Proving Why Almost Every Number is Normal
Omhari Indoria, Utkarsh Bharadwaj, Kavadi Shyam Sundar, Harsh Kumar, Abhinav
October 2026
Definitions of a Normal Number in base \(r \geq 2\) and Lebsegue Measure
Borel’s Proof of Absolute Normality (with i.i.d assumption)
Hausdorff’s Proof of Simple Normality (without i.i.d assumption)
Generalisation of Hausdorff’s Proof to base \(r \geq 2\)
Simply Normal: For \(b \in \{0, \ldots, r - 1\} = \Sigma_{r}\), define \[p_{b, n, \alpha, r} = |\left\{j \in \mathbb{N}^{+} : j \leq n \text{ and } \alpha_{j, r} = b\right\}|\] which counts how often the digit \(b\) occurs among the first \(n\) digits of \(\alpha\). A number \(\alpha \in \mathbb{R}\) is called Simply Normal in base \(r\) if \[ \lim_{n \to \infty} \dfrac{p_{b, n, \alpha, r}}{n} = \dfrac{1}{r} \] holds for every digit \(b \in \Sigma_r\)
Normal and absolutely normal: The number \(\alpha\) is called normal in base r if \(r^m \alpha\) is simply normal in base \(r^n \; \forall n \in \mathbb{N}^{+}\) and all \(m \in N\). The number \(\alpha\) is called absolutely normal if it is normal in every integer base \(r \geq 2\)
Example: Champernowne number (in base 10) \[C_{10} = 0.1\,2\,3\,4\,5\,6\,7\,8\,9\,1\!0\,1\!1\,1\!2\,1\!3\ldots\] This number is normal is base 10 although is worthy to note that proving \(C_{10}\) absolutely normal reamins an open problem
Measure: It is a function \(\mu\) that takes a certain subset of a given set \(X\) and assigns it to a number on the non-negative extended real number line \(\left(i.e., [0, \infty]\right)\). It also satisfies the following two properties:
\(\mu (\emptyset) = 0\)
\(\sigma-\) additivity: If \(A_1, A_2, \ldots,\) are pairwise disjoint measurable sets, then \[\mu \left(\bigcup_{i = 1}^{\infty} A_i\right) = \sum_{i = 1}^{\infty} \mu (A_i)\]
Lebsegue Measure: It is a measure on \(\mathbb{R}\) which satisfies:
\(\lambda \left([a, b]\right) = b - a\) (likewise for open and half open intervals)
\(\lambda(A + c) = \lambda(A)\) for all measurable sets \(A\) and \(c \in \mathbb{R}\)
Null Set: A set \(A\) is called a Null set if \(\lambda(A) = 0\). Equivalently, \(A\) is a null set if, for every \(\varepsilon > 0\), one can find countably many intervals whose union covers \(A\) and whose total length is at most \(\varepsilon\).
Examples
Singleton Set
Countable Union of Singleton Set
Cantor Set
Émile Borel was French mathematician who first introduced and proved the results of normal of real numbers back in 1909. His proof is based on a quintessential assumption that the digits of a random number in \([0, 1)\) in base \(q\) behaves like i.i.d random variable, each uniform on \(\Sigma_q\).
Almost all real numbers in the interval \([0, 1)\) are absolutely normal
We’ll now spend the next couple of minutes trying to understand his proof
Fix a base \(q \geq 2\). Every \(x \in [0, 1)\) has a unique base-\(q\) expansion (except for a countable, hence null, set of terminating fractions): \[x = \sum_{i=1}^{\infty} \frac{x_i}{q^i}, \qquad x_i \in \Sigma_q = \{0, 1, \ldots, q-1\}\] Fix a digit \(d \in \Sigma_q\) and let \[S_n(x) = \text{number of times } d \text{ appears among } x_1, \ldots, x_n\] Write \(p = \frac{1}{q}\) and \(p' = 1 - \frac{1}{q}\).
Picking \(x \in [0,1)\) “at random” is like rolling a fair \(q\)-sided die forever, where \(x_i\) is the \(i\)-th roll. Then \(S_n\) counts how many times we saw the face \(d\) in \(n\) rolls, so we expect \(S_n \approx np\).
Goal: show \(\dfrac{S_n(x)}{n} \to \dfrac{1}{q}\) for almost every \(x\).
