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For $x<y$, finite additivity on the disjoint pieces of the half-line givesBy monotonicity of a measure and the Lebesgue measure of an interval,Exchanging the two points if necessary provesThus the cumulative measure of a finite-measure set is a Lipschitz function with Lipschitz constant at most one, hence is uniformly continuous: given $\epsilon >0$, the choice $\delta =\epsilon$ works simultaneously at every point.

To locate its endpoint limits, the measurable sets $E\cap (-\infty ,N]$ increase to $E$, so continuity from below of a measure gives $h(N)\to m(E)$. The measurable sets $E\cap (-\infty ,-N]$ decrease to the empty set and all have finite measure, so continuity from above of a measure gives $h(-N)\to 0$. Since $h$ is nondecreasing,For $0<c<m(E)$, choose $R>0$ with $h(-R)<c<h(R)$. The intermediate value theorem on $[-R,R]$ gives $t$ with $h(t)=c$. ThenThe endpoint values $c=0,m(E)$ can also be achieved by $A=\varnothing ,E$. This proves divisibility of Lebesgue measure for arbitrary finite-measure Lebesgue measurable sets on the real line, including unbounded ones. The non-atomic measure structure permits prescribed fractions of measure; a measure concentrated at one point would not have this property.

A subset $E\subseteq \mathbb{R}$ is a Lebesgue measurable set when it satisfies Carathéodory's criterion: for every $T\subseteq \mathbb{R}$,Suppose $m^{\ast }(A)=0$ and let $B\subseteq A$, allowing $B=A$. The monotonicity of Lebesgue outer measure follows directly from covers: every cover of a larger set covers a smaller set. Thus $m^{\ast }(B)=m^{\ast }(T\cap B)=0$, while $m^{\ast }(T\setminus B)\le m^{\ast }(T)$. ConsequentlyThe reverse inequality is countable subadditivity of Lebesgue outer measure applied to $T=(T\cap B)\cup (T\setminus B)$. Equality follows, even if $m^{\ast }(T)=\infty$. Therefore $A$ and every subset $B\subseteq A$ are Lebesgue measurable sets of measure zero. This is the completeness of Lebesgue measure, obtained without presupposing measurability of the outer-null set.

For a Lebesgue measurable function $f:\mathbb{R}\to \mathbb{R}$, the preimage $f^{-1}(U)$ of every open set $U$ is a Lebesgue measurable set. For a Borel measurable function, each such preimage must instead be a Borel set. Equivalently, both definitions can test all Borel target sets: the collectionis a sigma-algebra, since preimages preserve complements and countable unions. Containing all open sets therefore makes it contain the Borel sigma-algebra.

Now let $U$ be open. Since $h$ is a Borel measurable function, $h^{-1}(U)$ is Borel. Since $g$ is a Lebesgue measurable function, $g^{-1}(h^{-1}(U))$ is a Lebesgue measurable set. ButThus $h\circ g$ is a Lebesgue measurable function. The intermediate sigma-algebra is Borel; this is why Borel postcomposition preserves Lebesgue measurability.

The monotone convergence theorem states that, for nonnegative measurable $f_n$ on a Lebesgue measurable set $E$ with $f_n\uparrow f$ almost everywhere,allowing infinity. Fatou's lemma states that, for nonnegative measurable $f_n$,Now write $C={\sup }_n{\int }_E|g_n|<\infty$. A pointwise limit superior of measurable functions is measurable, and it agrees with $g$ off the exceptional null set. By completeness of Lebesgue measure, assigning any values on that null set preserves measurability. Therefore $g$ is measurable. Since $|g_n|\to |g|$ almost everywhere, Fatou's lemma givesThus $g$ is finite almost everywhere and is a Lebesgue integrable function. This is an almost-everywhere limit under a uniform L1 bound. The argument proves integrability of the limit, not convergence in L1 space.

The density of the rational numbers gives a countable way to separate two unequal values:Indeed, $f(x)<g(x)$ holds exactly when some rational number lies strictly between them. Every set inside the union is a Lebesgue measurable set, by the definition of a measurable function and the closure properties of a sigma-algebra. Since $\mathbb{Q}$ is countable, the union is measurable.

