Reference for Kronecker-Weyl theorem in full generality

Why not just prove the result from scratch? It's only a couple of pages and involves only basic Fourier analysis for locally compact abelian groups.

Let $$\mathbb{T}^n = \left\{(z_1,\ldots,z_n) \in \mathbb{C}^n : |z_l| = 1 \text{ for all $1 \leq l \leq n$}\right\}$$ be the $n$-torus. Let $t_1, \ldots, t_n$ be arbitrary real numbers, and let $H$ be the topological closure in $\mathbb{T}^n$ of the subgroup $$\widetilde{H} = \left\{\left(e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}\right) \in \mathbb{T}^n : y \in \mathbb{R}\right\}.$$ The Kronecker--Weyl theorem states that

(1) $H$ is a closed connected subgroup of $\mathbb{T}^n$, namely an $r$-dimensional subtorus of $\mathbb{T}^n$, where $0 \leq r \leq n$ is the dimension over $\mathbb{Q}$ of the span of $t_1, \ldots, t_n$, and that

(2) for any continuous function $h : \mathbb{T}^n \to \mathbb{C}$, we have that $$\lim_{Y \to \infty} \frac{1}{Y} \int^{Y}_{0}{h\left(e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}\right) \: dy} = \int_{H}{h(z) \: d\mu_H(z)},$$ where $\mu_H$ is the normalised Haar measure on $H$.

To prove this, we begin by observing that $H$ is a closed connected subgroup of $\mathbb{T}^n$ as it is the topological closure of $\widetilde{H}$, which is a subgroup of the compact abelian group $\mathbb{T}^n$, being the image of the continuous group homomorphism $\phi : \mathbb{R} \to \mathbb{T}^n$ given by $\phi(y) = \left(e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}\right)$.

Next we recall that a character $\chi : \mathbb{T}^n \to \mathbb{T}$ is of the form $$\chi(z_1,\ldots,z_n) = z_1^{k_1} \cdots z_n^{k_n}$$ for some $(k_1,\ldots,k_n) \in \mathbb{Z}^n$. Conversely, for any $(k_1,\ldots,k_n) \in \mathbb{Z}^n$, the function $\chi : \mathbb{T}^n \to \mathbb{T}$ defines a character of $\mathbb{T}^n$. In particular, the dual group of $\mathbb{T}^n$ is isomorphic to $\mathbb{Z}^n$, and hence every character $\chi : \mathbb{Z}^n \to \mathbb{T}$ is of the form $$\chi(k_1,\ldots,k_n) = z_1^{k_1} \cdots z_n^{k_n}$$ for some $(z_1,\ldots,z_n) \in \mathbb{T}^n$.

We claim that the annihilator $H^{\perp}$ of $H$ (namely the set of characters of $\mathbb{T}^n$ that are trivial on $H$) is isomorphic to $\left\{k \in \mathbb{Z}^n : t_1 k_1 + \cdots + t_1 k_n = 0\right\}$, and consequently $H$ is isomorphic to a torus $\mathbb{T}^r$ for some $0 \leq r \leq n$. Indeed, each character $\chi \in H^{\perp}$ is of the form $\chi(z_1,\ldots,z_n) = z_1^{k_1} \cdots z_n^{k_n}$ for some $(k_1,\ldots,k_n) \in \mathbb{Z}^n$ with the property that for all $y \in \mathbb{R}$, $$1 = \chi\left(e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}\right) = e^{2\pi i (t_1 k_1 + \cdots + t_n k_n) y},$$ and hence $t_1 k_1 + \cdots + t_n k_n = 0$. Conversely, if $t_1 k_1 + \cdots + t_n k_n = 0$, then the homomorphism $\chi(z_1,\ldots,z_n) = z_1^{k_1} \cdots z_n^{k_n}$ satisfies $\chi \vert_{H} = 1$. Now by construction, $H^{\perp}$ is isomorphic to $V \cap \mathbb{Z}^n$ for some vector subspace $V$ of $\mathbb{Q}^n$ of dimension $n - r$, where $r$ is the dimension over $\mathbb{Q}$ of the span of $t_1, \ldots, t_n$, and so $H^{\perp} \cong \mathbb{Z}^{n-r}$. Consequently, $\widehat{H} \cong \mathbb{Z}^n / H^{\perp} \cong \mathbb{Z}^r$, as the dual group of $\mathbb{T}^n$ is isomorphic to $\mathbb{Z}^n$, and hence $H \cong \mathbb{T}^r$, thereby proving (1).