Since the digits behave like i.i.d. rolls, \(S_n\) is Binomial: \[\mathbb{P}(S_n = k) = \binom{n}{k} p^{k} (p')^{n-k}\] Write \(k = np + s\), where \(s\) is the deviation from the expected count. Using Stirling’s approximation \(N! \sim N^N e^{-N}\sqrt{2\pi N}\) and \(\ln(1+u) \approx u - \frac{u^2}{2}\) for small \(u\), the linear terms cancel and what is left is \[\ln \mathbb{P}(np + s) \approx -\frac{s^2}{2npp'}\]
\[\mathbb{P}(np + s) \approx \frac{1}{\sqrt{2\pi npp'}} \, e^{-s^2/(2npp')}\] A bell curve centred at \(np\) with width about \(\sqrt{npp'}\).
Choose \(\varepsilon \in (0, \frac{1}{2})\) and look at deviations of size \(s = n^{1/2 + \varepsilon}\), slightly bigger than the typical \(\sqrt{n}\). Let \[E_n = \left\{x \in [0,1) : |S_n(x) - np| \geq n^{1/2 + \varepsilon}\right\}\] Then \(\dfrac{s^2}{2npp'} = \dfrac{n^{2\varepsilon}}{2pp'}\), so integrating the Gaussian tails gives \[\mu(E_n) \leq A\, e^{-c\, n^{2\varepsilon}}\] for positive constants \(A, c\) depending only on \(p\) and \(p'\).
\(e^{-c n^{2\varepsilon}}\) decays faster than any inverse polynomial of \(n\). In particular \[\sum_{n=1}^{\infty} \mu(E_n) < \infty\]
If \(\sum_{n} \mu(E_n) < \infty\), then \(\mu(\limsup_{n} E_n) = 0\). That is, almost every \(x\) lies in only finitely many of the \(E_n\).
Why? \(\limsup E_n \subseteq \bigcup_{k \geq n} E_k\) for every \(n\), so \[\mu(\limsup E_n) \leq \sum_{k \geq n} \mu(E_k) \xrightarrow{n \to \infty} 0\] since it is the tail of a convergent series.
Hence for almost every \(x\) there is an \(N_x\) with \(|S_n(x) - np| < n^{1/2+\varepsilon}\) for all \(n \geq N_x\). Dividing by \(n\): \[\left|\frac{S_n(x)}{n} - p\right| < \frac{1}{n^{1/2 - \varepsilon}} \xrightarrow{n \to \infty} 0 \quad \Longrightarrow \quad \lim_{n\to\infty}\frac{S_n(x)}{n} = \frac{1}{q}\]
We climb up in four small steps, each one using “countable intersections of full-measure sets are still of full measure”.
All digits at once: there are only \(q\) digits, so almost every \(x\) is simply normal in base \(q\).
Blocks: grouping digits into blocks of length \(k\) in base \(q\) is the same as using single digits in base \(Q = q^k\). Apply Steps 1–3 with \(Q\). Shifting by \(m\) places (\(q^m x\)) only drops the first \(m\) digits, which cannot change limiting frequencies. So almost every \(x\) is normal in base \(q\) for each \(k\) and \(m\).
All block lengths and shifts: \(k, m\) range over a countable set, so almost every \(x\) is normal in base \(q\).
All bases: \(\{2, 3, 4, \ldots\}\) is countable, so almost every \(x \in [0, 1)\) is absolutely normal. \(\qquad \blacksquare\)
Today, Steps 1–3 are replaced by a single sentence. View \(([0,1), \mathcal{B}, \mu)\) as a probability space and the digits \(x_i\) as i.i.d. variables. With the indicator \(I_i(x) = \mathbf{1}[x_i = d]\) we have \(\mathbb{E}[I_i] = \frac{1}{q}\) and \(S_n = \sum_{i=1}^{n} I_i\), so the Strong Law of Large Numbers gives \[\lim_{n \to \infty} \frac{S_n(x)}{n} = \frac{1}{q} \quad \text{for almost all } x\]
But Borel published in 1909, before the SLLN existed. By bounding the deviations by hand (Stirling) and using countable intersections, Borel effectively proved the SLLN for Bernoulli trials.
Question: can we avoid the i.i.d. assumption and use only counting? Hausdorff says yes.
