[proofplan] We work on the probability space $([0,1),\mathcal{B}([0,1)),\lambda)$, where $\lambda$ is the restriction of one-dimensional Lebesgue measure $\mathcal{L}^1$ to $[0,1)$, and $T_\alpha$ acts as addition by $\alpha$ modulo $1$. We first verify that $T_\alpha$ preserves $\lambda$ by decomposing it into two pure translations on a partition of $[0,1)$. The Fourier characters $e_n(x)=e^{2\pi i n x}$ diagonalize the Koopman operator $U_{T_\alpha}$, so an $L^2$ function invariant under $U_{T_\alpha}$ can have nonzero Fourier coefficients only at frequencies $n$ satisfying $e^{2\pi i n\alpha}=1$. If $\alpha$ is irrational, only the zero frequency survives, so all invariant $L^2$ functions are constants $\lambda$-a.e., and the spectral characterization of ergodicity then gives ergodicity. If $\alpha$ is rational with denominator $q$, the character $e_q$ is a nonconstant invariant function, so $T_\alpha$ is not ergodic. [/proofplan]

[step:Set up the Fourier basis on $L^2([0,1),\lambda;\mathbb{C})$]For each $n \in \mathbb{Z}$, define the Fourier character \begin{align*} e_n: [0,1) &\to \mathbb{C} \\ x &\mapsto e^{2\pi i n x}. \end{align*} By the [Fourier Orthonormal Basis Theorem for $L^2$ on the Unit Interval](/theorems/fourier-basis-l2-unit-interval), the family $(e_n)_{n\in\mathbb{Z}}$ is an orthonormal basis of the complex Hilbert space $L^2([0,1),\lambda;\mathbb{C})$. Thus every $f \in L^2([0,1),\lambda;\mathbb{C})$ has the expansion \begin{align*} f = \sum_{n\in\mathbb{Z}} \widehat f(n)\,e_n \end{align*} with convergence in $L^2([0,1),\lambda;\mathbb{C})$, where the Fourier coefficient function \begin{align*} \widehat f: \mathbb{Z} &\to \mathbb{C} \\ n &\mapsto \int_{[0,1)} f(x)\,\overline{e_n(x)}\,d\lambda(x) \end{align*} is defined by the Hilbert space inner product.[/step]

[guided]The natural coordinates for a rotation are the Fourier characters. For every integer $n \in \mathbb{Z}$, define \begin{align*} e_n: [0,1) &\to \mathbb{C} \\ x &\mapsto e^{2\pi i n x}. \end{align*} The [Fourier Orthonormal Basis Theorem for $L^2$ on the Unit Interval](/theorems/fourier-basis-l2-unit-interval) applies to the probability space $([0,1),\mathcal{B}([0,1)),\lambda)$ from the theorem hypothesis and states that $(e_n)_{n\in\mathbb{Z}}$ is an orthonormal basis of $L^2([0,1),\lambda;\mathbb{C})$. Therefore, for each $f \in L^2([0,1),\lambda;\mathbb{C})$, there is an $L^2$-convergent Fourier expansion \begin{align*} f = \sum_{n\in\mathbb{Z}} \widehat f(n)\,e_n, \end{align*} where the coefficient at frequency $n$ is \begin{align*} \widehat f(n)=\int_{[0,1)} f(x)\,\overline{e_n(x)}\,d\lambda(x). \end{align*} This basis is useful because rotations multiply each Fourier character by a scalar, so the invariant functions can be detected frequency by frequency.[/guided]

