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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.

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