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A conditional probability of an invariant probability measure given its invariant sigma-algebra. In a standard Borel probability system, almost every such probability is invariant and ergodic. Their integral recovers the original measure. The countable-test proof of ergodicity of conditional components explains why invariance alone is not the whole conclusion.

Use a standard Borel probability space so that conditional measures can be realized by measures. The disintegration theorem for a probability measure states the following. For a measurable map $\pi :X\to Y$ between standard Borel spaces and $\nu ={\pi }_{\ast }\mu$, there is a measurable probability kernel $y\mapsto {\mu }_y$, unique for $\nu$-almost every $y$, such thatMeasurability means $y\mapsto {\mu }_y(A)$ is measurable for every Borel set $A$. More generallyfor integrable $f$, with equality of the last expression almost everywhere. The ${\mu }_y$ are the conditional measures of a factor. This also defines conditional measures for a countably generated sub-sigma-algebra, or a measurable partition represented by such a map. Sigma-algebras are considered modulo null sets where necessary.

For an invariant probability measure, disintegrate over the invariant sigma-algebra $\mathcal{I}$. On a standard probability space it can be represented, modulo null sets, by a countably generated measurable factor $\pi$. One justification for this countability is the separability of $L^2$: a countable dense family of $\mathcal{I}$-measurable functions generates $\mathcal{I}$ modulo null sets. The resulting conditional measures are the ergodic components:Their invariance is one of the permitted facts. It remains to prove that almost every component is ergodic, rather than simply invariant.

Choose a countable generating algebra $\mathcal{A}$ generating the Borel sigma-algebra of $X$. Apply the Birkhoff ergodic theorem simultaneously to its indicators. For $\mu$-almost every $z$ and all $A\in \mathcal{A}$,Disintegration transfers this common full-measure set to ${\mu }_y$-almost every $z$, for $\nu$-almost every $y$. On that fiber the conditional expectation is constant and equals ${\mu }_y(A)$. Thus, simultaneously for every $A\in \mathcal{A}$,Fix such a $y$, for which ${\mu }_y$ is also invariant. Applying the Birkhoff ergodic theorem to the system with measure ${\mu }_y$ identifies these limits withFinite linear combinations of the algebra's indicators are dense in $L^2({\mu }_y)$, by the Monotone class theorem. Since conditional expectation is an $L^2$ contraction, its projection onto invariant functions consequently sends every $L^2({\mu }_y)$ function to its constant integral. In particular, for any ${\mu }_y$-invariant set $D$,and so ${\mu }_y(D)\in \{0,1\}$. This proves almost every component is ergodic. The countable-test proof of ergodicity of conditional components avoids taking an invalid intersection of uncountably many full-measure sets.

For rotation by $1/5$, take the quotient map $\pi (x)=5x(\mathrm{mod}1)$. Its fibers are exactly the five-point orbits. The ergodic components of a rational circle rotation are thereforeEquivalently the component through $x$ is $\frac{1}{5}{\sum }_{j=0}^4{\delta }_{x+j/5}$. To check the disintegration, for bounded measurable $f$ use the change of variables on the five consecutive fifth-intervals:The rotation cyclically permutes the five atoms, preserving their equal masses. A set invariant modulo ${\mu }_y$ must contain all or none of this finite orbit, since every atom has positive mass. Thus each ${\mu }_y$ is ergodic. Finally, the invariant sigma-algebra is the pullback under $\pi$: a function invariant under translation by $1/5$ is constant on each orbit and is a measurable function of $5x(\mathrm{mod}1)$. This confirms that these are the conditional measures over $\mathcal{I}$, not merely some decomposition into invariant measures.