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Basic properties

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The statement of the theorem

Let μ\mu be a measure. If E1E_{1} and E2E_{2} are measurable sets with E1⊆E2E_{1}\subseteq E_{2} then μ(E1)≤μ(E2).\mu (E_{1})\leq \mu (E_{2}). For any countablesequence E1,E2,E3,…E_{1},E_{2},E_{3},\ldots of (not necessarily disjoint) measurable sets EnE_{n} in Σ:\Sigma : μ(⋃i=1∞Ei)≤∑i=1∞μ(Ei).\mu \left(\bigcup _{i=1}^{\infty }E_{i}\right)\leq \sum _{i=1}^{\infty }\mu (E_{i}). If E1,E2,E3,…E_{1},E_{2},E_{3},\ldots are measurable sets that are increasing (meaning that E1⊆E2⊆E3⊆…E_{1}\subseteq E_{2}\subseteq E_{3}\subseteq \ldots ) then the union of the sets EnE_{n} is measurable and μ(⋃i=1∞Ei) = lim⁡i→∞μ(Ei)=sup⁡i≥1μ(Ei).\mu \left(\bigcup _{i=1}^{\infty }E_{i}\right)~=~\lim _{i\to \infty }\mu (E_{i})=\sup _{i\geq 1}\mu (E_{i}). If E1,E2,E3,…E_{1},E_{2},E_{3},\ldots are measurable sets that are decreasing (meaning that E1⊇E2⊇E3⊇…E_{1}\supseteq E_{2}\supseteq E_{3}\supseteq \ldots ) then the intersection of the sets EnE_{n} is measurable; furthermore, if at least one of the EnE_{n} has finite measure then μ(⋂i=1∞Ei)=lim⁡i→∞μ(Ei)=inf⁡i≥1μ(Ei).\mu \left(\bigcap _{i=1}^{\infty }E_{i}\right)=\lim _{i\to \infty }\mu (E_{i})=\inf _{i\geq 1}\mu (E_{i}). This property is false without the assumption that at least one of the EnE_{n} has finite measure. For instance, for each n∈N,n\in \mathbb {N} , let En=[n,∞)⊆R,E_{n}=[n,\infty )\subseteq \mathbb {R} , which all have infinite Lebesgue measure, but the intersection is empty.