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A function $f:\mathbb R \to \mathbb C$ with period 1 can be identitied with a function defined on the 1-dimensional torus $\mathbb T = \mathbb R / \mathbb Z$, the latter being continuous if and only if $f$ is. The Lebesgue measure on $\mathbb R$ is a Haar measure on the additive topological group $\mathbb R$ and the normed Haar measure on the compact topological group $\mathbb T$ is such that if $f:\mathbb T \to \mathbb C$ is continuous, then $\int_\mathbb T f(t) dt = \int_0^1 f(x) dx$ (indeed, the functional defined in this way is easily shown to satisfy the definition of a Haar measure on $\mathbb T$ and its value on the constant function 1 is 1). My purpose is to generalize the equality just written to arbitrary functions f, in the sense that the left side is defined (i.e. f is integrable) if and only if the right side is (i.e. $f$ as a periodic function on the real line is integrable on the interval [0,1] which means that the product $f$ times the characteristic function of this interval is integrable on $\mathbb R$), and then the equality holds.

Motivation: in his book "An Introduction to Harmonic Analysis", Yitzhak Katznelson uses in chapter 1 the formula as a definition of integrability and the integral for functions defined on the torus (instead of using Haar measure which he has not yet introduced at that point) ... and I want to make sure that this is compatible with what results from the use of the Haar measure on the torus.

What I have done so far: I can show that, $h$ being the quotient map $\mathbb R \to \mathbb T$, the measure of a subset $A$ of $\mathbb R$ is equal to that of $h(A)$ if $A$ ia an interval and as a consequence also if $A$ is any open set. Moreover, if $A$ is a null set (i.e. has measure 0) then so is $h(A)$ and if $B$ is a null set in $\mathbb T$, then so is $h^{-1} (B)$.

My idea is then to use the density of the continuous functions with compact support (which are all the continuous ones in the case of the torus) in $L^1$ to obtain the desired conclusion. I have not yet verified my guess that this is easy in one direction. But in the other one, there are complications due to the facts that on the real line such functions are not periodic and periodic functions are not integrable except when they are = zero a.e. And it might help to reduce the case of complex functions to that of real ones and this to that of positive functions, which I hope is not too difficult. And then, if $f$ is continuous with compact support, the translates of $f$ by an integer are the terms of a series converging absolutely (by what I mean that the series of norms for uniform convergence converges) in every compact subset of $\mathbb R$, the sum being continous and periodic.

Ulysse Keller
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  • Let $\lambda'$ be the restriction of the Lebesgue measure $\lambda$ from $\mathbb R$ to $[0,1)$ and $\mu$ be the Haar measure on $\mathbb T$. The map $f\mapsto hf$ is a bijection between complex-valued functions on $[0,1)$ and $\mathbb T$. It seems that to answer your question we have to check that this map is also a bijection between Lebesgue integrable functions $f$ on $[0,1)$ and integrable with respect to the Haar measure functions on $\mathbb T$, and $\int fd\lambda'=\int hfd\mu$, and this check is straighforward. – Alex Ravsky Aug 29 '23 at 10:29
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    What you call hf is in fact - I bet - f $\circ$ h - or more precisely its restriction to [0,1[ ... plus you make me think of the saying that mathematicians say something is obvious when they in fact don't know how to prove it (but of course think it's true). Your argument seems to use in a subtle way what is to be proved; where enters the fact that $\mu$ is Haar measure on the torus? – Ulysse Keller Aug 29 '23 at 20:52
  • Oops, the map indeed has to be defined differently. Thanks for noticing this and sorry for the error. I have to disagree with your interpretation of the usage the word "obvious" by mathematicians. I think that when a mathematician (well, at least a good one) says that a claim is obvious then it means rather that the claim very easily follows from the previous or the well-known facts, and, of course, the claim is already checked (proved) to be true. – Alex Ravsky Aug 30 '23 at 07:13
  • Next, "straightforward" is not the same as "obvious". The former can mean, for instance, to be proved without problems and refer, for instance, to two pages of routine calculations. – Alex Ravsky Aug 30 '23 at 07:13
  • I did not check my claim about the properties of the map. But I tried to formulate a claim, which should answer your question. I expect it will be true after I fix the error with the map definition and then it can be proved straightforwardly, based on the properties of the map and of both measures $\lambda$ and $\mu$. If the proof of fixed conjecture would answer your question then I can try to provide the proof. But, of course, you are free to formulate an other claim, which has to proved to answer your question. – Alex Ravsky Aug 30 '23 at 07:13

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Once the integration of continuous functions w.r.t. the measure is fixed, the measure is uniquely determined if you want the measure to be regular. (Which both of your measures are.) Basically for an open set $U$, $\mu(U)$ must be the supremum of the integrals of all continuous functions taking values between 0 and 1 and with compact support contained in $U$. Clearly this supremum is bounded above by $\mu(U)$. The other direction follows from inner regularity, since for any compact set $K \subseteq U$ you can find a continuous functions taking value between 0 and 1, with support contained in $U$, and taking value 1 on $K$. The integral of such a continuous function is larger than $\mu(K)$. Since $\mu(U)$ is the supremum of $\mu(K)$ for all such $K$, we get the desired result. (This is just the uniqueness part of Riesz representation theorem.)