For (2), we require the Fourier transform $\widehat{h} : \mathbb{Z}^n \to \mathbb{C}$ of a continuous function $h : \mathbb{T}^n \to \mathbb{C}$, defined by $$\widehat{h}(\chi) = \int_{\mathbb{T}^n}{h(z) \overline{\chi(z)} \: dz}.$$ The Poisson summation formula for a closed subgroup of $\mathbb{T}^n$ states that $$\int_{H}{h(z) \: d\mu_H(z)} = \int_{H^{\perp}}{\widehat{h}(\chi) \: d\mu_{H^{\perp}}(\chi)},$$ where $\mu_H$ is the normalised Haar measure on $H$ and $\mu_{H^{\perp}}$ is the induced Haar measure on $H^{\perp}$, which is the counting measure as $H^{\perp}$ is discrete. Now if $h : \mathbb{T}^n \to \mathbb{C}$ is a trigonometric polynomial, which is to say a function of the form $$h(z) = \sum_{k \in \mathbb{Z}^n} c_k z_1^{k_1} \cdots z_n^{k_n}$$ for $z = (z_1, \ldots, z_n) \in \mathbb{T}^n$, where all but finitely many of the coefficients $c_k \in \mathbb{C}$ are zero, we claim that $$\lim_{Y \to \infty} \frac{1}{Y} \int^{Y}_{0}{h\left(e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}\right) \, dy} = \int_{H}{h(z) \: d\mu_H(z)},$$ where $\mu_H$ is the normalised Haar measure on $H$. From this, we may easily obtain the result in the general case where $h$ is merely a continuous function by the density of the trigonometric polynomials in the space of continuous complex-valued functions on $\mathbb{T}^n$ with regards to the supremum norm. This yields (2).

To prove the claim, we let $\chi : \mathbb{T}^n \to \mathbb{T}$ be a character corresponding to $\widetilde{k} \in \mathbb{Z}^n$. Then $$\widehat{h}(\chi) = \int_{\mathbb{T}^n}{h(z) \overline{\chi(z)} \, dz} = \int_{\mathbb{T}} \hspace{-0.1cm} \cdots \hspace{-0.1cm} \int_{\mathbb{T}}{\sum_{k \in \mathbb{Z}^n} c_k z_1^{k_1} \cdots z_n^{k_n} \overline{z_1^{\widetilde{k_1}} \cdots z_n^{\widetilde{k_n}}} \, dz_1 \cdots dz_n}.$$ We may interchange the order of summation and integration, as there are only finitely many nonzero members in this sum, and evaluate this integral in order to find that $\widehat{h}(\chi) = c_{\widetilde{k}}$. Recalling that $H^{\perp}$ is isomorphic to $\left\{k \in \mathbb{Z}^n : t_1 k_1 + \cdots + t_n k_n \in \mathbb{Z}\right\}$, so that the Haar measure $\mu_{H^{\perp}}$ on $H^{\perp}$ is simply the counting measure, we therefore obtain by the Poisson summation formula that $$\int_{H}{h(z) \: d\mu_H(z)} = \sum_{\substack{k \in \mathbb{Z}^n \\ t_1 k_1 + \cdots + t_n k_n = 0}} c_k.$$ On the other hand, \begin{align*} \lim_{Y \to \infty} \frac{1}{Y} \int^{Y}_{0}{h\left(e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}\right) \: dy} & = \lim_{Y \to \infty} \frac{1}{Y} \sum_{k \in \mathbb{Z}^n} c_k \int^{Y}_{0}{e^{2\pi i (t_1 k_1 + \cdots + t_n k_n) y} \: dy} \\ & = \sum_{\substack{k \in \mathbb{Z}^n \\ t_1 k_1 + \cdots + t_n k_n = 0}} c_k \end{align*} as required, where we justify the interchanging of order of summation and integration by noting that there being only finitely many nonzero members in this sum.

Maybe my answer is beside the point, because the question speaks about the "limit distribution of $t(\theta_1,\dots,\theta_d)$" (for real $t$?). Shouldn't it be $n(\theta_1,\dots,\theta_d)$ for $n=1,2,\ldots$? See https://mathworld.wolfram.com/Kronecker-WeylTheorem.html. The long answer "from scratch" by Peter Humphries proves a different theorem that seems to be more in line with real parameters $t$. Also, the sequence $n(\theta_1,\dots,\theta_d)$ can fill a disconnected manifold, which wouldn't properly be called a "subtorus".

If the question is indeed about the sequence $n(\theta_1,\dots,\theta_d)$, the book by Kuipers and Niederreiter (Uniform Distribution of Sequences, 1974), contains the generic version of what seems to be the Kronecker-Weyl Theorem as Example 6.1 on p. 48. The proper condition is the real numbers $1,\theta_1,\dots,\theta_d$ are linearly independent over the rationals.

The notes on p.51 mention that

a discussion of the exceptional case in this example was also carried out by Weyl,

referring to his classical paper: Hermann Weyl. Über die Gleichverteilung von Zahlen mod. Eins. Mathematische Annalen, 77:313–352, 1916. DOI:10.1007/BF01475864

Indeed, §5 of that paper, "Die Ausnahmefälle" (the exceptional cases), contains a Theorem 18 (pp. 340-341). It deals with the more general case where each coordinate is not just a linear function of $n$ but an arbitrary polynomial. The conclusion is that the points cover a finite number of affine-linear $r$-dimensional manifolds, (possibly with different integer multiplicities), all these manifolds are parallel, and each of them is filled with uniform density. The theorem specifies how to determine $r$ and the multiplicities.

