# Borel measure

(Redirected from Borel measurable)
In mathematics, specifically in measure theory, a Borel measure is defined as follows: let X be a locally compact Hausdorff space, and let $\mathfrak{B}(X)$ be the smallest σ-algebra that contains the open sets of X; this is known as the σ-algebra of Borel sets. Any measure μ defined on the σ-algebra of Borel sets is called a Borel measure. Some authors require in addition that μ(C) < ∞ for every compact set C. If a Borel measure μ is both inner regular and outer regular, it is called a regular Borel measure (some authors also require it to be tight). If μ is both inner regular and locally finite, it is called a Radon measure. Note that a locally finite Borel measure automatically satisfies μ(C) < ∞ for every compact set C.
The real line $\mathbb R$ with its usual topology is a locally compact Hausdorff space, hence we can define a Borel measure on it. In this case, $\mathfrak{B}(\mathbb R)$ is the smallest σ-algebra that contains the open intervals of $\mathbb R$. While there are many Borel measures μ, the choice of Borel measure which assigns $\mu([a,b])=b-a$ for every interval $[a,b]$ is sometimes called "the" Borel measure on $\mathbb R$. In practice, even "the" Borel measure is not the most useful measure defined on the σ-algebra of Borel sets; indeed, the Lebesgue measure $\lambda$ is an extension of "the" Borel measure which possesses the crucial property that it is a complete measure (unlike the Borel measure). To clarify, when one says that the Lebesgue measure $\lambda$ is an extension of the Borel measure $\mu$, it means that every Borel-measurable set E is also a Lebesgue-measurable set, and the Borel measure and the Lebesgue measure coincide on the Borel sets (i.e., $\lambda(E)=\mu(E)$ for every Borel measurable set).