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Every subset of a Lebesgue-null set is Lebesgue measurable and has measure zero.

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.

A Euclidean body is a nonempty bounded measurable subset of ${\mathbb{R}}^n$ of positive Lebesgue measure. Projection inequalities are unchanged by modifying a body on a set of measure zero; geometric equality statements commonly impose additional regularity.

Let $h:\mathbb{R}\to \mathbb{R}$ be continuous, strictly increasing, and satisfy Lusin condition N. Thendefines a measure on the Lebesgue measurable sets, and $\nu \ll \lambda$. Moreover,for $a\le b$.

For finitely many Lebesgue measurable sets $A_i\subseteq [0,1]$ and nonnegative demands $b_i$, pairwise disjoint measurable subsets $B_i\subseteq A_i$ with $\lambda (B_i)=b_i$ exist exactly when $\lambda ({\bigcup }_{i\in I}{A}_i)\ge {\sum }_{i\in I}{b}_i$ for every index subset $I$. Necessity is additivity and monotonicity of Lebesgue measure. For sufficiency, partition the set union into membership cells $E_J$, send flow from a source through demand vertices $i$ to cells with $i\in J$, then to a sink. Source capacities are $b_i$, cell capacities are $\lambda (E_J)$, and intermediate capacities are $D={\sum }_i{b}_i$. Every cut of a flow network has capacity at least $D$ by the assumed inequalities. The max-flow min-cut theorem provides the allocations, and divisibility of Lebesgue measure turns each cell allocation into disjoint pieces. If $D=0$, all $B_i$ can simply be empty.

A subset of $X$ is measurable relative to a sigma-algebra $\mathcal{F}$ when it belongs to $\mathcal{F}$. A measure defined on $\mathcal{F}$ assigns sizes to these sets. Measurability depends on the chosen sigma-algebra; for example, a Lebesgue measurable set need not be a Borel set.

First take nonempty bounded open sets $A,B\subset {\mathbb{R}}^n$. Their Minkowski sum is open and hence a Lebesgue measurable set. Apply the Prékopa–Leindler inequality to their indicator functions and the indicator function of $(1-\theta )A+\theta B$. This gives the multiplicative formTo obtain the usual additive Brunn–Minkowski inequality, put $a={\lambda }_n(A{)}^{1/n}$, $b={\lambda }_n(B{)}^{1/n}$, $A_0=A/a$ and $B_0=B/b$. Both normalized open sets have Lebesgue measure one. For $\theta =b/(a+b)$,The multiplicative bound and the scaling rule for Lebesgue measure implyReplacing $A,B$ by $A\cap B(0,R),B\cap B(0,R)$ and using continuity from below extends this form of the Brunn–Minkowski inequality to all nonempty open sets, including those of infinite Lebesgue measure. This open-set form is sufficient here and avoids any measurability qualification for sums of arbitrary measurable sets.

Write $D={\sum }_{i=1}^n{b}_i$ and let $\lambda$ denote Lebesgue measure. Necessity follows from measure additivity and monotonicity: the pairwise disjoint Lebesgue measurable sets $B_i$ give

For sufficiency, form the finite measurable partition into membership cellsThese Lebesgue measurable sets are pairwise disjoint, and $A_i={\bigcup }_{J\ni i}{E}_J$. Empty cells may be retained. Construct a flow network with a source, one vertex for each index $i$, one vertex for each cell $E_J$, and a sink. Give the source-to-$i$ edge capacity $b_i$, the $i$-to-$J$ edge capacity $D$ whenever $i\in J$, and the $J$-to-sink edge capacity $\lambda (E_J)$.

Consider any cut of a flow network, and let $I$ be its index vertices on the source side. If an index-to-cell edge crosses the cut of a flow network, its capacity alone is $D$. Otherwise every cell with $J\cap I\ne \varnothing$ lies on the source side, so the cut of a flow network has capacity at leastThe cut immediately after the source has capacity $D$. The max-flow min-cut theorem, valid for finite flow networks with real capacities, therefore supplies a flow of value $D$. Its cell allocations $f_{iJ}\ge 0$ satisfy

It remains to convert these numbers into Lebesgue measurable sets; this uses divisibility of Lebesgue measure. Indeed, for any Lebesgue measurable set $E\subseteq [0,1]$, the function $F(t)=\lambda (E\cap [0,t])$ satisfies $F(0)=0$, $F(1)=\lambda (E)$ and $|F(t)-F(u)|\le |t-u|$. Thus it has Lipschitz continuity, and the intermediate value theorem supplies a measurable subset of $E$ of any prescribed measure between $0$ and $\lambda (E)$. Successively apply this to the remaining portion of each $E_J$, choosing disjoint pieces $E_{iJ}$ of measure $f_{iJ}$; choose the empty set when $f_{iJ}=0$. ThenThis proves the measurable Hall theorem. The splitting step is essential: an arbitrary measure with an atom of a measure would not support the same conclusion. Here inclusion allows equality; even if strict inclusion is required, deleting one point of each nonempty $A_i$ from every $B_j$ preserves all measures and ensures $B_i\ne A_i$.