[step:Verify that $T_\alpha$ preserves Lebesgue measure on $[0,1)$]Set $\beta := \alpha - \lfloor \alpha \rfloor \in [0,1)$. For every $x \in [0,1)$, $x + \alpha$ and $x + \beta$ differ by the integer $\lfloor \alpha \rfloor$, so $\lfloor x+\alpha \rfloor = \lfloor x+\beta \rfloor + \lfloor \alpha \rfloor$ and hence $T_\alpha(x) = x + \beta - \lfloor x+\beta\rfloor$. We may therefore assume $\alpha = \beta \in [0,1)$ throughout this step. If $\beta = 0$, then $T_\alpha$ is the identity, which satisfies $T_\alpha^{-1}(A) = A$ for every Borel $A \subseteq [0,1)$, and so preserves $\lambda$. Assume now $\beta \in (0,1)$. Partition $[0,1)$ as \begin{align*} [0,1) = [0, 1-\beta) \sqcup [1-\beta, 1). \end{align*} For $x \in [0,1-\beta)$ we have $0 \leq x+\beta < 1$, so $\lfloor x+\beta\rfloor = 0$ and $T_\alpha(x) = x+\beta$. For $x \in [1-\beta, 1)$ we have $1 \leq x+\beta < 2$, so $\lfloor x+\beta\rfloor = 1$ and $T_\alpha(x) = x+\beta-1$. Define the two restrictions \begin{align*} T_1: [0,1-\beta) &\to [\beta,1), & x &\mapsto x+\beta, \\ T_2: [1-\beta,1) &\to [0,\beta), & x &\mapsto x+\beta-1. \end{align*} Each $T_i$ is a Borel-measurable bijection between half-open intervals, and each is the restriction of a translation of $\mathbb{R}$. Let $A \in \mathcal{B}([0,1))$. Split $A = (A \cap [\beta,1)) \sqcup (A \cap [0,\beta))$. Then \begin{align*} T_\alpha^{-1}(A) = T_1^{-1}(A \cap [\beta,1)) \sqcup T_2^{-1}(A \cap [0,\beta)). \end{align*} The preimage $T_1^{-1}(B)$ of any Borel $B \subseteq [\beta,1)$ equals $B - \beta$, and $T_2^{-1}(B')$ for any Borel $B' \subseteq [0,\beta)$ equals $B' + (1-\beta)$. By translation invariance of $\mathcal{L}^1$, \begin{align*} \lambda(T_\alpha^{-1}(A)) &= \lambda\big((A \cap [\beta,1)) - \beta\big) + \lambda\big((A \cap [0,\beta)) + (1-\beta)\big) \\ &= \lambda(A \cap [\beta,1)) + \lambda(A \cap [0,\beta)) \\ &= \lambda(A). \end{align*} Therefore $T_\alpha$ is a [measure-preserving transformation](/page/Measure-Preserving%20Transformation) of $([0,1),\mathcal{B}([0,1)),\lambda)$.[/step]

[guided]Both the Koopman operator and the spectral characterization of ergodicity used below require $T_\alpha$ to preserve $\lambda$. We verify this directly rather than invoking it as a known fact. Replace $\alpha$ by its fractional part $\beta := \alpha - \lfloor \alpha\rfloor \in [0,1)$. The two rotations $T_\alpha$ and $T_\beta$ coincide on $[0,1)$, because shifting $\alpha$ by the integer $\lfloor \alpha\rfloor$ shifts $\lfloor x+\alpha\rfloor$ by the same integer, so $x+\alpha-\lfloor x+\alpha\rfloor = x+\beta-\lfloor x+\beta\rfloor$. Hence it suffices to treat $\alpha = \beta \in [0,1)$. If $\beta = 0$, then $T_\alpha$ is the identity, and every set is its own preimage, so $T_\alpha$ preserves $\lambda$. Now suppose $\beta \in (0,1)$. The point $1-\beta$ is where the rotation crosses the upper boundary and wraps around. Split $[0,1)$ at this point: \begin{align*} [0,1) = [0,1-\beta) \sqcup [1-\beta,1). \end{align*} For $x \in [0,1-\beta)$ the sum $x+\beta$ stays in $[0,1)$, so no wrap occurs and $T_\alpha(x) = x+\beta$. For $x \in [1-\beta,1)$ the sum $x+\beta$ lies in $[1,2)$, so the floor equals $1$ and $T_\alpha(x) = x+\beta-1$. The rotation therefore decomposes into two pure translations: \begin{align*} T_1: [0,1-\beta) &\to [\beta,1), & x &\mapsto x+\beta, \\ T_2: [1-\beta,1) &\to [0,\beta), & x &\mapsto x+\beta-1. \end{align*} Now fix a Borel set $A \subseteq [0,1)$. To compute $\lambda(T_\alpha^{-1}(A))$, split $A$ at the boundary $\beta$: \begin{align*} A = (A \cap [\beta,1)) \sqcup (A \cap [0,\beta)). \end{align*} Points in $T_\alpha^{-1}(A \cap [\beta,1))$ come from $T_1$, because the codomain of $T_1$ is $[\beta,1)$, and points in $T_\alpha^{-1}(A \cap [0,\beta))$ come from $T_2$. Therefore \begin{align*} T_\alpha^{-1}(A) = T_1^{-1}(A \cap [\beta,1)) \sqcup T_2^{-1}(A \cap [0,\beta)). \end{align*} Each piece is a translate of a Borel subset of $[0,1)$, and translation preserves one-dimensional Lebesgue measure. Therefore \begin{align*} \lambda(T_\alpha^{-1}(A)) = \lambda(A \cap [\beta,1)) + \lambda(A \cap [0,\beta)) = \lambda(A). \end{align*} This is exactly the statement that $T_\alpha$ is $\lambda$-preserving on $\mathcal{B}([0,1))$. We use this fact below to define the [Koopman operator](/page/Koopman%20Operator) on $L^2$ and to apply the [Spectral Characterization of Ergodicity](/theorems/spectral-characterization-ergodicity).[/guided]

