Sigma-finite uniqueness theorem for measures

If two measures agree on a generating $\pi$-system and the space is a countable union of members having finite common measure, then they agree on the generated $\sigma$-algebra.

Lebesgue measure on the real numbers extends interval length: $\lambda ([a,b])=b-a$ for $a\le b$. Its restriction to the Borel sigma-algebra is the unique translation-invariant sigma-finite measure on that sigma-algebra normalized by $\lambda ((0,1])=1$. The full Lebesgue measure is the completion of a measure of this restriction; its domain is the Lebesgue measurable sets. Translation invariance follows by comparing a translated Borel measure on half-open intervals and applying the uniqueness theorem for measures. For the construction and interval properties, see MIT lecture 8.

Every singleton has Lebesgue measure zero: $\{x\}\subseteq [x-\epsilon ,x+\epsilon ]$ gives $0\le \lambda (\{x\})\le 2\epsilon$ for every $\epsilon >0$. Enumerating the countable set $\mathbb{Q}\cap [0,1]$ and applying countable additivity yieldsThus a dense set can be a null set, and a union of null sets indexed by an uncountable set can have positive Lebesgue measure. The distinction explains both the Dirichlet function example and why almost everywhere is weaker than everywhere.

The Lebesgue inner measure approximates a subset of the real line from within by compact sets, while Lebesgue outer measure approximates it from outside by countable interval covers. For a finite closed interval $I$ containing $E$,Indeed, every compact $K\subseteq E$ gives $m^{\ast }(I\setminus E)\le m(I)-m(K)$. Conversely, an open cover of $I\setminus E$ with total length arbitrarily close to its outer measure has compact complement in $I$ lying inside $E$, giving the reverse inequality in the limit.

For $m^{\ast }(E)<\infty$, equality of inner and outer measures holds precisely when $E$ is Lebesgue measurable. For an unbounded set, measurability is equivalently checked on all bounded intersections $E\cap [-n,n]$. Equality $m_{\ast }(E)=m^{\ast }(E)=\infty$ alone is insufficient: adjoining a bounded nonmeasurable subset to a disjoint half-line makes both quantities infinite without making the union measurable.

Translating a Lebesgue measurable set by a real number preserves measurability and its Lebesgue measure. Translation invariance of Lebesgue outer measure and Carathéodory's criterion give the measurability assertion and equality. Applying the result to indicator functions, then simple functions, and finally increasing simple approximations gives ${\int }_{\mathbb{R}}u(x-t)dx={\int }_{\mathbb{R}}u(x)dx$ for every nonnegative Lebesgue measurable function. For a Lebesgue integrable function, apply the identity to its positive part and negative part.

For a Lebesgue measurable set $A$ in an open set $\Omega \subseteq {\mathbb{R}}^n$, its essential closure consists of the points whose every sufficiently small ball has positive Lebesgue measure intersection with $A$. Equivalently,where the right side is the support of a distribution. To see the equivalence, a ball on which $A$ is null gives zero pairings with all test functions. If $A$ has positive measure in a ball, cover that ball by smaller compactly contained balls and choose a nonnegative cutoff function positive on one meeting $A$ in positive measure; its pairing is positive. Thus essential closure is relatively closed and unchanged by modifying $A$ on a Lebesgue null set. For an open set $A$ it equals its relative closure.

Target: Singular support of an indicator function. Parent: Singular support. Intended file: distribution-theory.bigb.

For any circle null set and any positive sequence ${ϵ}_j$, take infinitely many countable covers with summable total lengths. Regard their arcs as a two-index array. Choose increasing finite-square cutoffs so the length sum outside each square is smaller than the next desired bound. The finite square annuli then give finite arc families with the stated small lengths. Every point belongs to some arc in every row, hence to infinitely many annular families. This handles dense and nonclosed null sets without claiming that they have finite small covers.

An open subset of the circle can be covered by countably many closed subarcs with disjoint interiors. Grouping a fixed enumeration into finite batches gives compact sets covering every point of the open set. By taking each finite batch long enough, the remaining total measure can be forced below any prescribed positive sequence tending to zero. This controls the measure of subsequent batches without assuming that a prescribed null subset is itself a countable union of compact null sets.

The Lebesgue measure of a measurable set is the supremum of the measures of its compact subsets. In particular, an open set of finite positive measure contains a compact subset with at least half its measure. Together with a finite subcover, this turns a finite Wiener covering lemma estimate into an estimate for arbitrary open superlevel sets.

An uncountable subset $S$ of the real numbers whose intersection with every null set for Lebesgue measure is countable. Testing only Borel sets suffices because every Lebesgue-null set has a Borel null superset. No measurability of $S$ is assumed.

If $E\subseteq [0,1]$ is a Lebesgue measurable set and $0\le b\le \lambda (E)$, some measurable subset $B\subseteq E$ has Lebesgue measure $b$. The function $F(t)=\lambda (E\cap [0,t])$ has Lipschitz continuity with constant $1$, starts at $0$ and ends at $\lambda (E)$; the intermediate value theorem proves the assertion. Applying the assertion successively to remaining portions of $E$ gives pairwise disjoint measurable pieces with any finite list of nonnegative measures whose sum is at most $\lambda (E)$. This is a concrete consequence of non-atomic measure structure, which fails for general measures containing an atom of a measure.

For a Lebesgue measurable set $E\subseteq \mathbb{R}$ of finite measure, the displayed function satisfies $|H_E(t)-H_E(s)|\le |t-s|$. Indeed, for $s<t$, finite additivity and monotonicity of a measure give $H_E(t)-H_E(s)=m(E\cap (s,t])\le t-s$. Thus $H_E$ is a nondecreasing Lipschitz function and is uniformly continuous even if $E$ is unbounded.

Continuity from above of a measure gives $H_E(t)\to 0$ as $t\to -\infty$, while continuity from below of a measure gives $H_E(t)\to m(E)$ as $t\to \infty$. Every strictly intermediate value is attained by the intermediate value theorem. The corresponding half-line cut of $E$ realizes any prescribed measure between zero and $m(E)$; the endpoints are realized by the empty set and $E$ itself.

Every subset $B$ of a Lebesgue null set is a Lebesgue measurable set and is itself a null set. In fact the same conclusion follows directly from $m^{\ast }(B)=0$, without assuming measurability in advance. For any $A\subseteq \mathbb{R}$, monotonicity of Lebesgue outer measure gives $m^{\ast }(A\cap B)=0$ and $m^{\ast }(A\setminus B)\le m^{\ast }(A)$. ThereforeThe reverse inequality follows from countable subadditivity of Lebesgue outer measure, applied to the union $A=(A\cap B)\cup (A\setminus B)$. This is Carathéodory's criterion, proving measurability. If $B$ is a subset of an outer-null $N$, its outer measure is zero by the same monotonicity. Thus the restriction of outer measure to measurable sets is a complete measure.