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Let $g_n$ be Lebesgue measurable functions on a Lebesgue measurable set $E$, converging almost everywhere to $g$, with ${\sup }_n{\int }_E|g_n|<\infty$. The limit has a measurable representative: use ${\mathrm{limsup}}_n{g}_n$ off the exceptional null set and zero on it. The completeness of Lebesgue measure makes any alternative values on that null set measurable as well. Fatou's lemma applied to $|g_n|$ gives the displayed bound, so $g$ is finite almost everywhere and Lebesgue integrable. A uniform L1 norm bound alone does not imply convergence in L1 space; the strict inequality in Fatou lemma supplies a shrinking-spike example.

If $g$ is a Lebesgue measurable function and $h$ is a Borel measurable function, then $h\circ g$ is a Lebesgue measurable function. For an open set $U$, $h^{-1}(U)$ is a Borel set. By measurability of $g$, $g^{-1}(h^{-1}(U))$ is a Lebesgue measurable set, and it equals $(h\circ g{)}^{-1}(U)$. This is composition of measurable functions with the intermediate Borel sigma-algebra stated explicitly. Merely knowing both functions are Lebesgue measurable would not ensure that the intermediate sigma-algebras match.

Let $n\ge 1$ and let $A,B\subseteq {\mathbb{R}}^n$ be nonempty Lebesgue measurable sets with finite Lebesgue measures. Assume that $A+B$ is also a Lebesgue measurable set; its measure may be infinite. ThenHere ${\lambda }_n$ denotes $n$-dimensional Lebesgue measure in Euclidean space. Nonempty compact sets and nonempty open sets of finite measure satisfy the measurability assumptions. Measurability of arbitrary summands alone does not guarantee measurability of their Minkowski sum. The theorem does not require convexity. See R. J. Gardner, The Brunn-Minkowski Inequality (2002), Theorem 4.1 and the following remarks.

The $n$th root converts a volume into a quantity that scales like a length: uniform dilation gives ${\lambda }_n(tA{)}^{1/n}=t{\lambda }_n(A{)}^{1/n}$ for $t>0$. In the Euclidean plane the theorem compares square roots of areas, not the areas themselves. For the two rectangles above it reads $3\ge \sqrt{2}+\sqrt{2}=2\sqrt{2}$. Equivalently their sum has area $9\ge 8$. For the two disks, it reads $3\sqrt{\pi }=\sqrt{\pi }+2\sqrt{\pi }$, an equality. These are exact geometric examples, not measured data.

Applying the theorem to the scaled sets gives, whenever that sum is a Lebesgue measurable set,For convex bodies, this means the root-volume of the interpolated shape lies above the straight line between the input root-volumes. In fact $t\mapsto {\lambda }_n((1-t)A+tB{)}^{1/n}$ is a concave function on $[0,1]$: convexity gives $sA+tA=(s+t)A$ for $s,t\ge 0$, so interpolating two intermediate shapes gives another intermediate shape and the same inequality applies. The beauty is that no particular boundary shape appears in the bound: possible pointwise displacements force a minimum amount of space.

For convex bodies $A,B$, equality holds exactly when they are homothetic convex bodies, meaning $B=v+sA$ for some vector $v$ and $s>0$. This equality qualification is part of the Brunn-Minkowski theorem for convex bodies. ^[ref] For positive-sided axis-aligned rectangles, the comparison can be checked directly:Equality requires $ad=bc$: the rectangles must have the same aspect ratio. Our $2\times 1$ and $1\times 2$ rectangles do not. All positive-radius disks are homothetic convex bodies, explaining their equality.

Do not apply that equality criterion without its hypotheses. If $A=\{a\}$, its Lebesgue measure is zero and $A+B$ is a translation of $B$, so equality holds even for nonconvex $B$. Also, changing a summand on a null set can change the measure of its Minkowski sum: in one dimension, $A=[0,1]\cup \{3\}$ and $B=[0,1]$ both have measure $1$, but $A+B=[0,2]\cup [3,4]$ has measure $3$. Removing the single extra point restores equality with sum measure $2$. Thus merely being equal almost everywhere to homothetic convex bodies is not a sufficient equality test for arbitrary sets.

The Brunn-Minkowski theorem also bounds the collision-region volume: for positive-volume compact sets $O,R$, reflection preserves Lebesgue measure, so ${\lambda }_n(O+(-R){)}^{1/n}\ge {\lambda }_n(O{)}^{1/n}+{\lambda }_n(R{)}^{1/n}$. In the toy collision example this gives area at least $(2+\sqrt{2}{)}^2=6+4\sqrt{2}\approx 11.657$, compared with the exact area $12$. A lower bound alone does not identify safe reference positions; the actual Minkowski sum supplies those.

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.

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.

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 $A$ 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.

A Euclidean body is a nonempty bounded Lebesgue measurable set in ${\mathbb{R}}^n$ of positive Lebesgue measure. Projection inequalities require measurable coordinate projections, as for a compact set or a bounded Borel set. Modifying a body on a set of volume zero can change its coordinate-projection measures, so the actual projection sets matter. Geometric equality statements commonly impose additional regularity.

For any $E\subseteq \mathbb{R}$ with positive Lebesgue outer measure and any $0<\lambda <1$, some bounded nonempty open interval $I$ satisfies the displayed inequality. The set $E$ need not be a Lebesgue measurable set.