[step:Diagonalize the Koopman operator on the Fourier characters]Because $T_\alpha$ preserves $\lambda$ by the previous step, the [Koopman operator](/page/Koopman%20Operator) \begin{align*} U_{T_\alpha}: L^2([0,1),\lambda;\mathbb{C}) &\to L^2([0,1),\lambda;\mathbb{C}) \\ f &\mapsto f\circ T_\alpha \end{align*} is well-defined: if $f = g$ $\lambda$-a.e. on $[0,1)$, then $\{x : f(T_\alpha x) \neq g(T_\alpha x)\} \subseteq T_\alpha^{-1}(\{f \neq g\})$ has $\lambda$-measure zero by $\lambda$-preservation, so $f\circ T_\alpha = g\circ T_\alpha$ $\lambda$-a.e. Moreover, the change-of-variables identity \begin{align*} \int_{[0,1)} |f\circ T_\alpha|^2\,d\lambda = \int_{[0,1)} |f|^2\,d\lambda, \end{align*} which is the pushforward identity for the measure-preserving map $T_\alpha$, shows that $f\circ T_\alpha \in L^2$ with the same $L^2$ norm. In particular $U_{T_\alpha}$ is an isometry. For every $n \in \mathbb{Z}$ and every $x \in [0,1)$, \begin{align*} (U_{T_\alpha}e_n)(x) &= e_n(T_\alpha x) \\ &= e^{2\pi i n(x+\alpha-\lfloor x+\alpha\rfloor)} \\ &= e^{2\pi i n\alpha}\,e^{2\pi i n x}\,e^{-2\pi i n\lfloor x+\alpha\rfloor} \\ &= e^{2\pi i n\alpha}\,e_n(x), \end{align*} because $n\lfloor x+\alpha\rfloor \in \mathbb{Z}$ and hence $e^{-2\pi i n\lfloor x+\alpha\rfloor}=1$. Thus $e_n$ is an eigenfunction of $U_{T_\alpha}$ with eigenvalue $e^{2\pi i n\alpha}$.[/step]

[guided]The [Koopman operator](/page/Koopman%20Operator) is the linear translation of a measure-preserving map into an operator on $L^2$. Define \begin{align*} U_{T_\alpha}: L^2([0,1),\lambda;\mathbb{C}) &\to L^2([0,1),\lambda;\mathbb{C}) \\ f &\mapsto f\circ T_\alpha. \end{align*} Two checks make this well-defined on $L^2$ equivalence classes, and both rely on the measure preservation established in the previous step. First, if $f = g$ $\lambda$-a.e., then $\{x : f(T_\alpha x) \neq g(T_\alpha x)\}$ is contained in $T_\alpha^{-1}(\{f \neq g\})$, whose $\lambda$-measure equals $\lambda(\{f \neq g\}) = 0$ by $\lambda$-preservation. So the equivalence class of $f \circ T_\alpha$ depends only on the equivalence class of $f$. Second, the change-of-variables identity for the $\lambda$-preserving map $T_\alpha$ gives \begin{align*} \int_{[0,1)} |f\circ T_\alpha|^2\,d\lambda = \int_{[0,1)} |f|^2\,d\lambda, \end{align*} which shows $f \circ T_\alpha \in L^2$ and that $U_{T_\alpha}$ is an isometry on $L^2([0,1),\lambda;\mathbb{C})$. Now compute its action on one Fourier character. For $n \in \mathbb{Z}$ and $x \in [0,1)$, \begin{align*} (U_{T_\alpha}e_n)(x) &= e_n(T_\alpha x) \\ &= e^{2\pi i n(x+\alpha-\lfloor x+\alpha\rfloor)} \\ &= e^{2\pi i n\alpha}\,e^{2\pi i n x}\,e^{-2\pi i n\lfloor x+\alpha\rfloor}. \end{align*} The last factor equals $1$ because both $n$ and $\lfloor x+\alpha\rfloor$ are integers. Hence \begin{align*} (U_{T_\alpha}e_n)(x)=e^{2\pi i n\alpha}\,e_n(x). \end{align*} So every Fourier character is an eigenfunction of $U_{T_\alpha}$, with eigenvalue $e^{2\pi i n\alpha}$ at frequency $n$.[/guided]

