A measure tells us how large a set is, but dynamics asks a sharper question: what happens to size when a system evolves? If $T: E \to E$ moves every point of a [measure space](/page/Measure%20Space) $(E, \mathcal E, \mu)$ to its next position, then the image of a set can be distorted, folded, or spread out. The central question is whether the statistical size of every observable event is unchanged by this motion. A measure-preserving transformation is the answer: it is a measurable self-map whose inverse images have the same measure as the original sets.

The inverse image is the right object because events are tested by asking where they came from. If $A \in \mathcal E$ is an event after one step, then $T^{-1}(A)$ is the event that the starting point lands in $A$ after one step. Preserving measure means that the probability, mass, or volume of seeing $A$ after applying the system is the same as seeing $A$ before applying it. This turns a deterministic map into a statistical symmetry.

[example: A Rotation That Preserves Arc Length] Let $E=[0,1)$, $\mathcal E=\mathcal B([0,1))$, and $\mu=\mathcal L^1|_{[0,1)}$. Fix $\alpha\in[0,1)$ and define \begin{align*} T(x)=x+\alpha \pmod 1. \end{align*} We compute the preimage of a half-open interval $A=[a,b)\subsetneq[0,1)$, where $0\leq a<b\leq 1$. If $a\geq \alpha$, then \begin{align*} T^{-1}(A)=[a-\alpha,b-\alpha) \end{align*} because $x+\alpha\in[a,b)$ exactly when $x\in[a-\alpha,b-\alpha)$. Hence \begin{align*} \mu(T^{-1}(A))=(b-\alpha)-(a-\alpha)=b-a=\mu(A). \end{align*} If $b\leq \alpha$, then $x+\alpha\pmod 1\in[a,b)$ exactly when $x+\alpha-1\in[a,b)$, so \begin{align*} T^{-1}(A)=[1+a-\alpha,1+b-\alpha). \end{align*} Therefore \begin{align*} \mu(T^{-1}(A))=(1+b-\alpha)-(1+a-\alpha)=b-a=\mu(A). \end{align*} It remains to handle the case $a<\alpha<b$, where the interval crosses the cut point after shifting backward. Then $x+\alpha\pmod 1\in[a,b)$ holds either when $x+\alpha\in[\alpha,b)$ or when $x+\alpha-1\in[a,\alpha)$, so \begin{align*} T^{-1}(A)=[0,b-\alpha)\cup[1+a-\alpha,1). \end{align*} The two intervals are disjoint, and their total length is \begin{align*} (b-\alpha)+\bigl(1-(1+a-\alpha)\bigr)=(b-\alpha)+(\alpha-a)=b-a. \end{align*} Thus $\mu(T^{-1}(A))=\mu(A)$ for every half-open interval $[a,b)\subsetneq[0,1)$. Since the half-open intervals generate $\mathcal B([0,1))$, the equality extends to all Borel sets by the standard uniqueness principle for measures generated by intervals. The rotation can move every point, but it leaves the Lebesgue distribution of points unchanged. [/example]

The rotation example is deliberately simple, but it already contains the main lesson. A transformation may preserve measure without fixing sets pointwise. It may also fail to be invertible as a function while still preserving measure through preimages. The theory is therefore about measurable dynamics rather than geometric rigidity: the important question is what happens to measured events under time evolution.

## Definition

### Preservation by Preimages

Before defining the transformation, recall that this is a child concept of measure. The ambient structure is already a measure space: a set equipped with a $\sigma$-algebra of measurable events and a countably additive measure. The new ingredient is a measurable self-map that acts on that measured world without changing the measure of any event when pulled back.

[definition: Measure-Preserving Transformation] Let $(E, \mathcal E, \mu)$ be a measure space. A measurable map $T: E \to E$ is a measure-preserving transformation if \begin{align*} \mu(T^{-1}(A)) = \mu(A) \end{align*} for every $A \in \mathcal E$. [/definition]

The definition uses preimages, not images. Images of measurable sets need not be measurable under an arbitrary measurable map, and even when they are measurable, image size is not the right invariant for non-injective maps. Preimages always respect the $\sigma$-algebra: if $A$ is measurable and $T$ is measurable, then $T^{-1}(A)$ is measurable.

Many applications use total mass $1$, because the measure describes a probability distribution on possible states. It is useful to name this common case separately: the same equation is now read as stationarity of probabilities under time evolution.

[definition: Probability-Preserving Transformation] Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space. A measurable map $T: \Omega \to \Omega$ is a probability-preserving transformation if \begin{align*} \mathbb P(T^{-1}(A)) = \mathbb P(A) \end{align*} for every $A \in \mathcal F$. [/definition]

This is the same definition with the letter $\mathbb P$ emphasizing probability. In much of ergodic theory, by convention, the abbreviation m.p.t. refers to a probability-preserving transformation unless a more general measure space is explicitly stated.

### Pushforwards and Observable Tests

The next task is to express the same preservation law as an equality between measures. That reformulation is useful because dynamics often starts with a map and asks which measures remain fixed by it, and because later examples will move between set-level preservation and distribution-level invariance.

[definition: Pushforward Measure] Let $(E, \mathcal E)$ and $(F, \mathcal F)$ be measurable spaces, let $\mu$ be a measure on $(E, \mathcal E)$, and let $T: E \to F$ be measurable. The pushforward measure $T_\#\mu$ on $(F, \mathcal F)$ is defined by \begin{align*} (T_\#\mu)(B) = \mu(T^{-1}(B)) \end{align*} for every $B \in \mathcal F$. [/definition]

Once pushforwards are available, there are two apparently different ways to say that no mass changes under the dynamics: every event has the same measure as its preimage, or the whole measure is unchanged after being pushed forward by the map. The useful point is that these are not separate requirements. They are the same condition, and that equivalence lets later arguments switch between event-level tests and measure-level fixed points without adding any new hypothesis.

This raises the practical test for invariance: when a transformation is given, should one verify preservation by checking all measurable events, or by checking a single equality of measures? The following result supplies exactly that bridge, turning the setwise preservation condition into a pushforward fixed-point condition.

[quotetheorem:8380]

This characterisation is often the most compact way to remember the definition. It also makes measure preservation look like a fixed-point condition: the measure $\mu$ is fixed by the operation of pushing measures forward along $T$.

A tempting but wrong alternative would be to ask for $\mu(T(A))=\mu(A)$ for measurable sets $A$. That condition is unreliable: images may fail to be measurable, and non-injective maps can preserve preimage-measure while collapsing many points onto the same image event. The next example makes the failure visible and prepares the observable viewpoint: if images are the wrong primitive, then events and functions pulled back along $T$ become the natural tests.