In 1914, Felix Hausdorff gave a proof that is purely combinatorial: it never mentions random variables, only lengths of intervals.
Take an irrational \(x \in (0,1)\) with binary expansion \[x = \frac{x_1}{2} + \frac{x_2}{2^2} + \frac{x_3}{2^3} + \cdots, \qquad x_n \in \{0, 1\}\] (rationals are countable, hence null). Among the first \(n\) digits let \(p\) be the number of zeros and \(q = n - p\) the number of ones.
The set of \(x \in (0,1)\) for which \(\lim_{n \to \infty} \frac{p}{n} = \frac{1}{2}\) has Lebesgue measure 1.
The \(x\)’s whose expansion starts with prescribed digits \(a_1, \ldots, a_n\) form an interval \[\frac{a_1}{2} + \cdots + \frac{a_n}{2^n} < x < \frac{a_1}{2} + \cdots + \frac{a_n + 1}{2^n}\] of length \(\dfrac{1}{2^n}\).
There are \(\binom{n}{p}\) ways to place \(p\) zeros among \(n\) positions, so the set where exactly \(p\) of the first \(n\) digits are zero is a union of \(\binom{n}{p}\) disjoint intervals, of total measure \(\binom{n}{p}\dfrac{1}{2^n}\).
For \(\varepsilon > 0\) let \(M(n, \varepsilon) = \left\{x : \left|\frac{p}{n} - \frac{1}{2}\right| \geq \varepsilon\right\}\). Its measure is \[m(n, \varepsilon) = \sum_{\substack{p = 0 \\ |p/n - 1/2| \geq \varepsilon}}^{n} \binom{n}{p} \frac{1}{2^n}\]
We want to show \(m(n, \varepsilon)\) is small, without a bell curve.
Start from the binomial identity \(\sum_{p=0}^{n} \binom{n}{p} x^p y^q = (x+y)^n\) (with \(q = n-p\)) and apply the operator \(x\frac{\partial}{\partial x} - y\frac{\partial}{\partial y}\) repeatedly. Each application multiplies the \(p\)-th term by \((p - q)\). With \(u = x+y\) and \(v = x-y\): \[\begin{aligned} \sum_{p} \tbinom{n}{p} (p-q)^2\, x^p y^q &= n u^n + n(n-1)u^{n-2}v^2 \\ \sum_{p} \tbinom{n}{p} (p-q)^4\, x^p y^q &= n(3n-2)u^n + n(n-1)(6n-8)u^{n-2}v^2 \\ &\quad + n(n-1)(n-2)(n-3)u^{n-4}v^4 \end{aligned}\] Now set \(x = y = \frac{1}{2}\), so \(u = 1\) and \(v = 0\). Every term with \(v\) vanishes!
\[\sum_{p=0}^{n} \binom{n}{p}\frac{1}{2^n}(p-q)^4 = 3n^2 - 2n < 3n^2\]
Since \(p - q = 2p - n\), dividing by \(n^4\) gives \[\sum_{p=0}^{n} \binom{n}{p}\frac{1}{2^n}\left(\frac{p}{n} - \frac{1}{2}\right)^4 < \frac{3}{16 n^2}\] On the bad set \(\left|\frac{p}{n} - \frac{1}{2}\right| \geq \varepsilon\), so \(\left(\frac{p}{n} - \frac{1}{2}\right)^4 \geq \varepsilon^4\). Keeping only those terms: \[\varepsilon^4 \cdot m(n, \varepsilon) \leq \sum_{p} \binom{n}{p}\frac{1}{2^n}\left(\frac{p}{n} - \frac{1}{2}\right)^4 < \frac{3}{16n^2}\]
\[m(n, \varepsilon) < \frac{3}{16\, \varepsilon^4\, n^2}\]
Why the fourth power? A second power would give a bound of order \(\frac{1}{n}\), and \(\sum \frac{1}{n}\) diverges. Order \(\frac{1}{n^2}\) is summable, and that is exactly what we need next.