Here is a statement of this theorem (with different notation), specialized and reformulated for the case of an arithmetic progression $n(\theta_1,\dots,\theta_d)$ as opposed to arbitrary polynomials. The multiplicities are then not necessary.

Let $\vec\theta=(\theta_1,\dots,\theta_d)$. Let $C$ be the set of vectors $\vec x\in\mathbb R^d$ such that $$\langle \vec a,\vec x\rangle\equiv b \pmod 1$$ for all integer vectors $\vec a\in\mathbb Z^d$ and rational numbers $b$ for which the equation $$\langle \vec a,\vec\theta\rangle= b$$ holds. Then all numbers $b$ appearing in these equations have a least common denominator $g$. The sequence $n\vec\theta$ is uniformly distributed modulo 1 in the disjoint union of parallel subspaces $C\cup 2C\cup \dots \cup gC$.

The denominator $g$ is the smallest number $g\ge1$ for which $gC$ contains an integer point (or equivalently, for which $gC$ modulo 1 contains the origin and is therefore equivalent to its corresponding linear subspace $C-C$). If rational dependencies exist only among the numbers $\theta_1,\dots,\theta_d$ and not with the number 1, then the right-hand side $b$ is always $0$, and we set $g=1$.

It is clear that $n\vec\theta$ cycles through the $g$ sets $C,2C,3C,...$. Thus it suffices to look at the generating vector $g\vec\theta$ and prove that its multiples are uniformly distributed modulo one in $gC=C-C$. It is an exercise to reduce this case to the independent case, along the lines of the reduction that I gave for the continuous version. (In fact many of the arguments in that proof appear in Weyl's proof already.)

Let me try to reinstate honor to the solution that proposed the basis change, by reducing the general case to the independent ("generic") case via a basis change, as opposed to proving it from scratch. This time I am treating the continuous version. (as in Peter Humphries' solution)

Let $\vec\theta=(\theta_1,\dots,\theta_k)$. Let $S$ be the subspace of vectors $\vec x\in\mathbb R^k$ such that $$\langle \vec a,\vec x\rangle=0$$ for all rational vectors $\vec a\in\mathbb Q^k$ for which $$\langle \vec a,\vec\theta\rangle=0$$ The integer points in $S$ form a lattice $L$ (a discrete subgroup of $\mathbb R^n$), which can be written as $L=\{\,g_1\vec b_1+\dots+g_r\vec b_r\mid g_i\in\mathbb Z\,\}$ for some generating basis vectors $\vec b_1,\dots,\vec b_r\in\mathbb{Z}^k$. Since $S$ is defined by rational equations, this basis spans $S$.
Let $(\theta'_1,\dots,\theta_r')$ be the coordinates of the point $\vec\theta\in S$ with respect to this basis: $$\vec\theta=\theta_1'\vec b_1+\dots+\theta_r'\vec b_r$$ Then $(\theta'_1,\dots,\theta_r')$ is independent over the rationals (see below (*) for a proof). Thus, we can apply the generic continuous Kronecker-Weyl Theorem, and $(t(\theta_1',\dots,\theta_r'))_{t\in \mathbb{R}}$ is uniformly distributed modulo 1 in the $r$-torus $[0,1)^r$. Transforming back to the original coordinates, this means that $(t\vec \theta)_{t\in \mathbb{R}}$ is uniformly distributed modulo $L$ in the fundamental region $$ F = \{\,\lambda_1\vec b_1+\dots+\lambda_r\vec b_r \mid 0\le\lambda_i<1\,\}$$ of the lattice.
Now we map $F$ back into the standard torus $[0,1)^k$ by taking all coordinates modulo 1. No two points of $F$ are mapped to the same point (otherwise we would have a nonzero integer point inside $F$), but "opposite" boundary points are mapped to the same point because they differ by a basis vector $b_i\in\mathbb{Z}^k$. So $F$ forms a nice $r$-dimensional subtorus of $[0,1)^k$.

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(*) Here is the proof that $(\theta'_1,\dots,\theta_r')$ is independent over the rationals. It is not so obvious as I thought. Consider a rational relation $c_1\theta_1'+\dots+c_r\theta'_r=0.$ We can choose inside $S$ a rational vector $\vec a$ such that $\langle \vec a,\vec b_i\rangle = c_i$ for $i=1,\dots,r$. (The vector $\vec a$ is uniquely determined by these equations.) Then $$\langle \vec a, \vec\theta\rangle= \langle \vec a, (\theta_1'\vec b_1+\dots+\theta_r'\vec b_r)\rangle =c_1\theta_1'+\dots+c_r\theta'_r=0 $$ Thus, by the definition of $S$, $\vec a$ should be orthogonal to $S$. Therefore, $\vec a=\vec 0$, and all $c_i$ are $0$.