[step:Translate invariance into Fourier coefficient equations]Let $f \in L^2([0,1),\lambda;\mathbb{C})$ satisfy $U_{T_\alpha}f=f$. Write its Fourier expansion as \begin{align*} f=\sum_{n\in\mathbb{Z}}\widehat f(n)\,e_n \end{align*} in $L^2([0,1),\lambda;\mathbb{C})$. The Koopman operator $U_{T_\alpha}$ is bounded (it is an isometry by the previous step) and $U_{T_\alpha}e_n=e^{2\pi i n\alpha}e_n$, so applying $U_{T_\alpha}$ term by term to the $L^2$-convergent series gives \begin{align*} U_{T_\alpha}f = \sum_{n\in\mathbb{Z}} e^{2\pi i n\alpha}\,\widehat f(n)\,e_n \end{align*} in $L^2([0,1),\lambda;\mathbb{C})$. Comparing Fourier coefficients in the orthonormal basis $(e_n)_{n\in\mathbb{Z}}$ with the expansion of $f$ gives, for every $n \in \mathbb{Z}$, \begin{align*} e^{2\pi i n\alpha}\,\widehat f(n)=\widehat f(n), \end{align*} equivalently \begin{align*} (e^{2\pi i n\alpha}-1)\,\widehat f(n)=0. \end{align*}[/step]

[guided]Suppose $f \in L^2$ is invariant under the rotation in the $L^2$ sense: $U_{T_\alpha}f=f$. Expand $f$ in the Fourier basis: \begin{align*} f=\sum_{n\in\mathbb{Z}}\widehat f(n)\,e_n \end{align*} with convergence in $L^2([0,1),\lambda;\mathbb{C})$. Because $T_\alpha$ preserves $\lambda$ (verified two steps above), the Koopman operator $U_{T_\alpha}$ is an isometry on $L^2([0,1),\lambda;\mathbb{C})$, hence bounded and continuous. Bounded linear operators commute with $L^2$-convergent series, so \begin{align*} U_{T_\alpha}f = \sum_{n\in\mathbb{Z}}\widehat f(n)\,U_{T_\alpha}e_n. \end{align*} Substituting the eigenvalue computation from the previous step, \begin{align*} U_{T_\alpha}f = \sum_{n\in\mathbb{Z}}e^{2\pi i n\alpha}\,\widehat f(n)\,e_n. \end{align*} Since $U_{T_\alpha}f=f$, we have two Fourier expansions of the same $L^2$ function: \begin{align*} \sum_{n\in\mathbb{Z}}e^{2\pi i n\alpha}\,\widehat f(n)\,e_n = \sum_{n\in\mathbb{Z}}\widehat f(n)\,e_n. \end{align*} Uniqueness of Fourier coefficients in an orthonormal basis then gives, for each integer $n$, \begin{align*} e^{2\pi i n\alpha}\,\widehat f(n)=\widehat f(n), \end{align*} or equivalently \begin{align*} (e^{2\pi i n\alpha}-1)\,\widehat f(n)=0. \end{align*} This equation is the entire spectral content of the proof: an invariant function can only have Fourier mass at frequencies whose eigenvalue is $1$.[/guided]

