[proofplan] We first put the doubling map on the precise probability space $X=[0,1)$ with Lebesgue measure and verify that it preserves that measure. Then we invoke the $L^2$ fixed-function criterion for ergodicity. A fixed function is expanded against the Fourier basis $e_n(x)=\exp(2\pi i n x)$; invariance under the doubling map forces the coefficient identity $\widehat f(2n)=\widehat f(n)$. Parseval's theorem then forces every nonzero Fourier coefficient to vanish, so every fixed $L^2$ function is constant, which is exactly ergodicity by the criterion. [/proofplan] [step:Verify that the doubling map preserves Lebesgue measure] Let $X := [0,1)$, let $\mathcal A$ denote the Lebesgue $\sigma$-algebra on $X$, and let $\mu := \mathcal L^1\!\restriction_X$ be one-dimensional Lebesgue measure restricted to $X$. We use the representative \begin{align*} T:X &\to X \\ x &\mapsto \begin{cases} 2x, & 0 \leq x < \frac{1}{2},\\ 2x-1, & \frac{1}{2} \leq x < 1. \end{cases} \end{align*} Define the two inverse-branch maps \begin{align*} S_0:X &\to \left[0,\frac{1}{2}\right) \\ y &\mapsto \frac{y}{2}, \end{align*} and \begin{align*} S_1:X &\to \left[\frac{1}{2},1\right) \\ y &\mapsto \frac{y+1}{2}. \end{align*} For every $A \in \mathcal A$, \begin{align*} T^{-1}(A) = S_0(A) \cup S_1(A), \end{align*} and the union is disjoint because the ranges of $S_0$ and $S_1$ are disjoint. Since $S_0$ and $S_1$ are affine maps with scaling factor $\frac{1}{2}$, the scaling and translation invariance of Lebesgue measure gives \begin{align*} \mu(S_0(A)) = \frac{1}{2}\mu(A), \qquad \mu(S_1(A)) = \frac{1}{2}\mu(A). \end{align*} Therefore \begin{align*} \mu(T^{-1}(A)) = \mu(S_0(A))+\mu(S_1(A)) = \mu(A). \end{align*} Thus $T$ is measurable and measure preserving on $(X,\mathcal A,\mu)$. [/step] [step:Reduce ergodicity to the absence of nonconstant fixed functions] Since $\mu(X)=1$ and $T$ is measure preserving, the hypotheses of the fixed-function criterion in [Equivalence of Ergodicity Conditions](/theorems/3444) are satisfied. Define the Koopman operator \begin{align*} U_T:L^2(X,\mathcal A,\mu;\mathbb C) &\to L^2(X,\mathcal A,\mu;\mathbb C) \\ h &\mapsto h \circ T. \end{align*} The criterion says that $T$ is ergodic if and only if every $h \in L^2(X,\mathcal A,\mu;\mathbb C)$ satisfying $U_T h=h$ is $\mu$-almost everywhere constant. It remains to prove this fixed-function statement. Let \begin{align*} f:X &\to \mathbb C \end{align*} be a measurable representative of an element of $L^2(X,\mathcal A,\mu;\mathbb C)$ such that $f \circ T = f$ in $L^2(X,\mathcal A,\mu;\mathbb C)$, equivalently $\mu$-almost everywhere. [guided] The fixed-function criterion converts ergodicity into an $L^2$ statement. We have already verified that $T$ preserves the probability measure $\mu$, so [Equivalence of Ergodicity Conditions](/theorems/3444) applies to the probability-preserving system $(X,\mathcal A,\mu,T)$. It says that $T$ is ergodic exactly when the only fixed vectors of the Koopman operator are the almost everywhere constant functions. Define \begin{align*} U_T:L^2(X,\mathcal A,\mu;\mathbb C)&\to L^2(X,\mathcal A,\mu;\mathbb C)\\ h&\mapsto h\circ T. \end{align*} Thus it is enough to take an arbitrary measurable representative \begin{align*} f:X&\to\mathbb C \end{align*} of an $L^2$ class satisfying $f\circ T=f$ $\mu$-almost everywhere and prove that $f$ is $\mu$-almost everywhere constant. The remaining steps do this by Fourier coefficients. [/guided] [/step] [step:Compare Fourier coefficients using the identity $e_n \circ T=e_{2n}$] For each $n \in \mathbb Z$, define \begin{align*} e_n:X &\to \mathbb C \\ x &\mapsto \exp(2\pi i n x), \end{align*} where $i \in \mathbb C$ is the imaginary unit. Define the Fourier coefficient map \begin{align*} \widehat f:\mathbb