Let \(M(\varepsilon) = \limsup_{n} M(n, \varepsilon)\), the \(x\) with \(|\frac{p}{n} - \frac{1}{2}| \geq \varepsilon\) for infinitely many \(n\). Define \(S(n, \varepsilon) = \bigcup_{k \geq n} M(k, \varepsilon)\). Then \(M(\varepsilon) = \bigcap_{n=1}^{\infty} S(n, \varepsilon)\) and \[\lambda\big(S(n, \varepsilon)\big) \leq \sum_{k=n}^{\infty} m(k, \varepsilon) \leq \frac{3}{16\varepsilon^4}\sum_{k=n}^{\infty}\frac{1}{k^2} \xrightarrow{n \to \infty} 0\] because \(\sum \frac{1}{k^2}\) converges. Hence \(\lambda(M(\varepsilon)) = 0\).
The set where \(\lim \frac{p}{n} \neq \frac{1}{2}\) is \(M = \bigcup_{k=1}^{\infty} M(1/k)\), a countable union of null sets, hence null. Its complement has measure \(1\). \(\qquad \blacksquare\)
Notice that no probability and no i.i.d. assumption was used, only lengths of intervals.
Almost all real numbers in \(\mathbb{R}\) are simply normal for base 2.
Why start on \([0,1)\)? It has measure exactly \(1\), so lengths act like probabilities and each \(n\)-digit string corresponds to an interval of length \(2^{-n}\). On \(\mathbb{R}\) (infinite measure) there is no uniform probability.
Proof idea:
Write \(x = m + y\) with \(m = \lfloor x \rfloor \in \mathbb{Z}\) and \(y \in [0,1)\). The integer part is a finite prefix, so \(x\) is simply normal iff \(y\) is.
The non-normal numbers in \([m, m+1)\) are \(E_m = N^c + m\), so by translation invariance \(\mu(E_m) = \mu(N^c) = 0\).
\(E = \bigcup_{m \in \mathbb{Z}} E_m\) and countable subadditivity give \(\mu(E) \leq \sum_m \mu(E_m) = 0\). \(\qquad \blacksquare\)
Hausdorff only treated base \(2\). Following Edmund Weitz (2021), we now extend the same idea to any base \(r \geq 2\). The plan:
Two small lemmas on null sets.
Normality and words (patterns of digits).
The Main Lemma: Hausdorff’s fourth-moment computation for base \(r\).
Conclude: almost all numbers are normal, then absolutely normal.
In base \(r\), a given digit \(b\) appears with probability \(\frac{1}{r}\) and any other digit with probability \(\frac{r-1}{r}\). So we redo the base-\(2\) computation with \(\left(\frac{1}{2}, \frac{1}{2}\right)\) replaced by \(\left(\frac{1}{r}, \frac{r-1}{r}\right)\).
Every countable set is a null set.
Proof. Enumerate \(A = \{a_0, a_1, \ldots\}\) and fix \(\varepsilon > 0\). Cover \(a_n\) by an interval \(I_n\) of length \(\varepsilon/2^{n+1}\). Then \(A \subseteq \bigcup_n I_n\) and \(\sum_n \lambda(I_n) = \sum_n \frac{\varepsilon}{2^{n+1}} = \varepsilon\). As \(\varepsilon\) is arbitrary, \(\lambda(A) = 0\).
A countable union of null sets is a null set.
Proof. Let \(A_1, A_2, \ldots\) be null and \(\varepsilon > 0\). Cover \(A_k\) by intervals \(\{I_{k,j}\}_j\) of total length \(\leq \varepsilon/2^k\). All the \(I_{k,j}\) together cover \(\bigcup_k A_k\), with total length \(\leq \sum_k \varepsilon/2^k = \varepsilon\). \(\qquad \blacksquare\)
A finite string \(w = b_1 b_2 \cdots b_n\) of base-\(r\) digits is an \(r\)-word, with length \(|w| = n\). Let \[p_{w, n, \alpha, r} = \left|\left\{j \in \mathbb{N}^{+} : j + |w| - 1 \leq n \text{ and } \alpha_{[j,\, j+|w|-1],\, r} = w\right\}\right|\] count the (possibly overlapping) occurrences of \(w\) among the first \(n\) digits of \(\alpha\).
If \(r^m\alpha\) is simply normal in base \(r^n\) for all \(n \in \mathbb{N}^{+}\) and all \(m < n\), then \(\alpha\) is normal in base \(r\).
Idea: for \(m \geq n\) write \(m = kn + c\) with \(0 \leq c < n\). Multiplying by \(r^{kn}\) only drops the first \(k\) blocks, so \(r^m\alpha\) and \(r^c\alpha\) have the same base-\(r^n\) digits from some point on, and \(c < n\) is covered by the hypothesis.
