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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 law of ${\sum }_{j\ge 1}2{\omega }_j{3}^{-j}$, for independent fair binary digits ${\omega }_j$, is supported on the Cantor set. It is invariant and ergodic under $T_3$, as the image of a Bernoulli shift. Each permitted ternary cylinder of length $n$ has measure $2^{-n}$, giving entropy rate $\log 2$. The Host equidistribution theorem makes almost every point a normal number in base $2$, while its ternary digits omit $1$.

Form Champernowne constant by using concatenation to join the decimal representations of the positive integers $1,2,3,\dots$ after the decimal point. Champernowne proved in 1933 that this particular real number is a normal number in base ten. The construction is striking because its rule is completely explicit, while all finite words have the prescribed limiting digit block frequencies. ^[ref] Copeland and Erdős explicitly recall this result at the start of their own paper. ^[ref]

The first $100000$ digits of Champernowne constant have the following exact counts, in digit order $0,1,\dots ,9$:For the $99999$ adjacent pairs fully inside that prefix, the words $00,01,12,99$ occur $642,1753,2088,753$ times respectively. Thus, for example, $F_{100000}(12)=2088/99999$, which is about $0.02088$, rather than $0.01$. The prefix ends partway through the decimal representation of $22222$; the many leading ones in earlier positive integers help explain the visible imbalance.

These are finite counts, not a proof of the normal number theorem: a finite prefix can be followed by an all-zero tail. The theorem for Champernowne constant concerns the infinite limit, and establishes base-ten normality, rather than absolute normality. Nor do computations prove normality for familiar constants: the normality of Pi and Euler's number remains unproved, as discussed by Bailey and Borwein. ^[ref]

For a finite prefix of $N\ge r$ digits in positional notation, the digit block frequency $F_N(w)$ of a length-$r$ word $w$ is its number of overlapping occurrences divided by $N-r+1$, the number of available starting positions. All such digit block frequencies sum to one. The infinite limit used for a normal number can instead divide counts by $N$: the factor $(N-r+1)/N$ tends to one for fixed $r$.

For the six binary digits $010101$, the single-digit counts are three zeros and three ones. Its five adjacent pairs are $01,10,01,10,01$, givingThis example explains both the overlapping convention and the end correction in a finite sample. The corresponding infinite binary expansion has pair frequencies $0,1/2,1/2,0$, rather than the four values $1/4$ required of a normal number in base two.

A real number $x$ is a normal number in base $K$ exactly when its iterates under the integer multiplication map on the circle, $T_K(x)=\{Kx\}$, form an equidistributed sequence. Braces denote the fractional part. Multiplication by $K$ shifts the positional notation one digit to the left:A length-$r$ word with base-$K$ integer value $a$ specifies the half-open interval $[a/K^r,(a+1)/K^r)$, whose length is $K^{-r}$. Visiting that interval is exactly encountering that word at the next digit position. These digit intervals form arbitrarily fine grids, so their correct limiting visiting frequencies imply equidistribution on every interval. This converts a question about digits into one about a dynamical system. Bailey and Crandall develop this connection using equidistribution of the fractional parts of $K^{n}x$. ^[ref]

In base ten the word $12$ corresponds to $[0.12,0.13)$. For Champernowne constant, counting the visits $T_{10}^n(x)$ for $0\le n\le 99998$ gives precisely the $2088$ occurrences of $12$ in its first $100000$ digits. For $x=1/3$, the base-two integer multiplication map on the circle instead alternates between $1/3$ and $2/3$; it never visits $[0,1/4)$, the interval for $00$. This two-point orbit is not an equidistributed sequence.

A real number is normal in an integer base $K\ge 2$ when every finite word of digits occurs in its positional notation with the same limiting digit block frequency as every other word of the same length. Write its fractional part as $x=0.d_1{d}_2\dots$ in base $K$, using the expansion that is not eventually all $K-1$ digits when two expansions are possible. For every length $r\ge 1$ and every word $w\in \{0,\dots ,K-1{\}}^r$, the requirement isOccurrences overlap: the word $11$ occurs twice in $111$. There are $K^r$ possible words of length $r$, so $K^{-r}$ is the equal share for each. In a decimal representation, this means frequency $1/10$ for each digit, $1/100$ for each pair, and $1/1000$ for each triple, continuing through every finite length. The definition agrees with that used by Copeland and Erdős in their work on explicit normal numbers. ^[ref]

The appeal is that a precise, deterministic real number can have the same limiting digit block frequencies as repeated draws from a uniform distribution on a finite set, with independence between draws. Champernowne constant is an explicit example. Normality is useful for studying digit distribution, but balanced digits alone establish only a simply normal number, and a finite sample cannot establish any of these infinite limits.