Z &\to \mathbb C \\ n &\mapsto \int_X f(x)\overline{e_n(x)} \, d\mu(x). \end{align*} The integral is finite because $f \in L^2(X,\mu)$, $|e_n|=1$, and the [Cauchy-Schwarz Inequality](/theorems/432) gives \begin{align*} \int_X |f(x)\overline{e_n(x)}| \, d\mu(x) \leq \left(\int_X |f(x)|^2 \, d\mu(x)\right)^{1/2} \left(\int_X |e_n(x)|^2 \, d\mu(x)\right)^{1/2} < \infty. \end{align*} For every $x \in X$ and every $n \in \mathbb Z$, the definition of $T$ gives \begin{align*} e_n(T(x)) = e_{2n}(x). \end{align*} Therefore, for each $n \in \mathbb Z$, \begin{align*} \widehat f(2n) &= \int_X f(x)\overline{e_{2n}(x)} \, d\mu(x)\\ &= \int_X f(T(x))\overline{e_{2n}(x)} \, d\mu(x)\\ &= \int_X f(T(x))\overline{e_n(T(x))} \, d\mu(x)\\ &= \int_X \bigl(f\overline{e_n}\bigr)(T(x)) \, d\mu(x)\\ &= \int_X f(x)\overline{e_n(x)} \, d\mu(x)\\ &= \widehat f(n). \end{align*} The second equality uses $f\circ T=f$ $\mu$-almost everywhere, and the fifth equality uses the change-of-variables formula for measure-preserving maps applied to the integrable function $f\overline{e_n}$. [guided] For each integer $n$, define the Fourier basis function \begin{align*} e_n:X&\to\mathbb C\\ x&\mapsto \exp(2\pi i n x). \end{align*} The Fourier coefficient \begin{align*} \widehat f(n)=\int_X f(x)\overline{e_n(x)}\,d\mu(x) \end{align*} is well-defined: $f\in L^2$, $|e_n|=1$, $\mu(X)=1$, and [Cauchy-Schwarz Inequality](/theorems/432) gives \begin{align*} \int_X |f(x)\overline{e_n(x)}|\,d\mu(x) \leq \left(\int_X |f(x)|^2\,d\mu(x)\right)^{1/2} \left(\int_X |e_n(x)|^2\,d\mu(x)\right)^{1/2} <\infty. \end{align*} The doubling map sends the frequency $n$ to frequency $2n$ because \begin{align*} e_n(T(x))=\exp(2\pi i n(2x \pmod 1))=\exp(2\pi i(2n)x)=e_{2n}(x). \end{align*} Using $f\circ T=f$ $\mu$-almost everywhere and then measure preservation for the integrable function $f\overline{e_n}$, \begin{align*} \widehat f(2n) &=\int_X f(x)\overline{e_{2n}(x)}\,d\mu(x)\\ &=\int_X f(T(x))\overline{e_n(T(x))}\,d\mu(x)\\ &=\int_X (f\overline{e_n})(T(x))\,d\mu(x)\\ &=\int_X f(x)\overline{e_n(x)}\,d\mu(x)\\ &=\widehat f(n). \end{align*} This coefficient identity is the whole dynamical content of the proof. [/guided] [/step] [step:Use Parseval to eliminate all nonzero Fourier coefficients] The trigonometric system $(e_m)_{m\in\mathbb Z}$ is an orthonormal basis for $L^2(X,\mathcal A,\mu;\mathbb C)$. Hence [Parseval's Identity](/theorems/434) gives \begin{align*} \sum_{m\in\mathbb Z} |\widehat f(m)|^2 = \int_X |f(x)|^2 \, d\mu(x) <\infty. \end{align*} Thus $|\widehat f(m)| \to 0$ as $|m|\to\infty$. Fix $n \in \mathbb Z\setminus\{0\}$. Iterating the identity $\widehat f(2m)=\widehat f(m)$ gives \begin{align*} \widehat f(2^k n)=\widehat f(n) \end{align*} for every integer $k\geq 0$. Since $|2^k n|\to\infty$ as $k\to\infty$, Parseval's summability implies \begin{align*} |\widehat f(n)| = \lim_{k\to\infty}|\widehat f(2^k n)| = 0. \end{align*} Therefore $\widehat f(n)=0$ for every $n\neq 0$. [/step] [step:Identify the fixed function as constant and conclude ergodicity] Let $c := \widehat f(0) \in \mathbb C$, and define \begin{align*} g:X &\to \mathbb C \\ x &\mapsto f(x)-c. \end{align*} The Fourier coefficients of $g$ all vanish: for $n\neq 0$ this follows from $\widehat f(n)=0$, and for $n=0$ it follows from the definition of $c$. Applying [Parseval's Identity](/theorems/434) to $g$ gives \begin{align*} \int_X |g(x)|^2 \, d\mu(x) = \sum_{n\in\mathbb Z} |\widehat g(n)|^2 = 0. \end{align*} Hence $g=0$ $\mu$-almost everywhere, so $f=c$ $\mu$-almost everywhere. Thus every $L^2$ fixed function of $U_T$ is almost everywhere constant. By [Equivalence of Ergodicity Conditions](/theorems/3444), the doubling map $T(x)=2x \pmod 1$ is ergodic with respect to Lebesgue measure on $[0,1)$. [/step]