[step:Use irrationality to force invariant functions to be constant]Assume $\alpha \notin \mathbb{Q}$. If $n \in \mathbb{Z}\setminus\{0\}$ and $e^{2\pi i n\alpha}=1$, then $n\alpha \in \mathbb{Z}$, so \begin{align*} \alpha=\frac{n\alpha}{n}\in \mathbb{Q}, \end{align*} a contradiction. Hence $e^{2\pi i n\alpha}\neq 1$ for every $n\in\mathbb{Z}\setminus\{0\}$. The coefficient equation from the previous step then forces \begin{align*} \widehat f(n)=0 \end{align*} for every $n\in\mathbb{Z}\setminus\{0\}$, so \begin{align*} f=\widehat f(0)\,e_0 \end{align*} in $L^2([0,1),\lambda;\mathbb{C})$. Since $e_0(x)=1$ for every $x\in[0,1)$, every $U_{T_\alpha}$-invariant function in $L^2([0,1),\lambda;\mathbb{C})$ is constant $\lambda$-almost everywhere. The [Spectral Characterization of Ergodicity](/theorems/spectral-characterization-ergodicity) states that a measure-preserving transformation of a probability space is [ergodic](/page/Ergodicity) if and only if every $U_T$-invariant function in $L^2$ is constant almost everywhere. Its hypotheses are satisfied: $([0,1),\mathcal{B}([0,1)),\lambda)$ is a probability space because $\lambda([0,1)) = 1$, and $T_\alpha$ preserves $\lambda$ by the step "Verify that $T_\alpha$ preserves Lebesgue measure on $[0,1)$". Therefore $T_\alpha$ is ergodic when $\alpha$ is irrational.[/step]

[guided]Now assume $\alpha$ is irrational. The coefficient equation from the previous step forces every Fourier coefficient $\widehat f(n)$ to vanish unless the eigenvalue $e^{2\pi i n\alpha}$ equals $1$. We pin down exactly when that can happen. If $n \neq 0$ and $e^{2\pi i n\alpha}=1$, then the exponent $n\alpha$ must be an integer. Therefore \begin{align*} \alpha=\frac{n\alpha}{n}. \end{align*} The numerator is an integer and the denominator is a nonzero integer, so this equation would force $\alpha\in\mathbb{Q}$, contradicting the hypothesis $\alpha\notin\mathbb{Q}$. Hence, for every nonzero integer $n$, \begin{align*} e^{2\pi i n\alpha}\neq 1. \end{align*} Returning to \begin{align*} (e^{2\pi i n\alpha}-1)\,\widehat f(n)=0, \end{align*} we conclude \begin{align*} \widehat f(n)=0 \end{align*} for every $n\neq 0$. Only the zero Fourier mode remains: \begin{align*} f=\widehat f(0)\,e_0 \end{align*} in $L^2([0,1),\lambda;\mathbb{C})$. Since $e_0$ is the constant function $1$, this says that $f$ is constant $\lambda$-almost everywhere. We now invoke the [Spectral Characterization of Ergodicity](/theorems/spectral-characterization-ergodicity). Its hypotheses are (i) a probability space and (ii) a measure-preserving transformation. We verify both: $([0,1),\mathcal{B}([0,1)),\lambda)$ is a probability space because $\lambda([0,1)) = 1$, and the step "Verify that $T_\alpha$ preserves Lebesgue measure on $[0,1)$" established that $T_\alpha$ preserves $\lambda$. The theorem then states that ergodicity is equivalent to: every $U_{T_\alpha}$-invariant function in $L^2$ is constant almost everywhere. We have proved precisely that condition. Therefore $T_\alpha$ is [ergodic](/page/Ergodicity) for irrational $\alpha$.[/guided]

