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Use finite cutoff windows to avoid subtracting two infinite masses. Write $D$ for the unit disc, $D_U={\mathbf{L}}_D\setminus {\mathbf{L}}_U$ on loops surrounding $0$, and $A(U)=\nu (D_U)$. For $0<r<1$, let $H_r=D_{rD}$. These increase to all loops in ${\mathbf{L}}_D$ surrounding $0$ as $r\downarrow 0$.

On the finite-mass windows, the domain-containment events form a pi-system. Indeed, if $U,V$ are simply connected and contain $0$, a simple loop surrounding $0$ that lies in both has its filled interior in both. It therefore lies in the component $W$ of $U\cap V$ containing $0$, which is simply connected. Thus, on the support of $\nu$,Inclusion-exclusion for the finite deficits givesTaking $V=rD$ and subtracting from $\nu (H_r)$ gives the useful recovery formulaThe assumed generation of the loop sigma-field by domain-containment events and the uniqueness theorem for measures now determine the finite restriction to $H_r$. Increasing these windows recovers ${\nu }_D$; conformal transport recovers its restriction to any simply connected domain containing the marked point. If a simply connected domain omits $0$, no loop in it can surround $0$.

The argument uses finite annular deficits. This is the usual local-finiteness convention for these loop measures and is explicitly ensured by the finite logarithmic formula assumed in the continuation. Bare sigma-finiteness must not be used as permission for formal $\infty -\infty$ subtraction. This is the recovery of a loop measure from conformal deficits.

On a fixed underlying set, a pi-system is a nonempty family closed under finite intersections. A Dynkin system, also called a d-system, contains the whole underlying set, is closed under complements relative to it, and under countable disjoint unions. Equivalently it is closed under differences of nested members and increasing countable unions. A sigma-algebra contains the whole set, is closed under complements and arbitrary countable unions, hence also countable intersections. A Dynkin system need not be a sigma-algebra unless additional intersection closure is available.

Let $V$ be the closure of the simple predictable processes in the Hilbert space $L^2(\mathcal{P},\mu )$. Common refinement of deterministic partitions makes these processes a vector space. To prove density of simple predictable processes for finite measures, suppose $g\in V^{\perp }$. Since $\mu$ is a finite measure, the Cauchy-Schwarz inequality gives $g\in L^1(\mu )$, sodefines a finite signed measure. The indicator of each predictable generating rectangle, including each time-zero rectangle, is a simple predictable process; hence $\nu$ vanishes on these rectangles.

Finite intersections of the rectangles are again rectangles or empty. Add the whole space to this pi-system. Its $\nu$-mass is zero because $1_{\Omega \times [0,n]}$ is simple and converges to $1$ in $L^2(\mu )$ by the dominated convergence theorem. The zero sets of $\nu$ form a lambda-system now that its total mass is zero. The Pi-lambda theorem therefore gives $\nu (E)=0$ for every $E\in \mathcal{P}$. Taking the sets where $g>0$ and $g<0$ gives $g=0$ $\mu$-almost everywhere.

Thus $V^{\perp }=\{0\}$ and the orthogonal complement criterion for a closed subspace of a Hilbert space yieldsOmitting the time-zero term would make this conclusion false when $\mu$ is concentrated at time zero.

If $\theta$ is measure-preserving, equality holds on every measurable set and hence on the specified $\pi$-system. Conversely, let $\mathcal{D}=\{A\in \mathcal{E}:{\mu }_{\theta }(A)=\mu (A)\}$. Both finite measures have equal total mass. Therefore $E\in \mathcal{D}$, complements of members remain in $\mathcal{D}$ by subtracting from the finite total mass, and disjoint countable unions remain in $\mathcal{D}$ by countable additivity. Thus $\mathcal{D}$ is a Dynkin system.

It contains the generating pi-system $\mathcal{A}$, so Dynkin lemma gives $\mathcal{E}=\sigma (\mathcal{A})\subseteq \mathcal{D}$. ConsequentlyThis is exactly the measure-preserving property.

The map swaps each adjacent pair of binary digits: $({\omega }_1,{\omega }_2,{\omega }_3,{\omega }_4,\dots )\mapsto ({\omega }_2,{\omega }_1,{\omega }_4,{\omega }_3,\dots )$. Canonical expansions with no eventual all-ones tail remain canonical under this swap, and ${\theta }^2=\mathrm{id}$.

Under Lebesgue measure, any specified values of $m$ distinct binary digits have probability $2^{-m}$. Indeed they describe a union of dyadic intervals at the largest of the specified positions, with that total length. The preimage of a cylinder prescribing its first $m$ output digits prescribes $m$ distinct input digits and therefore has the same measure. The half-open dyadic intervals, with the empty set, form a pi-system generating the Borel sigma-algebra. Apply the preceding uniqueness result to obtainThe choice of binary expansion matters at dyadic endpoints for the map's pointwise definition; the chosen convention removes that ambiguity and those endpoints are null in the measure calculation.

The product sigma-algebra isThe product measure ${\mu }_1\otimes {\mu }_2$ is the probability measure on this sigma-algebra satisfyingThe measurable rectangles form a pi-system generating ${\mathcal{A}}_1\otimes {\mathcal{A}}_2$. Any two probability measures agreeing on them therefore agree on the generated sigma-algebra by the pi-lambda theorem, proving uniqueness.

On a probability space carrying an infinite IID sequence $(X_n)$ with law $m$, map $\omega$ to $(X_1(\omega ),X_2(\omega ),\dots )$. Its pushforward measure $\mu$ satisfiesfor every cylinder set. Cylinder sets form a pi-system generating $\sigma (\mathcal{C})$, so the sigma-finite uniqueness theorem for measures proves uniqueness. This is the countable product measure $m^{\otimes \mathbb{N}}$.

For a cylinder $A=A_1\times \cdots \times A_N\times \mathbb{R}\times \cdots$,The same generating-class argument extends this equality to every measurable set, so $\theta$ is a measure-preserving transformation.

If $A={\theta }^{-1}A$, then $A={\theta }^{-n}A$ for every $n$, so membership in $A$ is independent of the first $n$ coordinates. Thus $A$ belongs to the tail sigma-algebra. The zero-one law gives $\mu (A)\in \{0,1\}$, proving that the shift is ergodic.