Lebesgue–Rokhlin probability space

In probability theory, a standard probability space, also called Lebesgue–Rokhlin probability space or just Lebesgue space (the latter term is ambiguous) is a probability space satisfying certain assumptions introduced by Vladimir Rokhlin in 1940. Informally, it is a probability space consisting of an interval and/or a finite or countable number of atoms.

The theory of standard probability spaces was started by von Neumann in 1932 and shaped by Vladimir Rokhlin in 1940. Rokhlin showed that the unit interval endowed with the Lebesgue measure has important advantages over general probability spaces, yet can be effectively substituted for many of these in probability theory. The dimension of the unit interval is not an obstacle, as was clear already to Norbert Wiener. He constructed the Wiener process (also called Brownian motion) in the form of a measurable map from the unit interval to the space of continuous functions.

Short history

The theory of standard probability spaces was started by von Neumann in 1932 and shaped by Vladimir Rokhlin in 1940. For modernized presentations see , , and .

Nowadays standard probability spaces may be (and often are) treated in the framework of descriptive set theory, via standard Borel spaces, see for example . This approach is based on the isomorphism theorem for standard Borel spaces . An alternate approach of Rokhlin, based on measure theory, neglects null sets, in contrast to descriptive set theory.
Standard probability spaces are used routinely in ergodic theory,"Ergodic theory on Lebesgue spaces" is the subtitle of the book .

Definition

One of several well-known equivalent definitions of the standardness is given below, after some preparations. All probability spaces are assumed to be complete.

Isomorphism

An isomorphism between two probability spaces $\textstyle (\Omega_1,\mathcal_1,P_1)$, $\textstyle (\Omega_2,\mathcal_2,P_2)$ is an invertible map $\textstyle f : \Omega_1 \to \Omega_2$ such that $\textstyle f$ and $\textstyle f^$ both are (measurable and) measure preserving maps.

Two probability spaces are isomorphic if there exists an isomorphism between them.

Isomorphism modulo zero

Two probability spaces $\textstyle (\Omega_1,\mathcal_1,P_1)$, $\textstyle (\Omega_2,\mathcal_2,P_2)$ are isomorphic $\textstyle \operatorname \, 0$ if there exist null sets $\textstyle A_1 \subset \Omega_1$, $\textstyle A_2 \subset \Omega_2$ such that the probability spaces $\textstyle \Omega_1 \setminus A_1$, $\textstyle \Omega_2 \setminus A_2$ are isomorphic (being endowed naturally with sigma-fields and probability measures).

Standard probability space

A probability space is standard, if it is isomorphic $\textstyle \operatorname \, 0$ to an interval with Lebesgue measure, a finite or countable set of atoms, or a combination (disjoint union) of both.

See , , and . See also , and . In the measure is assumed finite, not necessarily probabilistic. In atoms are not allowed.

Examples of non-standard probability spaces

A naive white noise

The space of all functions $\textstyle f : \mathbb \to \mathbb$ may be thought of as the product $\textstyle \mathbb^\mathbb$ of a continuum of copies of the real line $\textstyle \mathbb$. One may endow $\textstyle \mathbb$ with a probability measure, say, the standard normal distribution $\textstyle \gamma = N(0,1)$, and treat the space of functions as the product $\textstyle (\mathbb,\gamma)^\mathbb$ of a continuum of identical probability spaces $\textstyle (\mathbb,\gamma)$. The product measure $\textstyle \gamma^\mathbb$ is a probability measure on $\textstyle \mathbb^\mathbb$. Naively it might seem that $\textstyle \gamma^\mathbb$ describes white noise.

However, the integral of a white noise function from 0 to 1 should be a random variable distributed ''N''(0, 1). In contrast, the integral (from 0 to 1) of $\textstyle f \in \textstyle (\mathbb,\gamma)^\mathbb$ is undefined. ''ƒ'' also fails to be almost surely measurable, and the probability of ''ƒ'' being measurable is undefined. Indeed, if ''X'' is a random variable distributed (say) uniformly on (0, 1) and independent of ''ƒ'', then ''ƒ''(''X'') is not a random variable at all (it lacks measurability).

A perforated interval

Let $\textstyle Z \subset (0,1)$ be a set whose inner Lebesgue measure is equal to 0, but outer