[step:Use rationality to construct a nonconstant invariant eigenfunction]Assume $\alpha\in\mathbb{Q}$. Choose integers $p\in\mathbb{Z}$ and $q\in\mathbb{N}$ with $\alpha=p/q$. Then $q\alpha=p\in\mathbb{Z}$, so \begin{align*} e^{2\pi i q\alpha}=e^{2\pi i p}=1. \end{align*} The eigenvalue computation from the Koopman step gives \begin{align*} U_{T_\alpha}e_q=e^{2\pi i q\alpha}\,e_q=e_q, \end{align*} so $e_q \in L^2([0,1),\lambda;\mathbb{C})$ is $U_{T_\alpha}$-invariant. The function $e_q$ is not constant $\lambda$-almost everywhere: it is continuous on $[0,1)$ with \begin{align*} e_q(0)=1 \qquad\text{and}\qquad e_q\!\left(\tfrac{1}{2q}\right)=e^{\pi i}=-1. \end{align*} The preimages $U_+ := e_q^{-1}(\{z \in \mathbb{C} : |z-1| < 1/2\})$ and $U_- := e_q^{-1}(\{z \in \mathbb{C} : |z+1| < 1/2\})$ are open subsets of $[0,1)$ containing $0$ and $1/(2q)$ respectively, hence each has positive $\lambda$-measure. The sets $U_+$ and $U_-$ are disjoint because $|z-1|<1/2$ and $|z+1|<1/2$ cannot hold simultaneously. So no constant $c \in \mathbb{C}$ can agree with $e_q$ on a set of full $\lambda$-measure: either $|c-1| \geq 1/2$, in which case $e_q$ disagrees with $c$ on $U_+$, or $|c+1| \geq 1/2$, in which case $e_q$ disagrees with $c$ on $U_-$. Hence $e_q$ is not $\lambda$-a.e. constant. By the [Spectral Characterization of Ergodicity](/theorems/spectral-characterization-ergodicity), applied to the probability space $([0,1),\mathcal{B}([0,1)),\lambda)$ and the $\lambda$-preserving transformation $T_\alpha$ (preservation verified two steps above), the existence of a nonconstant $U_{T_\alpha}$-invariant $L^2$ function shows that $T_\alpha$ is not ergodic. Combined with the irrational case, this proves that $T_\alpha$ is ergodic with respect to $\lambda$ if and only if $\alpha\notin\mathbb{Q}$.[/step]

[guided]Now assume $\alpha$ is rational. Choose integers $p\in\mathbb{Z}$ and $q\in\mathbb{N}$ such that \begin{align*} \alpha=\frac{p}{q}. \end{align*} Then $q\alpha=p$ is an integer, so \begin{align*} e^{2\pi i q\alpha}=e^{2\pi i p}=1. \end{align*} The Fourier character at frequency $q$ is \begin{align*} e_q: [0,1) &\to \mathbb{C} \\ x &\mapsto e^{2\pi i qx}. \end{align*} Using the eigenvalue computation from the Koopman step, \begin{align*} U_{T_\alpha}e_q=e^{2\pi i q\alpha}\,e_q=e_q, \end{align*} so $e_q$ is an invariant $L^2$ function. We must check that this invariant function is not constant almost everywhere. Evaluate it at two points: \begin{align*} e_q(0)=1 \qquad\text{and}\qquad e_q\!\left(\tfrac{1}{2q}\right)=e^{2\pi i q/(2q)}=e^{\pi i}=-1. \end{align*} Because $e_q$ is continuous, the preimages $U_+ := e_q^{-1}(\{z : |z-1| < 1/2\})$ and $U_- := e_q^{-1}(\{z : |z+1| < 1/2\})$ are open subsets of $[0,1)$ containing $0$ and $1/(2q)$ respectively. Nonempty open subsets of $[0,1)$ have positive $\lambda$-measure, so $\lambda(U_+) > 0$ and $\lambda(U_-) > 0$. The sets $U_+$ and $U_-$ are disjoint because the discs $\{|z-1|<1/2\}$ and $\{|z+1|<1/2\}$ are disjoint in $\mathbb{C}$. So no constant $c \in \mathbb{C}$ can agree with $e_q$ on a set of full $\lambda$-measure: either $|c-1| \geq 1/2$ (and $e_q$ disagrees with $c$ on $U_+$) or $|c+1| \geq 1/2$ (and $e_q$ disagrees with $c$ on $U_-$). Either way, $e_q$ differs from $c$ on a set of positive $\lambda$-measure, so $e_q$ is not constant $\lambda$-almost everywhere. The [Spectral Characterization of Ergodicity](/theorems/spectral-characterization-ergodicity), applied to $([0,1),\mathcal{B}([0,1)),\lambda)$ and the $\lambda$-preserving transformation $T_\alpha$, says that ergodicity is equivalent to all invariant $L^2$ functions being constant $\lambda$-a.e. We have produced a nonconstant invariant function, so $T_\alpha$ is not ergodic when $\alpha$ is rational. Together with the irrational case, this proves the stated equivalence.[/guided]