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An absolutely normal number is a normal number in every integer base $K\ge 2$. This requires every finite word in every such positional notation to have its appropriate limiting digit block frequency. Knowing base-ten normality alone does not establish absolute normality; the base is part of the statement.

Borel's 1909 result says that absolute normality holds almost everywhere for Lebesgue measure: within $[0,1)$ the exceptions form a Lebesgue null set. In particular, the set of absolutely normal numbers has Lebesgue measure one there. This is a statement about the size of an exceptional set, not a test for a specified constant. ^{[ref] [ref]}

A modern route uses the Birkhoff ergodic theorem and ergodicity of integer multiplication on the circle. For each fixed base and digit interval, the time proportion equals its Lebesgue measure almost everywhere. The bases and finite words form countable sets, so a countable intersection of these measure-one sets still has Lebesgue measure one. For an absolutely normal number, the word $11$ therefore has frequency $1/4$ in base two, $1/9$ in base three, and $1/100$ in base ten. The same symbols represent different intervals in different bases.

The Cantor set is obtained from $[0,1]$ by repeatedly deleting the open middle third of every remaining interval. Its first stages retain $[0,1/3]\cup [2/3,1]$, then four intervals of length $1/9$, then eight of length $1/27$. After level $n$, there are $2^n$ retained intervals of length $3^{-n}$. The Cantor set is the intersection of all these stages, rather than the set of endpoints visible in any finite drawing.

Equivalently,Thus its points admit a base-three expansion using only zero and two. The word “admit” handles numbers with two expansions: $1/3=0.1_3=0.02222{\dots }_3$ is retained as an endpoint. Every infinite binary choice determines a point of $C$, and differing choices give differing points, so $C$ is uncountable.

Its total retained interval length is $(2/3{)}^n\to 0$, so $C$ has zero Lebesgue measure. It is nevertheless nonempty, compact, and uncountable, and has Hausdorff dimension $\log 2/\log 3$: the covers and mass bounds are worked out in Hausdorff dimension of the Cantor set. This separates counting, ordinary length, and fine-scale geometric size in one concrete example.

The self-similarity is exact:The Cantor distribution puts half the probability into each copy at every split. It has a continuous cumulative distribution function, the Cantor function, while all its mass lies on a Lebesgue null set. This makes the construction useful in measure theory as well as fractal geometry: it provides a concrete model in which several plausible meanings of “small” disagree.

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.

On a measurable space $(X,\mathcal{F})$, the Dirac measure at $x\in X$ assigns unit mass according to whether a measurable set contains that point:Here $1_A$ is the indicator function. This is a probability measure, since ${\delta }_x(X)=1$. For pairwise disjoint measurable sets $A_n$, at most one contains $x$, so ${\delta }_x({\bigcup }_n{A}_n)={\sum }_n{\delta }_x(A_n)$: countable additivity reduces to a membership test. The definition works even when the singleton $\{x\}$ is not a measurable set. ^[ref]

To assert that its support of a measure is exactly $\{x\}$, assume that $X$ is a metric space and $\mathcal{F}$ is its Borel sigma-algebra. The singleton $\{x\}$ is then a closed set and a Borel set, with ${\delta }_x(\{x\})=1$. Every open neighbourhood of $x$ has mass one; every other point $y$ has an open ball of radius less than its distance from $x$, with mass zero. Thus $\mathrm{supp}{\delta }_x=\{x\}$. An abstract measurable space has no specified topology, so this topological conclusion is not part of the bare definition. Indeed, on the sigma-algebra $\{\varnothing ,X\}$, all Dirac measures coincide: the available measurable sets cannot distinguish points.

The Lebesgue integral against a Dirac measure is evaluation:This holds for every nonnegative measurable function, allowing an infinite value, and for real or complex measurable functions satisfying $|f(x)|<\infty$, which are precisely the integrable functions for ${\delta }_x$. For a nonnegative simple function, only the cell containing $x$ contributes to its integral; increasing simple functions and the monotone convergence theorem extend the identity. Splitting into positive and negative parts, and then real and imaginary parts, gives the remaining cases. This is why a Dirac measure turns integration into a point query.

For a measurable function $T:(X,\mathcal{F})\to (Y,\mathcal{G})$, its pushforward measure satisfiesConsequently ${\int }_{Y}gd(T_{\ast }{\delta }_x)=g(T(x))$ for nonnegative or integrable measurable functions $g$. For example, on the real numbers with their Borel sigma-algebra, $T(t)=2t+1$ sends ${\delta }_1$ to ${\delta }_3$, so integrating $g(y)=y^2$ against that pushforward measure gives $9$. ^[ref]

Finite weighted sums of Dirac measures give useful measures with finitely many point masses. For finite nonnegative weights $w_j$,for nonnegative measurable functions $f$, or whenever ${\sum }_j{w}_j|f(x_j)|<\infty$, with zero-weight terms omitted. Repeated locations have their weights added. On the real numbers, this is the finite version of the measure underlying the left-continuous cumulative function of an atomic measure; a positive-weight singleton is an atom of a measure. Unit weights at the distinct points of a finite set recover counting measure on that set. ^[ref]

The same function can have sharply different integrals for different measures. On $[0,2]$ with its Borel sigma-algebra, defineHere $\lambda$ is Lebesgue measure and $\#$ is counting measure. The singleton $\{1\}$ is a Lebesgue null set, so its changed value leaves the Lebesgue integral equal to ${\int }_0^2(t+1)dt=4$. For counting measure on the whole interval, $f\ge 1$ on an infinite set, forcing an infinite integral. On the smaller measure space $D=\{0,1,2\}$ with its power set and counting measure, the restricted function instead has integral $1+5+3=9$.