Émile Borel introduced normal numbers and established their prevalence in his 1909 paper applying probability to number theory; the stronger all-base statement is discussed under absolutely normal number. ^[ref]

Use the probability-system convention $\mu (X)=1$. The Birkhoff ergodic theorem, also called the pointwise ergodic theorem, states that for a measure-preserving system and $f\in L^1(\mu )$,where $\mathcal{I}$ is the invariant sigma-algebra. The limit is integrable and has the same integral as $f$. On a probability space the convergence also holds in $L^1$, as in the allowed mean ergodic theorem. If $T$ is an ergodic transformation, $\mathcal{I}$ is trivial modulo null sets, giving

The integer multiplication map on the circle preserves Lebesgue measure: for any integrable $g$ on $[0,1)$,To prove the ergodicity of integer multiplication on the circle, suppose $u\in L^2(m)$ satisfies $u\circ T_K=u$. Let $c_j$ be its Fourier coefficients in the Fourier basis $e_j(x)=e^{2\pi ijx}$. Since $e_j\circ T_K=e_{Kj}$, the Fourier coefficients of $u\circ T_K$ at index $j$ are zero if $K$ does not divide $j$, and are $c_{j/K}$ otherwise. This identity holds for all $L^2$ functions by approximation with trigonometric polynomials and the isometry $u\mapsto u\circ T_K$. Invariance givesEvery nonzero integer $j$ can be divided by $K$ only finitely often. Thus $c_j=0$ for every $j\ne 0$, and completeness of the Fourier basis makes $u$ constant almost everywhere. Applying this to the indicator function of an invariant set gives measure zero or one, so

A normal number in base $K$ has every word $w=(w_1,\dots ,w_r)$ of $r$ base-$K$ digits occurring with limiting overlapping frequency $K^{-r}$. Use the expansion that is not eventually equal to $K-1$ when there are two expansions. The word $w$ corresponds to the half-open intervalA word starting at position $n+1$ occurs exactly when $T_K^{n}x\in I_w$. The Birkhoff ergodic theorem, applied to $1_{I_w}$, gives frequency $m(I_w)=K^{-r}$ almost everywhere. There are countably many pairs $(K,w)$, so their full-measure sets have a full-measure intersection. In particular,These are absolutely normal numbers, so existence follows as well. This interval description also proves normality and equidistribution under integer multiplication: the base-$K$ intervals form arbitrarily fine grids, so their frequencies imply the correct frequency for every interval by approximation from inside and outside.

For the growth assertion, put $g=|f|$. For every $\epsilon >0$, the Tonelli theorem gives the useful summability boundSince $T$ is a measure-preserving transformation, $\mu (g\circ T^n>\epsilon n)=\mu (g>\epsilon n)$. The first Borel-Cantelli lemma shows that $g(T^{n}x)>\epsilon n$ occurs only finitely often almost everywhere. Intersecting the resulting full-measure sets for $\epsilon =1,1/2,1/3,\dots$ proves the linear growth bound for integrable observables, $f(T^{n}x)/n\to 0$. Multiplication by $n^{1-a}\le 1$ then gives

The threshold is sharp. For $0<a<1$, choose $p$ with $1<p<1/a$, and take the Bernoulli shift on $X=(0,1{)}^{\mathbb{Z}}$ with the product measure of independent uniform coordinates. The left shift preserves that measure because it preserves the probability of every finite-coordinate event. Define $f(x)=x_0^{-1/p}$; it is integrable becauseThe variables $f(T^{n}x)=x_n^{-1/p}$ are independent. For any fixed $M\ge 1$,The probability sum diverges, so the second Borel-Cantelli lemma makes these events occur infinitely often almost surely. Intersecting over positive integer $M$ even yields ${\mathrm{limsup}}_{n}f(T^{n}x)/n^a=\infty$. For $a\le 0$, the constant observable $f=1$ already fails to give limit zero. Thus the sharpness of the linear growth bound for integrable observables gives

The Rudolph measure rigidity theorem has an essential ergodicity hypothesis. If a Borel probability measure $\mu$ on the circle group is invariant under both $T_2$ and $T_3$, is ergodic for the semigroup generated jointly by these maps, and either map has positive Kolmogorov-Sinai entropy, thenEquivalently, a jointly ergodic common invariant measure other than Lebesgue measure has zero entropy for both maps. Joint ergodicity means that every set invariant modulo $\mu$ under both maps has measure zero or one. Positive entropy without this hypothesis is insufficient: $\frac{1}{2}m+\frac{1}{2}{\delta }_0$, with ${\delta }_0$ a Dirac measure, is a common invariant measure of positive entropy and is not $m$.

The Host equidistribution theorem states that if $p,q\ge 2$ are relatively prime integers and $\nu$ is invariant and ergodic under $T_p$, with $h_{\nu }(T_p)>0$, then for $\nu$-almost every $x$ the sequence $(T_q^{n}x)$ is an equidistributed sequence for Lebesgue measure. Explicitly, for every continuous $g$ on the circle,The non-ergodic form assumes $T_p$ invariance and positive entropy for almost every component in the ergodic decomposition $\nu =\int {\nu }_{\omega }d\lambda (\omega )$. Applying the ergodic theorem of Host on each such component gives the same almost-everywhere conclusion for $\nu$. More generally, its conclusion holds on the part supported on positive-entropy components. A positive value of $h_{\nu }(T_p)$ alone does not eliminate zero-entropy components.