Writing $\mathcal{L}$ for the collection of Lebesgue measurable sets, saying that $\mathcal{L}$ is a sigma-algebra means that $\mathbb{R}\in \mathcal{L}$, that $\mathbb{R}\setminus A\in \mathcal{L}$ whenever $A\in \mathcal{L}$, and that ${\bigcup }_{n=1}^{\infty }{A}_n\in \mathcal{L}$ whenever every $A_n\in \mathcal{L}$. These properties also imply closure under countable intersections and differences.

A real-valued function $f$ is a Lebesgue measurable function precisely when the inverse image of every Borel set is in $\mathcal{L}$. Equivalently, it suffices to test the half-lines:The equivalence follows because these half-lines generate the Borel sigma-algebra, and inverse images preserve complements and countable unions.

True. Suppose instead that $B=A_1\times A_2$ were a Lebesgue measurable set in the plane. The section consequence of Fubini's theorem says that there is a Lebesgue null set $N\subseteq \mathbb{R}$ such thatis a Lebesgue measurable set for every $x\notin N$. This version applies to completed Lebesgue measure; it can be obtained on bounded rectangles and then extended by a countable exhaustion, so no finite-measure assumption on $B$ is needed.

For every $x\in A_1$ the section is exactly $A_2$, which is not measurable. Thus $A_1\subseteq N$. By completeness of Lebesgue measure, every subset of $N$ is measurable, contradicting the hypothesis on $A_1$. Therefore

False. Take $E=\{0\}$ and any non-measurable set $A\subseteq \mathbb{R}$. The vertical line $\{0\}\times \mathbb{R}$ is a planar Lebesgue null set: each segment $\{0\}\times [-N,N]$ can be enclosed in rectangles of arbitrarily small area, and the line is a countable union of these segments. Consequently its subset $E\times A$ is a Lebesgue measurable set of planar measure zero, by completeness of Lebesgue measure.

A nonempty factor can therefore have too little Lebesgue measure to transmit the other factor's failure of measurability. Its exceptional vertical section at $x=0$ is consistent with Lebesgue measurable sections being guaranteed only almost everywhere.

This requires an argument using Lebesgue outer measure, since $E$ need not be a Lebesgue measurable set. Some bounded portion $F=E\cap [-N,N]$ has $a=m^{\ast }(F)>0$: otherwise countable subadditivity would give $m^{\ast }(E)=0$. Also $a\le 2N<\infty$.

Suppose, for a contradiction, that every nonempty bounded open interval $I$ satisfies $m^{\ast }(E\cap I)\le \lambda |I|$. Because $a/\lambda >a$, the definition of Lebesgue outer measure gives a countable cover $F\subseteq {\bigcup }_j{I}_j$ by nonempty bounded open intervals withUsing countable subadditivity again,a contradiction. Hence at least one interval in such a cover satisfiesThis high-density interval for outer measure applies even to nonmeasurable sets; the argument does not require the Lebesgue density theorem.

Use Lebesgue measure restricted to the Lebesgue measurable set $E$. For $1\le p<\infty$, start with the measurable real-valued functions $f$ on $E$ satisfying ${\int }_E|f{|}^{p}dm<\infty$. Define an equivalence relation by $f\sim g$ when $f=g$ almost everywhere, and let $L^p(E)$ consist of their equivalence classes. Addition and scalar multiplication are defined using representatives. The Lp norm isFor $p=\infty$, use measurable functions bounded almost everywhere, apply the same identification, and define the L-infinity norm byThese expressions are independent of the representative. For finite $p$, zero norm is equivalent to $f=0$ almost everywhere: each set $\{|f|>1/n\}$ must have measure zero, and their union is $\{f\ne 0\}$. The same implication follows from the essential supremum for $p=\infty$. Absolute homogeneity is immediate; Minkowski inequality proves the triangle inequality and closure under addition. Thus these are normed vector spaces. Using complex-valued representatives gives their complex versions by the same construction.