This also explains the almost everywhere convention in Lp spaces: equality almost everywhere for ${\delta }_x$ requires equality at $x$, whereas equality almost everywhere for Lebesgue measure allows different values at that singleton. A point value therefore cannot be read unambiguously from an equivalence class in an Lp space formed using Lebesgue measure.

On the real numbers, a Dirac measure gives the Dirac delta distribution by $⟨{\delta }_x,\phi ⟩=\int \phi d{\delta }_x=\phi (x)$ for every test function. The name Dirac delta function does not mean an ordinary function that is zero away from $x$ and infinitely high at $x$. Such a nonnegative measurable function still has Lebesgue integral zero, because its nonzero values occur on a Lebesgue null set. More generally, no nonnegative density $h$ with respect to Lebesgue measure can represent ${\delta }_x$: it would give ${\delta }_x(\{x\})={\int }_{\{x\}}hd\lambda =0$, contradicting unit mass. The measure and its associated distribution act on different kinds of inputs, namely measurable sets and test functions, but express the same point evaluation. ^[ref]

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 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.

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.

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.

Assign any value, for example zero, at $(0,0)$, where the displayed quotient is undefined. A single point is a Lebesgue null set, so this choice does not affect Lebesgue integrability. Elsewhere the function is continuous, hence it is Lebesgue measurable on the plane after this extension.

In polar coordinates, $x^2+y^2=r^2$ and the area element is $rdrd\theta$. Applying the nonnegative change of variables formula to the absolute value givesFor $0<r\le 1$, $|\sin (r^2)|\le r^2$, so the radial integrand is bounded by $r^{3-2\alpha }$. This is integrable at zero when $\alpha <2$. For $r\ge 1$, $|\sin (r^2)|\le 1$, giving the integrable tail bound $r^{1-2\alpha }$ when $\alpha >1$. In particular,ThereforeEquivalently, the substitution rule $u=r^2$ reduces its exact L1 norm toThis is radial sine-power kernel integrability, reducing a two-dimensional question to the sine power kernel integrability thresholds at zero and infinity.

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.

Let $(f_n)$ be a Cauchy sequence in the L2 space $L^2(\mathbb{R})$, choosing measurable representatives. There is a subsequence $(f_{n_k})$ withWrite $d_k=f_{n_{k+1}}-f_{n_k}$ and $S_N={\sum }_{k=1}^N|d_k|$. Repeated Minkowski inequality gives $\Vert S_N{\Vert }_2\le {\sum }_{k=1}^N{2}^{-k}\le 1$. As $S_N^2$ increases pointwise, the monotone convergence theorem givesHence $S={\sum }_k|d_k|$ is finite almost everywhere and represents an L2 function. Definewhere the series converges, and set $f=0$ on the exceptional Lebesgue null set. It is a measurable function, and $|f|\le |f_{n_1}|+S$ almost everywhere, so Minkowski inequality gives $f\in L^2(\mathbb{R})$.

For fixed $k$, Fatou's lemma applied to squares of finite tails and then Minkowski inequality giveThus the subsequence converges to $f$ in norm. Given $\epsilon >0$, choose a sufficiently late $n_k$ with $\Vert f-f_{n_k}{\Vert }_2<\epsilon /2$ and with $\Vert f_n-f_{n_k}{\Vert }_2<\epsilon /2$ for every sufficiently large $n$, using the original Cauchy sequence property. The triangle inequality then gives $\Vert f_n-f{\Vert }_2<\epsilon$. ThereforeThis proves the needed instance of completeness of Lp spaces without assuming the result being proved.

True. Replace the dominating representative by $|g|$ if necessary, and take the countable union of the exceptional Lebesgue null sets so that every bound holds off a single null set. Since the interval has Lebesgue measure one, Cauchy-Schwarz inequality givesThus $g^2$ is a Lebesgue integrable function. The squared functions satisfy $|f_n{|}^2\to 0$ pointwise and $|f_n{|}^2\le |g{|}^2$ almost everywhere. The dominated convergence theorem therefore yieldsThis uses the Lp inclusion on a finite measure space to turn the assumed fourth-power bound into the second-power domination actually needed.

A Lebesgue null set $E\subset \mathbb{R}$ can, for every $ϵ>0$, be covered by countably many open intervals whose total length is less than $ϵ$. Equivalently, its Lebesgue outer measure is zero.

If $E_j$ is null, choose for it such a cover of total length below $ϵ2^{-j}$, $j\ge 1$. The union of these countably many countable covers covers ${\bigcup }_j{E}_j$ and has total length below $ϵ$. Since $ϵ$ is arbitrary, a countable union of null sets is null.

For positive integers $n,m$, let $E_{n,m}=\{x:f(x)\ge 1/n\}\cap [-m,m]$. This measurable set has finite measure, and $f\ge n^{-1}{1}_{E_{n,m}}$. Monotonicity of the Lebesgue integral givesso every $E_{n,m}$ is a Lebesgue null set. The set where $f>0$ is the countable union of these sets over $n,m$. Part (a) makes it null, henceThe bounded truncation ensures that only integrable indicator functions are used with the stated comparison rule.