subjet of a future post.
Unboundedness on every interval. Since \(G\) is dense in \(\mathbb{R}\), every interval contains points of \(G\) with arbitrarily large \(|n|\) and \(|m|\) values. So \(f\) takes values \(n \in \mathbb{Z}\) of arbitrary magnitude on every interval, and is therefore unbounded on every interval — what we will simply call nowhere bounded. The same holds for \(g\). I will leave it to the reader (for now at least; I wil liekly cover this in a future post) to show that \(\mathbb{Q}\)-linear dependence of the paiwrise incomensurable fundamental periods of \(f, g, h\) is also automatically gives nowhere boudnedness of the functions.
Almost-everywhere constancy and measurability. Both \(f\) and \(g\) vanish outside \(G\), and \(G\) is countable, hence has Lebesgue measure zero. So \(f, g\) are equal to zero almost everywhere — and in particular, they are Lebesgue measurable. This may seem to drain the example of interest from a measure-theoretic standpoint: \(f\) and \(g\) are equivalent (a.e.) to the zero function, which is plainly periodic with \(P = \mathbb{R}\). But the equivalence class of the zero function under “equality almost everywhere” includes many functions with wildly different pointwise behavior — and this example shows that interesting structural phenomena can hide entirely within a single a.e.-equivalence class. The lesson is that periodicity is a fundamentally pointwise notion: two functions equivalent almost everywhere can have very different period modules. From a measure-theoretic standpoint, \(E_1\) may be invisible; from a pointwise standpoint, it is the seed of everything that follows.
6.2 Breaking measurability: \(E_1'\)
Now we ask: can we modify \(E_1\) to remove the a.e.-constancy while preserving the algebraic features (strict periodicity of \(f, g, h\) with fundamental periods \(p, q, p+q\), and incommensurability of \(f, g\))? The answer is yes, assuming the existence of a Vitali-type selector for \(\mathbb{R}/G\), which is strictly weaker than a Hamel basis for \(\mathbb{R}\) over \(\mathbb{Q}\).
Construction \(E_1'\). Let \(\{y_\alpha\}_{\alpha \in I}\) be a complete set of representatives of cosets of \(\mathbb{R}/G\) — that is, every \(x \in \mathbb{R}\) has a unique decomposition \(x = y_\alpha + mp + nq\) with \(y_\alpha\) a representative and \(m, n \in \mathbb{Z}\). (We take the representative of the coset \(G\) itself to be \(0\).) On the coset \(G\) (i.e., where \(y_\alpha = 0\)), define \(f, g\) exactly as in \(E_1\): \[ f(mp + nq) := n, \qquad g(mp + nq) := -m. \] On every other coset \(y_\alpha + G\) with \(y_\alpha \neq 0\), define \(f\) and \(g\) as the constant functions equal to \(y_\alpha\): \[ f(y_\alpha + mp + nq) := y_\alpha, \qquad g(y_\alpha + mp + nq) := y_\alpha. \]
The assertion that these representatives exist is what uses a weak instance of the Axiom of Choice a selector for \(\mathbb{R}/G\), which is at the level of Vitali sets and strictly weaker than Hamel.
Verification. Let us check that \(E_1'\) inherits all the algebraic properties of \(E_1\).
Periodicity of \(f\). For \(x \in y_\alpha + G\) with \(y_\alpha \neq 0\), \(f\) is constant on the coset, so any \(t \in G\) trivially preserves \(f\) on this coset. On \(G\) itself, the analysis from \(E_1\) gives \(P_f \cap G = p\mathbb{Z}\). Now suppose \(t \notin G\). Then for \(x = 0 \in G\), \(f(0) = 0\), while \(0 + t = t\) lies in some coset \(y_\beta + G\) with \(y_\beta \neq 0\), so \(f(t) = y_\beta \neq 0\). Contradiction, so no \(t \notin G\) is a period. Hence \(P_f = p\mathbb{Z}\), exactly as in \(E_1\).
The same reasoning shows \(P_g = q\mathbb{Z}\), and \(P_{f+g}\) also turns out to equal \((p+q)\mathbb{Z}\) — verification mirrors the \(E_1\) case, accounting for the new behavior on cosets outside \(G\), where \(f + g \equiv 2y_\alpha\) (constant) and so trivially periodic with period \(p+q\).
So \(E_1'\) shares with \(E_1\) the strict periodicity of \(f, g, h\), the incommensurability of \(f, g\), the \(\mathbb{Q}\)-span property, and unboundedness (on every interval, since \(G\) is dense and on \(G\) the function is the same as in \(E_1\)).
Nonconstancy and non-measurability. But unlike \(E_1\), the function \(f\) is no longer zero almost everywhere: on each coset \(y_\alpha + G\) with \(y_\alpha \neq 0\), \(f\) takes the value \(y_\alpha\), and these cosets together cover all of \(\mathbb{R}\) except \(G\) (which has measure zero). So \(f\) takes a different nonzero value on each of uncountably many disjoint cosets — it is nowhere a.e. constant.
To see that \(f\) is non-measurable, consider the auxiliary function \(\varphi(y_\alpha + mp + nq) := y_\alpha\), defined on all cosets including \(G\) itself. The function \(f\) differs from \(\varphi\) only on \(G\) (a measure-zero set), since on \(G\), \(f\) takes the values \(\{n : n \in \mathbb{Z}\}\) while \(\varphi \equiv 0\) there. Hence \(f\) is measurable iff \(\varphi\) is.
