User:Aravind V R/sandbox1/Measure/Notes

σ-finite measure
a positive (or signed) measure μ defined on a &sigma;-algebra Σ of subsets of a set X is called finite if μ(X) is a finite real number (rather than ∞). The measure μ is called σ-finite if X is the countable union of measurable sets with finite measure. A set in a measure space is said to have σ-finite measure] if it is a countable union of sets with finite measure.

Dirac measure
A Dirac measure is a measure δx on a set X (with any &sigma;-algebra of subsets of X) defined for a given x ∈ X and any (measurable) set A ⊆ X by
 * $$\delta_{x} (A) = 1_A(x)= \begin{cases} 0, & x \not \in A; \\ 1, & x \in A. \end{cases}$$

where $$1_A$$ is the indicator function of $$A$$. The Dirac measure is a probability measure

Counting measure
The counting measure $$\mu$$ on this measurable space $$(X,\Sigma)$$ is the positive measure $$\Sigma\rightarrow[0,+\infty]$$ defined by

\mu(A)=\begin{cases} \vert A \vert & \text{if } A \text{ is finite}\\ +\infty & \text{if } A \text{ is infinite} \end{cases} $$ for all $$A\in\Sigma$$, where $$\vert A\vert$$ denotes the cardinality of the set $$A$$. The counting measure on $$(X,\Sigma)$$ is σ-finite if and only if the space $$X$$ is countable.

Complete measure
Complete measure (or, more precisely, a complete measure space) is a measure space in which every subset of every null set is measurable (having measure zero). More formally, (X, Σ, μ) is complete if and only if


 * $$S \subseteq N \in \Sigma \mbox{ and } \mu(N) = 0\ \Rightarrow\ S \in \Sigma.$$

Borel measure
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. A Borel measure is any measure μ defined on the σ-algebra of Borel sets. 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 $$\mathbb R$$.

Lebesgue measure
Given a subset $$E\subseteq\mathbb{R}$$, with the length of an (open, closed, semi-open) interval $$I = [a,b]$$ given by $$l(I)=b - a$$, the Lebesgue outer measure $$\lambda^*(E)$$ is defined as
 * $$\lambda^*(E) = \operatorname{inf} \left\{\sum_{k=1}^\infty l(I_k) : {(I_k)_{k \in \mathbb N}} \text{ is a sequence of open intervals with } E\subseteq \bigcup_{k=1}^\infty I_k\right\}$$.

The Lebesgue measure of E is given by its Lebesgue outer measure $$\lambda(E)=\lambda^*(E)$$ if, for every $$ A\subseteq\mathbb{R}$$,
 * $$\lambda^*(A) = \lambda^*(A \cap E) + \lambda^*(A \cap E^c) $$.

The Borel measure agrees with the Lebesgue measure on those sets for which it is defined; however, there are many more Lebesgue-measurable sets than there are Borel measurable sets. The Borel measure is translation-invariant, but not complete.

Urysohn's lemma
Urysohn's lemma states that a topological space is normal if and only if any two disjoint closed subsets can be separated by a function.

Egorov's theorem
Let (fn) be a sequence of M-valued measurable functions, where M is a separable metric space, on some measure space (X,Σ,μ), and suppose there is a measurable subset A of finite μ-measure such that (fn) converges μ-almost everywhere on A to a limit function f. The following result holds: for every ε > 0, there exists a measurable subset B of A such that μ(B) < ε, and (fn) converges to f uniformly on the relative complement A \ B.