David Gao
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  • This is the first time I start a bounty in this system and also the first time I got no answer nor any comment before I started the bounty. Now to this first (partial) answer, I don't really see how it advances me, excuse me that I can't tell better. The next of my comments will explain why – Ulysse Keller Aug 29 '23 at 11:58
  • David Gao (correctly) says that there are two measures involved: the Haar measures on the 1-dim. torus and the real line. But then the letter $\mu$ is used (without even mentioning that it is a measure but that is clear, and when a unique measure is discussed it is usually denoted by this letter) and it is not clear which of the two is meant. In fact a third measure could even be considered in this context but I think it wouldn't simplify the argument here to introduce it. May-be both, one after the other? (continued) – Ulysse Keller Aug 29 '23 at 12:12
  • But the two measures are related by the integral equality valid for continuous functions and to be generalized ... The whole thing isn't trivial as it might seem at first sight. The measures can't be considered as equal, even under the identification of functions used. Periodic functions on the real line are generally not integrable for Lebesgue measure: when measurable and bounded they are integrable on [0,1[ as well as [0,1] and therefore on every interval obtained by translation e.g. by an integer (periodicity of the funct. & transl. invar. of the measure) with same integral. (continued) – Ulysse Keller Aug 29 '23 at 12:28
  • The integral on the whole real line would then be a sum of infinitely many equal numbers. But as functions on the torus (by identification), they are integrable when measurable and bounded. Now the case of an open set U isn't a problem for me: as I said I solved it even if I didn't explain it in detail: it is a special case by considering the characteristic function of U. And I could also handle null sets. (If someone is interested, I can give details.) Please look at my ideas about how one could continue ... – Ulysse Keller Aug 29 '23 at 12:45
  • @UlysseKeller What I meant to say is as long as two regular measures give the same integral on continuous functions (which you have asserted is the case), then the two measures must be the same and hence any Borel measurable functions would give the same integral under these measures. If you want to be precise here, we need to first sort out what your equality for continuous functions even means, since $f$ cannot simultaneously be a function on $\mathbb{R}$ and $\mathbb{T}$... – David Gao Aug 29 '23 at 15:11
  • In that case what one can do is to note that your integral of $f$ w.r.t. the Haar measure on $\mathbb{T}$ can be understood as follows: considering the natural one-to-one map $\mathbb{T} \rightarrow [0, 1)$, then the integral of $f$, regarded as a function on the circle w.r.t. the Haar measure is the same as the integral of $f$, restricted to $[0, 1)$, w.r.t. the pushforward of the Haar measure. You then have two regular measures on $[0, 1]$ (by declaring ${1}$ has zero measure), namely this pushforward of Haar measure and the restriction of the Lebesgue measure... – David Gao Aug 29 '23 at 15:16
  • You can then apply the argument above to show that the two measures coincide as they give the same integral on continuous functions. Technically there is a small issue that a continuous function on $[0, 1]$ could have different values at two endpoints, but you can just subtract a linear term and calculate by hand that the integrals of a linear function under the two measures are the same. Does this solve your problem? – David Gao Aug 29 '23 at 15:19
  • David, to begin with: of course a function cannot be defined at the same time on the reals and the elements of the torus. But I announced an identification to begin with (in the question text) which is - yes - an abuse of language - see the wikipedia article https://en.wikipedia.org/wiki/Abuse_of_notation which also tells what are the reasons that kind of justify their use ... for one who knows what he is doing – Ulysse Keller Aug 29 '23 at 21:27
  • Now I need more time to judge if your answer and your comments following mine are helpful. A may-be important difficulty is that what you call the natural one-to-one map from the torus to [0,1[ is not a homeomorphism (clearly a topological circle and an interval of the real line are "different" things: in fact, that map isn't continuous - only its inverse is) so I doubt at first sight the regularity of the "pushforward of Haar measure" (extended or not to the closed interval) – Ulysse Keller Aug 29 '23 at 21:48
  • Your idea of making continuous functions on [0,1] have equal values in 0 and 1 by subtracting a linear one is probably useful in many circumstances ... I had it myself BUT if I remember well, I found it finally NOT useful in this case ... at least under my line of argument. But this might be different under another line of argument. I'll first try to continue on my line (as drafted towards the end of my Q:) If it leads to nothing I'll look wether your line might be made to work – Ulysse Keller Aug 29 '23 at 22:02