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.
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.)