# 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 $\|\nu \|=|\nu |(X).$
If Σ is a sigma-algebra, then the space $ca(\Sigma )$ is defined as the subset of $ba(\Sigma )$ consisting of countably additive measures. The notation ba is a mnemonic for bounded additive and ca is short for countably additive.
If X is a topological space, and Σ 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.

## 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 Σ 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(2$^N$), is often denoted as simply $ba$ and is isomorphic to the dual space of the ℓ$^∞$ space.

### Dual of B(Σ)

Let B(Σ) be the space of bounded Σ-measurable functions, equipped with the uniform norm. Then ba(Σ) = B(Σ)* is the continuous dual space of B(Σ). This is due to Hildebrandt and Fichtenholtz & Kantorovich. 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, and is often used to define the integral with respect to vector measures, and especially vector-valued Radon measures.
The topological duality ba(Σ) = B(Σ)* is easy to see. There is an obvious algebraic duality between the vector space of all finitely additive measures σ on Σ 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 σ is continuous in the sup-norm if σ is bounded, and the result follows since a linear form on the dense subspace of simple functions extends to an element of B(Σ)* if it is continuous in the sup-norm.

### Dual of L$^∞$(μ)

If Σ is a sigma-algebra and μ is a sigma-additive positive measure on Σ then the Lp space L$^∞$(μ) endowed with the essential supremum norm is by definition the quotient space of B(Σ) by the closed subspace of bounded μ-null functions:
    $N_{\mu }:=\{f\in B(\Sigma ):f=0\ \mu {\text{-almost everywhere}}\}.$
The dual Banach space L$^∞$(μ)* 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 Σ that are absolutely continuous with respect to μ (μ-a.c. for short).
When the measure space is furthermore sigma-finite then L$^∞$(μ) is in turn dual to L$^1$(μ), which by the Radon–Nikodym theorem is identified with the set of all countably additive μ-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 μ-a.c. bounded measures inside the space of all finitely additive μ-a.c. bounded measures.