A set $E\subseteq \mathbb{R}$ is Lebesgue measurable if it satisfies the Caratheodory criterionfor every $A\subseteq \mathbb{R}$, where $m^{\ast }$ is Lebesgue outer measure. Equivalently, $E$ belongs to the completion of the Borel sigma-algebra with respect to Lebesgue measure.

For every positive integer $k$, additivity gives $f(kx)=kf(x)$. HenceConversely, if $z\in f^{-1}(B_{k\epsilon })$, write $z=k(z/k)$. Thenso $z/k\in f^{-1}(B_{\epsilon })$. Thus

To prove continuity, fix $\epsilon >0$ and putThese sets are measurable and ${\mathbb{R}}^n={\bigcup }_{k\ge 1}{E}_k$, so at least one $E_k$ has positive Lebesgue measure. Part (ii) gives $E_k=k{E}_1$, and therefore $E_1$ also has positive measure. By the Steinhaus theorem, $E_1-E_1$ contains an open neighbourhood $U$ of zero. Part (i) givesThis proves continuity at zero. Finally,as $h\to 0$, so $f$ is continuous at every $x$. Thus every such map is an instance of the theorem that a measurable additive function is continuous.

Yes. The regularity of Lebesgue measure says that for every Lebesgue measurable set $A\subseteq [0,1]$ and every $ϵ>0$, there is an open set $O\supseteq A$ withWrite $O\cap [0,1]$ as a countable disjoint union of intervals. Since its total measure is at most one, a finite subunion $B$ can be chosen withThen $B\in \mathcal{B}$ and $m(A△B)<ϵ$. Hence every Lebesgue measurable set lies in $\mathcal{C}$, while $\mathcal{C}\subseteq {\mathcal{L}}_{[0,1]}$ by definition. Therefore

Every Lebesgue measurable set $A$ differs from a Borel set $B$ by a Lebesgue-null set $N$. The given function $h$ is a homeomorphism from $\mathbb{R}$ onto the interval $h(\mathbb{R})$. Hence $h(B)$ is Borel relative to $h(\mathbb{R})$ and therefore Lebesgue measurable in $\mathbb{R}$. Moreover,and $h(N)$ is null by Lusin condition N. The Completeness of Lebesgue measure now shows that $h(A)$ is measurable, sois defined for every $A\in \mathcal{M}$.

Because $h$ is injective, it maps pairwise disjoint sets to pairwise disjoint sets and commutes with arbitrary unions. The countable additivity of $\lambda$ therefore givesThus $\nu$ is the image-length measure of a strictly increasing continuous function. If $\lambda (A)=0$, the null-set hypothesis gives $\nu (A)=\lambda (h(A))=0$, so

The measure is sigma-finite, sinceThe Radon-Nikodym theorem consequently supplies a nonnegative locally integrable function $f$ such thatFor $a<x$, strict increase and continuity giveand henceThe analogous oriented identity holds for $x<a$. By differentiation of an indefinite Lebesgue integral,

Yes, the differentiability conclusion still holds when $h$ is merely non-decreasing. In fact, the Lebesgue theorem on differentiability of monotone functions says that every monotone function is differentiable almost everywhere; neither continuity nor the null-set hypothesis is needed for that conclusion. Flat intervals mean that the preceding image-set argument no longer gives disjoint images, so its natural replacement is the Lebesgue-Stieltjes measure determined by

Let $N_j(A)$ be the number of particles of gas $j$ in a bounded Lebesgue measurable set $A$. Because Lebesgue measure is diffuse, two independent spatial Poisson point processes have no common particle almost surely. Hence the mixture count isWriting ${\lambda }_j=z_j|A|$, independence gives the probability generating functionThus

If $A_1,\dots ,A_m$ are pairwise disjoint, the familyis independent: counts in disjoint sets are independent within each gas, and the two gases are independent. Therefore the sums $N(A_r)$ are independent. These two facts prove directly that the mixture is a Poisson point process. By the Superposition theorem for Poisson point processes, its activity is