Each coordinate projection $S_j$ is compact, hence a Lebesgue measurable set, because it is the continuous image of the compact set $K$. Its $(d-1)$-dimensional measure is finite. Choose any nonnegative integrable function $\eta$ on $\mathbb{R}$ with integral one, and defineIf $x\in K$, every projection ${\hat{x}}_j$ belongs to $S_j$. If $x\notin K$, $f(x)=0$. Thus for every $j$, $f(x)\le 1_{S_j}({\hat{x}}_j)=\int s_j({\hat{x}}_j,t)dt$, so the introductory estimate applies. Tonelli theorem gives $\Vert s_j{\Vert }_1={\lambda }_{d-1}(S_j)$, andSubstitute these values and raise to the $d$th power:This proves the geometric Loomis--Whitney inequality, including zero-volume sets or zero-measure projections, without any division by their measures.

The uniform covers theorem says that if a multiset $\mathcal{C}$ of coordinate subsets covers every coordinate exactly $k$ times, then a Euclidean body $S\subseteq {\mathbb{R}}^n$ satisfieswhere $S_A$ is its coordinate projection of a Euclidean body. Work with bounded Borel bodies, whose projections are Lebesgue measurable sets; an empty-coordinate projection has measure one.

Prove the theorem by induction on dimension, with the one-dimensional case immediate. Slice at the last coordinate, writing $S_t\subseteq {\mathbb{R}}^{n-1}$. The subsets $A\setminus \{n\}$ still cover each remaining coordinate $k$ times, so induction givesFor $A$ not containing $n$, the projected slice is contained in $S_A$, giving a constant bound. For $A$ containing $n$, its measure is the slice measure $f_A(t)$ of $S_A$. Exactly $k$ members of $\mathcal{C}$ contain $n$, counted with multiplicity. Integrate and apply Hölder's inequality to those $k$ factors:Fubini's theorem identifies $\int f_A=|S_A|$, completing the proof. Exact covering is needed for real volumes, which may be smaller than one.

The Bollobas--Thomason box theorem states that there is an axis-parallel box $B$ with $|B|=|S|$ and $|B_A|\le |S_A|$ for every nonempty coordinate subset $A$. Use the logarithmic linear program for the box theorem: if its side lengths are $b_i=e^{x_i}$, maximize ${\sum }_i{x}_i$ subject toThe linear program is feasible by choosing all $x_i$ sufficiently negative and is bounded above by the full-set constraint. Its dual minimizes ${\sum }_A{\lambda }_A\log |S_A|$ subject to ${\lambda }_A\ge 0$ and ${\sum }_{A\ni i}{\lambda }_A=1$ for each $i$: precisely a fractional uniform cover.

The dual feasible polytope is nonempty, bounded and defined by integer coefficients, so an optimal extreme point has rational coordinates. Multiply its weights by a common denominator $k$. The resulting integer uniform cover and the uniform covers theorem giveThus the dual optimum is at least $\log |S|$; weight one on $A=[n]$ attains equality. Strong duality gives a primal optimum ${\sum }_i{x}_i=\log |S|$. The corresponding box has the required volume and projected-volume bounds, proving the theorem.

For the Loomis–Whitney theorem, take the box supplied by the Bollobas--Thomason box theorem. Each side length occurs in exactly $n-1$ projections omitting one coordinate, soThis is the Loomis--Whitney inequality.

For the three projection areas of a volume-one body, necessity is $1=|S{|}^2\le abc$. Conversely suppose $abc\ge 1$ and setThe box $[0,x]\times [0,y]\times [0,z]$ has projected areas $a,b,c$ and volume $V=\sqrt{abc}\ge 1$. Inside it retain the three slabsFor $t>0$, each projection is the full corresponding rectangle, since the slab in the omitted coordinate supplies every point of that projection. The volume is $V[1-(1-t{)}^3]$. Chooseto make the volume one. This is a compact connected union of three positive-thickness boxes. Hence the complete answer is

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

The Lebesgue outer measure iswhere $|Q|$ is the product of its side lengths. The infimum ranges over countable covers; allowing other standard endpoint conventions gives the same outer measure by arbitrarily small enlargement. Lebesgue measure is its restriction to Lebesgue measurable sets.

Enlarged unit cubes give ${\mathcal{L}}^{\ast }([0,1{]}^n)\le 1$. For any countable open-box cover of this compact cube choose a finite subcover. A finite grid formed from its box endpoints and the cube endpoints partitions the cube into rectangular cells. The sum of the volumes of covering boxes is at least the sum of these cell volumes, which is one. Boundary faces have elementary zero volume and do not affect the finite volume comparison. Thus every covering sum is at least one, proving