 Ba Space
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Ba Space

In mathematics, the ba space $ba(\Sigma )$ of an algebra of sets $\Sigma$ is the Banach space consisting of all bounded and finitely additive signed measures on $\Sigma$ . The norm is defined as the variation, that is $\|\nu \|=|\nu |(X).$ (Dunford & Schwartz 1958, IV.2.15)

If ? is a sigma-algebra, then the space $ca(\Sigma )$ is defined as the subset of $ba(\Sigma )$ consisting of countably additive measures. (Dunford & Schwartz 1958, IV.2.16) The notation ba is a mnemonic for bounded additive and ca is short for countably additive.

If X is a topological space, and ? is the sigma-algebra of Borel sets in X, then $rca(X)$ is the subspace of $ca(\Sigma )$ consisting of all regular Borel measures on X. (Dunford & Schwartz 1958, IV.2.17)

## Properties

All three spaces are complete (they are Banach spaces) with respect to the same norm defined by the total variation, and thus $ca(\Sigma )$ is a closed subset of $ba(\Sigma )$ , and $rca(X)$ is a closed set of $ca(\Sigma )$ for ? the algebra of Borel sets on X. The space of simple functions on $\Sigma$ is dense in $ba(\Sigma )$ .

The ba space of the power set of the natural numbers, ba(2N), is often denoted as simply $ba$ and is isomorphic to the dual space of the l? space.

### Dual of B(?)

Let B(?) be the space of bounded ?-measurable functions, equipped with the uniform norm. Then ba(?) = B(?)* is the continuous dual space of B(?). This is due to Hildebrandt (1934) and Fichtenholtz & Kantorovich (1934). This is a kind of Riesz representation theorem which allows for a measure to be represented as a linear functional on measurable functions. In particular, this isomorphism allows one to define the integral with respect to a finitely additive measure (note that the usual Lebesgue integral requires countable additivity). This is due to Dunford & Schwartz (1958), and is often used to define the integral with respect to vector measures (Diestel & Uhl 1977, Chapter I), and especially vector-valued Radon measures.

The topological duality ba(?) = B(?)* is easy to see. There is an obvious algebraic duality between the vector space of all finitely additive measures ? on ? and the vector space of simple functions ($\mu (A)=\zeta \left(1_{A}\right)$ ). It is easy to check that the linear form induced by ? is continuous in the sup-norm iff ? is bounded, and the result follows since a linear form on the dense subspace of simple functions extends to an element of B(?)* iff it is continuous in the sup-norm.

### Dual of L?(?)

If ? is a sigma-algebra and ? is a sigma-additive positive measure on ? then the Lp space L?(?) endowed with the essential supremum norm is by definition the quotient space of B(?) by the closed subspace of bounded ?-null functions:

$N_{\mu }:=\{f\in B(\Sigma ):f=0\ \mu {\text{-almost everywhere}}\}.$ The dual Banach space L?(?)* is thus isomorphic to

$N_{\mu }^{\perp }=\{\sigma \in ba(\Sigma ):\mu (A)=0\Rightarrow \sigma (A)=0{\text{ for any }}A\in \Sigma \},$ i.e. the space of finitely additive signed measures on ? that are absolutely continuous with respect to ? (?-a.c. for short).

When the measure space is furthermore sigma-finite then L?(?) is in turn dual to L1(?), which by the Radon-Nikodym theorem is identified with the set of all countably additive ?-a.c. measures. In other words, the inclusion in the bidual

$L^{1}(\mu )\subset L^{1}(\mu )^{**}=L^{\infty }(\mu )^{*}$ is isomorphic to the inclusion of the space of countably additive ?-a.c. bounded measures inside the space of all finitely additive ?-a.c. bounded measures.