But \(\varphi\) has period module exactly equal to \(G\): every \(r \in G\) keeps \(x\) within its coset (where \(\varphi\) is constant), so \(r \in P_\varphi\); and any \(r \notin G\) moves \(x = 0\) out of \(G\), where \(\varphi \neq 0\), while \(\varphi(0) = 0\), so \(r \notin P_\varphi\). Hence \(P_\varphi = G\), which is dense in \(\mathbb{R}\). Yet \(\varphi\) is not constant a.e. (it takes a different value on each coset).
It is a classical result — going back at least to Mirotin’s work on a.e.-periodic functions — that any measurable periodic function with a dense period module must be constant almost everywhere. The contrapositive applies directly to \(\varphi\): since \(\varphi\) has a dense period module and is not a.e. constant, it cannot be measurable. And therefore neither is \(f\). The same argument shows \(g\) and \(h = f + g\) are non-measurable. (I have a fairly elementary proof of the dense-period-module result that I plan to write up in a future post — it has the bonus of yielding the ergodicity of irrational rotations as a corollary.)
So \(E_1'\) achieves what \(E_1\) could not: it preserves all the algebraic features of \(E_1\) — strict periodicity, incommensurability, \(\mathbb{Q}\)-span periodicity, unboundedness — while also answering Question 5 from Part I in the strongest possible sense: there exist incommensurable periodic functions with periodic sum that are nowhere a.e. constant, and as a consequence are non-measurable.
6.3 A failed naive extention
A reader paying close attention may wonder whether there is an even more natural variant of \(E_1'\): instead of making \(f, g\) constant on each coset \(y_\alpha + G\) with \(y_\alpha \neq 0\), simply repeat the original \(E_1\) formula on every coset, defining \(f(y_\alpha + mp + nq) := n\) and \(g(y_\alpha + mp + nq) := -m\) for all representatives \(y_\alpha\), including the nonzero ones. This seems, at first glance, like the most direct possible extension of \(E_1\).
This variant is more delicate than it appears. It is not hard to show that \(P_f \cap G = p\mathbb{Z}\) still holds, but the question of whether there exist additional periods \(r \notin G\) depends sensitively on the choice of representatives. In fact, one can show via transfinite recursion on a well-ordering of \(\mathbb{R}/G\) that there exist choices of representatives under which both \(f\) and \(g\) acquire common periods outside \(\mathbb{Q}_{p,q}\) — and so \(f\) and \(g\) end up commensurable. The Axiom of Choice, when invoked without structure, is a treacherous tool: distinct selectors yield functions with drastically different period modules, some preserving incommensurability and some destroying it.
Assuming a well order of the reals, one of these choices of representatives actually yields sometihng quite interesting: discontinous, nomeasurbale homomorphisms \(\mathbb{R}\to\mathbb{R}\). I will wrap up this post with the “construction” of such a homomorphism.
7 Preview of a discontinous, nomeasurable homomorphsim.
What we need is a choice of representatives that forms an additive group itself.
Let me first show you why such a choice of representatives, if it exists, gives us a homomorphism; actually, it gies us two complementary types of homomorphisms. I will then conclude the post with a brief sketch of how to prove the existence of such a choice of representatives under the assumption of a Hamel basis for \(\mathbb{R}\) over \(\mathbb{Q}\); the details of this proof will be the subject of a future post.
Ok, let \(\mathcal V\) be a vitali-type set of representatives that happens to form an additive subgroup of \(\mathbb{R}\), meaning in particular that the representative of \(G\) is \(0\) and that \(y_1, y_2 \in \mathcal V\) implies \(y_1 + y_2 \in \mathcal V\). The first, obvious homomorphsim then is the map that sends \(x\) to the representative of \(x + G\). The kernel of this homomorphism is exactly \(\mathcal V\). Now, let \(\varphi:G\to \mathbb{R}\) be any the function defined by \(\varphi(x) = y_\alpha\) for \(x \in y_\alpha + G\).
For this we will assume there exists a Hamel basis for \(\mathbb{R}\) over \(\mathbb{Q}\), which is a stronger assumption than the existence of a Vitali set, but still weaker than the full Axiom of Choice.
We will recursively well order both the cosets and their representatives First we set \(\gamma_0 = G\) we chose representative \(y_0 = 0\). For \(\gamma_1\) we choose any coset other than \(G\), and as its representative we choose any irrational \(y_1\in \gamma_1\); irrationality guarantees that \(ny_1\not\in G\cup \gamma_1\) for all integers \(n> 1\). So, for \(n<\omega\), let \(y_n = ny_1\). Whenever \(\theta>\omega\) is not a limit ordinal, \(\theta\) is equal to some $+n $ for a limit ordinal \(\lambda\) and some \(n\in \mathbb{N}\). Set \(y_\theta = ny_\lambda\) and \(\gamma_\theta = y_\theta + G\). Then for a limit ordinal \(\lambda\), choose any \[\gamma_\lambda \in (\mathbb{R}/G)\setminus\{\gamma_\theta\}_{\theta<\lambda}\] and for its representative, pick again any irrational $y__. $.