To deduce the joint version of the Rudolph measure rigidity theorem, suppose $h_{\mu }(T_2)>0$; if only $T_3$ has positive entropy, interchange the roles. Write the $T_2$ ergodic decomposition as $\mu =\int \nu d\tau (\nu )$. Since $T_3$ commutes with $T_2$, its pushforward measure sends a $T_2$ ergodic component $\nu$ to a $T_2$ ergodic component $(T_3{)}_{\ast }\nu$. On each component, $T_3$ is a factor of a measure-preserving system with fibres of size at most three. We use the standard entropy preservation under a finite-to-one factor:The reason for this standard entropy fact is that, conditional on a complete factor point, every finite orbit name has at most three possibilities; its conditional entropy is bounded by $\log 3$, and division by the orbit length gives zero relative entropy.

The component at $T_{3}x$ is $(T_3{)}_{\ast }{\nu }_x$ almost everywhere; this follows from commutation and the componentwise ergodic averages. The component entropy function $r(x)=h_{{\nu }_x}(T_2)$ is therefore invariant under both $T_2$ and $T_3$. Joint ergodicity makes it constant almost everywhere, and affinity of entropy under ergodic decomposition identifies the constant as $h_{\mu }(T_2)>0$. Thus almost every $T_2$ component has positive entropy, exactly the condition required in the non-ergodic Host equidistribution theorem. It follows that $\mu$-almost every point equidistributes for $m$ under $T_3$.

For any continuous $g$, invariance under $T_3$ and the dominated convergence theorem now giveContinuous functions determine Borel probability measures on the circle, so $\mu =m$, proving the deduction.

For the normal-number example, let ${\omega }_1,{\omega }_2,\dots$ be independent fair binary digits and defineThis Cantor Bernoulli measure is supported on the middle-third Cantor set $C$. If $S$ is the Bernoulli shift, then $T_3\circ \pi =\pi \circ S$ on the circle. Consequently $\nu$ is $T_3$ invariant and ergodic: a $T_3$ invariant event pulls back to an $S$ invariant event, which has probability zero or one.

Take the ternary digit measurable partition $\eta =\{[0,1/3),[1/3,2/3),[2/3,1)\}$. Its block partition of length $n$ has, up to null endpoints, $2^n$ positive-measure atoms under $\nu$, each of measure $2^{-n}$, and all other atoms have measure zero. HenceApply the Host equidistribution theorem with $p=3$, $q=2$. For $\nu$-almost every $x\in C$, the sequence $(T_2^{n}x)$ equidistributes for Lebesgue measure, so $x$ is a normal number in base $2$ by normality and equidistribution under integer multiplication.

The ternary expansion of $\nu$-almost every such $x$ contains only $0$ and $2$, so the frequency of digit $1$ is zero rather than $1/3$. The ambiguous ternary endpoints form a countable $\nu$ null set and can be removed. Therefore $x$ is not a normal number in base $3$. We have proved the stronger almost-everywhere existence statement

For the squared discrepancy of a digit frequency, write $d_N^2=(f_0-0.1{)}^2$. The plotted ordinates are $10^{-2}$, $4.00\times 10^{-4}$, $1.69\times 10^{-4}$, $4.90\times 10^{-5}$, $3.60\times 10^{-5}$ and $9.00\times 10^{-6}$ in increasing $N$. Under an independent, equiprobable-digit model, the zero count is binomial andThus the expected log-log slope is $-1$. If “relative” means division by the expected fraction before squaring, every ordinate is multiplied by 100 and the prediction becomes $9/N$; the slope is unchanged. These are nested prefixes of one digit sequence, so the points are correlated, rounded observations, not independent variance estimates. Their rough trend supports a reciprocal-power scale but does not prove independence or that $\pi$ is a normal number.

A real number is simply normal in base $K$ if each individual digit in its positional notation has limiting digit block frequency $1/K$. This is the length-one part of the definition of a normal number. It places no equal-frequency requirement on longer words.

For example, $x=0.010101{\dots }_2=1/3$ is a simply normal number in base two: each binary digit has frequency $1/2$. But $00$ and $11$ never occur, and $01$ and $10$ each have frequency $1/2$. Thus $x$ is not a normal number in base two. Even a sequence that looks more complicated, such as the deterministic Thue–Morse sequence, must be tested against the actual digit block frequency definition; visual complexity is no substitute for it.

The decimal representation $y=0.01234567890123456789\dots$ makes the same point in base ten. Every digit has frequency $1/10$, yet the only adjacent pairs areEach has limiting digit block frequency $1/10$; the other ninety pairs have frequency zero. A normal number in base ten requires $1/100$ for every one of the hundred pairs. This rational number is a simply normal number in base ten, but it is not a normal number in that base.