To prove this, first choose a bounded portion $F=E\cap [-N,N]$ with $0<a=m^{\ast }(F)<\infty$, using countable subadditivity. Choose a countable open-interval cover $(I_j)$ of $F$ with ${\sum }_j|I_j|<a/\lambda$. If $m^{\ast }(E\cap I_j)\le \lambda |I_j|$ for every $j$, thena contradiction. For measurable $E$ this gives the high-density interval in a positive-measure subset of the real line without requiring the full Lebesgue density theorem.

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 $a\le b$.

A Lebesgue measurable function $f:\mathbb{R}\to \mathbb{R}$ has a Lebesgue measurable set as the preimage of every open subset of $\mathbb{R}$. Equivalently, use every Borel set in the target: the sets whose preimages are measurable form a sigma-algebra containing all open sets. The source carries the Lebesgue sigma-algebra and the target its Borel sigma-algebra. A Borel measurable function on $\mathbb{R}$ instead requires the preimages to be Borel in the source as well. Restricting the source to a Lebesgue measurable set uses its induced sigma-algebra.

For a planar Lebesgue measurable set $B$, its vertical section $B_x$ is a one-dimensional Lebesgue measurable set for $x$ almost everywhere. Write $B△C\subseteq Z$ with $C$ a planar Borel set and $Z$ a Borel Lebesgue null set. Every section $C_x$ and $Z_x$ is Borel; Tonelli's theorem gives $\int m(Z_x)dx=m_2(Z)=0$. Thus $m(Z_x)=0$ for $x$ almost everywhere, and $B_x△C_x\subseteq Z_x$ proves the assertion by completeness of Lebesgue measure. No finite-measure condition on $B$ is needed.

Exceptional sections need not be measurable: $\{0\}\times A$ is a planar Lebesgue null set for every $A\subseteq \mathbb{R}$, even when $A$ is a non-measurable set. Conversely, if both $A_1$ and $A_2$ are non-measurable, $A_1\times A_2$ cannot be measurable. Otherwise every $x\in A_1$ would have to lie in the null exceptional set, making $A_1$ measurable by completeness of Lebesgue measure.

A subset $E\subseteq \mathbb{R}$ is a Lebesgue measurable set when it satisfies Carathéodory's criterion for Lebesgue outer measure: for every subset $A\subseteq \mathbb{R}$,These sets form the Lebesgue sigma-algebra, the completion of the Borel sigma-algebra with respect to Lebesgue measure. Equivalently, $E$ differs from a Borel set by a subset of a Borel null set.

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.

For every Lebesgue measurable set $A\subseteq {\mathbb{R}}^n$, its Lebesgue density is $1$ at almost every point of $A$ and $0$ at almost every point of its complement.

The sigma-algebra of all Lebesgue measurable sets, equivalently the completion of a measure of the Borel sigma-algebra under Lebesgue measure. It contains every subset of every Lebesgue null set. Applying Carathéodory's criterion to Lebesgue outer measure gives another construction of this same sigma-algebra.

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.

A subset $A$ of a measurable space $(X,\Sigma )$ is non-measurable when $A\notin \Sigma$. On the real line, unless another sigma-algebra is specified, this commonly means that $A$ is not a Lebesgue measurable set. By completeness of Lebesgue measure, such a set cannot be contained in a Lebesgue null set.

For an example, use the axiom of choice to select one representative from each equivalence class of $[0,1]$ under $x\sim y$ if $x-y\in \mathbb{Q}$, and call the resulting set $V$. The translates $V+q$, for $q\in \mathbb{Q}\cap [-1,1]$, are pairwise disjoint, their union contains $[0,1]$, and their union is contained in $[-1,2]$. If $V$ were a Lebesgue measurable set, translation invariance and countable additivity would make that union have measure zero when $m(V)=0$ and infinite measure when $m(V)>0$. Both conclusions contradict these inclusions. Thus $V$ is non-measurable.

Define two open sets byThe support of a distribution and its singular support areBoth are relatively closed because the removed sets are open. The locality of distributions gives $u{|}_{Z(u)}=0$. On an overlap of two open smoothness regions, the two smooth functions represent the same distribution, so their difference is zero almost everywhere and hence everywhere by continuity. They therefore join to one smooth function on $G(u)$. The same partition of unity argument shows that it represents $u$ there.

For $u=1_{\omega }$, openness of $\omega$ impliesIndeed every neighborhood of a point in this relative closure meets $\omega$ in an open set containing a nonnegative test function with positive integral; outside the closure the indicator function is identically zero near the point.

The singular support of an indicator function needs a measure-sensitive boundary. For a Lebesgue measurable set $A\subseteq \Omega$, define its essential closure byThen the general answer isOutside this intersection, some ball has either no points of $\omega$ or a complement of Lebesgue measure zero. The regular distribution is respectively the constant zero or one on that ball and is smooth. Conversely, at a point of the intersection, every sufficiently small ball has positive Lebesgue measure in both sets. If a smooth function represented the indicator function there, its values would be in $\{0,1\}$ everywhere: any other value, by continuity, would give a ball of positive Lebesgue measure on which equality almost everywhere fails. On a connected ball such a continuous two-valued function is constant, contradicting the two positive measures.

In particular, the answer need not be the full topological boundary. For $\Omega ={\mathbb{R}}^n$ and $\omega ={\mathbb{R}}^n\setminus \{0\}$, the indicator function is the constant one almost everywhere, so its singular support is empty although the boundary is $\{0\}$. The full relative boundary is the answer when every neighborhood of each boundary point contains a positive Lebesgue measure portion of the complement, as for an open set with smooth boundary.