Translation invariance of Lebesgue integral I am looking at the following statement: Let $u:\mathbb{R^3} \rightarrow \mathbb{S}^1$ and $z\in \mathbb{R}^3$ and $h>0$ small. Then the integral $\int_{\mathbb{T}^3} dx$ is translation invariant and therefore 
$\int_{\mathbb{T}^3} u(x)-u(x-\sqrt{h}z) dx = \int_{\mathbb{T}^3} u(x-\sqrt{h}z)-u(x-2\sqrt{h}z) dx$. 
I understand that the Lebesgue-measure is invariant under translation and this implies that the Lebesgue Integral is translation invariant as well, in the sense that for a measurable set $D \subset \mathbb{R}^3$ we have 
$\int_D u(x) dx = \int_{D-z} u(x+z) dx$. 
The torus is defined as the quotient of the cartesian plane with the identifications $(x,y,z)\text{~}(x+1,y,z)\text{~}(x,y+1,z)\text{~}(x,y,z+1)$. The thing I don't understand in the first statement is that we have both times the integral over $\mathbb{T}^3$. I thought that on the righthand side it should be $\mathbb{T}^3+\sqrt{h}z$ for $\sqrt{h}z \notin \mathbb{Z}^3$ after our definition of $\mathbb{T}^3$. Am I not understanding the notion of the three-torus correctly or where is my mistake?
 A: The torus $\mathbb{T}^n$ is mapped to itself upon a translation of it's native periodic coordinates. Think about the circle (less commonly noted as $\mathbb{T}^1$) for example- if you shift the angle coordinate by any amount, you will still traverse the whole circle on the interval $[0,2\pi)$. 
Of course, when a function is defined on the torus, in order for this to still hold true under integration, we need to equip the function with the same symmetry as the torus, in other words we must make it periodic in all directions with the same period as the torus. This makes sense if you want your function to be single-valued on the manifold.
Proof for the Riemann integral, with some simplifying assumptions
Here's a proof for the Riemann integral on $\mathbb{T}_1$ that generalizes well to higher dimensions:
Assume $f(x+1)=f(x)~\forall x$. Then we define the following Riemann integral
$$I(a)=\int_{0}^{1}f(x-a)dx=\int_{a}^{a+1}f(x)dx$$ 
We compute it's derivative and we find that:
$$\frac{dI(a)}{da}=f(a+1)-f(a)=0\iff I(a)=I(0)$$
and thus we proved 
$$\int_{\mathbb{T^1}}f(x)dx=\int_{\mathbb{T^1}}f(x-a)dx~ \forall a\in\mathbb{R}$$
Generalizing in higher dimensions is easy. Denoting $\mathbf{a}=(a_1,a_2,...a_n)$ and generalizing the periodicity property on $f$ accordingly, we define
$$I(\mathbf{a})=\int_{(0,1)^n} dx_1...dx_n~f(\mathbf{x+a})$$
Taking the gradient, we note that
$$\begin{align}\frac{\partial}{\partial a_i}I(\mathbf{a})&=\int_{(0,1)^n} dx_1...dx_n~\frac{\partial}{\partial a_i}f(\mathbf{x+a})\\&=\int_{(0,1)^n} dx_1...dx_n~\frac{\partial}{\partial x_i}f(\mathbf{x+a})\\&=F_i(a_i+1)-F_i(a_i)\end{align}$$
where we defined
$$F_i(t)=\int_{(0,1)^{n-1}}dx_1...dx_{i-1}dx_{i+1}...dx_nf(x_1+a_1,...,x_{i-1}+a_{i-1}, t,...,x_n+a_n)$$
Due to the periodicity condition in all coordinates, it is clear that
$$F_i(t+1)=F_i(t)~,~\forall i,t$$
and hence all first-order partial derivatives of $I$ identically vanish and the integral is shown to be a constant with respect to shift parameters. By setting $\mathbf{a}=0$ we conclude that
$$\int_\mathbb{T^n}d\mathbf{x}f(\mathbf{x+a})=\int_\mathbb{T^n}d\mathbf{x}f(\mathbf{x})$$
as desired. These ideas should have an analogue on the Lebesgue integral that I haven't been able to exactly pinpoint, but I hope they provide with a good starting point.
EDIT: Proof for Lebesgue integral, only assumes positivity and measurability
I came up with a Lebesgue-measure style outline of a proof for $\mathbb{T}^1$:
Consider the measure of the following set $\mu(x\in (0,T)|f(x)>t)$. If f is positive definite and measurable, the Lebesgue integral exists and is equal to $\int_{0}^{\infty}\mu(x\in (0,T)|f(x)>t)dt$.
However we note that if we shift the argument by an amount $a=mT+r, 0<r<T$ and assume $f$ is $T$-periodic, 
$$\begin{align}\mu(x\in (0,T)|f(x+a)>t)&=\mu(x\in (0,T)|f(x+r)>t)\\&=\mu(x\in (r,T)\cup(T,T+r)|f(x)>t)\\&=\mu(x\in (r,T)|f(x)>t)+\mu(x\in (T,T+r)|f(x)>t)\\&=\mu(x\in (r,T)|f(x)>t)+\mu(y\in (0,r)|f(y+T)>t)\\&=\mu(x\in (r,T)|f(x)>t)+\mu(x\in (0,r)|f(x)>t)\\&=\mu(x\in (0,T)|f(x)>t)\end{align}$$
and we are done. This can be readily generalized to higher dimensional analogues of the statement, since all we need to show is that what shifts do is to merely  rearrange the interval of integration (in your case the whole higher dimensional torus). In fact the previous proof goes through completely by just adding indices where appropriate, $a_i=m_iT+r_i$ and applying the one-dimensional procedure repeatedly!
Here's the outline of the proof for $\mathbb{T}^n$:
$$\begin{align}\mu((x_1,...,x_n)\in (0,T)^n|f(x_1+a_1, ...x_n+a_n)>t)&=\mu((x_1,...,x_n)\in (0,T)|f(x_1+r_1, ..., x_n+r_n)>t)\\&=\mu(x_1\in (0,T),(x_2,...x_n)\in(0,T)^{n-1}|f(x_1,x_2+r_2,...,x_n+r_n)>t)\\&=\mu((x_1,x_2)\in (0,T)^2,(x_3,...x_n)\in(0,T)^{n-2}|f(x_1,x_2+r_2,...,x_n+r_n)>t)=...\\...&=\mu((x_1,...,x_n)\in (0,T)^n|f(x_1, ...x_n)>t)\end{align}$$