If \(\alpha\) is normal in base \(r\), then for every \(r\)-word \(w\), \[\lim_{n \to \infty} \frac{p_{w, n, \alpha, r}}{n} = \frac{1}{r^{|w|}}\]
Idea of the proof. Let \(|w| = L\) and \(R = r^L\). The word \(w\) is a single digit \(d_w\) in base \(R\).
Grouping the digits of \(\alpha\) into blocks of length \(L\) finds the copies of \(w\) that start at positions \(1, L+1, 2L+1, \ldots\) Frequency: \(\frac{1}{R}\).
For \(c = 1, \ldots, L-1\), the number \(r^c\alpha\) is also simply normal in base \(R\), which finds copies starting at \(c+1, c+L+1, \ldots\) Frequency: \(\frac{1}{R}\) each.
Adding the \(L\) groupings: \(p_{w,n,\alpha,r} \approx L \cdot \frac{n/L}{r^L} = \frac{n}{r^L}\).
Takeaway: in a normal number every finite pattern (your birthday, a whole sentence in binary...) appears, and with the “fair” frequency.
If \(r \geq 2\) is an integer, there is a constant \(D > 0\) such that for all \(n \in \mathbb{N}^{+}\) \[\sum_{p=0}^{n} \binom{n}{p}\frac{(r-1)^{n-p}}{r^n}\left(\frac{p}{n} - \frac{1}{r}\right)^4 \leq \frac{D}{n^2}\]
What does it say? The sum is the fourth central moment (divided by \(n^4\)) of the number of times a digit appears: the fraction \(\frac{p}{n}\) is, on average, \(\frac{1}{r}\), and being \(\varepsilon\)-far from it is a rare event of size \(\sim \frac{1}{n^2}\).
Strategy: copy Hausdorff’s differentiation trick, now with the weights \(\left(\frac{1}{r}\right)^p\left(\frac{r-1}{r}\right)^{n-p}\).
Fix \(s = r - 1\) and \(n\). Define \(f_0(x,y) = \sum_{p=0}^{n}\binom{n}{p}x^{sp}y^{n-p} = (x^s + y)^n\) and \(f_{k+1} = x\frac{\partial f_k}{\partial x} - y\frac{\partial f_k}{\partial y}\). Differentiating term by term, induction shows \[f_k(x,y) = \sum_{p=0}^{n}\binom{n}{p}\big((s+1)p - n\big)^k x^{sp}y^{n-p}\] Computing \(f_4\) from \((x^s+y)^n\) is tedious but elementary (a computer algebra system helps). Now choose \[x = r^{-1/s}, \qquad y = \frac{r-1}{r} \quad\Longrightarrow\quad x^s = \frac{1}{r},\; s\,x^s = y\] so the troublesome terms vanish, just like \(v = 0\) in base \(2\), and \[f_4(x,y) = 3(r-1)^2 n^2 + (r^3 - 7r^2 + 12r - 6)\,n \leq C n^2\] With this choice \(x^{sp}y^{n-p} = \frac{(r-1)^{n-p}}{r^n}\), so the formula for \(f_4\) reads \(\sum_p \binom{n}{p}\frac{(r-1)^{n-p}}{r^n}(rp-n)^4 \leq Cn^2\). Dividing by \(r^4n^4\) gives the Lemma with \(D = C/r^4\). \(\qquad \blacksquare\)
If \(r \geq 2\) is any base, almost all numbers are simply normal in base \(r\).
Proof. Work in \([0,1)\), fix a digit \(b\) and let \(M_b(n, \varepsilon) = \left\{\alpha : \left|\frac{p_{b,n,\alpha,r}}{n} - \frac{1}{r}\right| \geq \varepsilon\right\}\). Numbers with exactly \(p\) copies of \(b\) among the first \(n\) digits have measure \(\binom{n}{p}\frac{(r-1)^{n-p}}{r^n}\). As in Hausdorff’s proof, using \(\varepsilon^4 \leq (\frac{p}{n} - \frac{1}{r})^4\) on the bad set and the Main Lemma: \[\lambda\big(M_b(n,\varepsilon)\big) \leq \frac{D}{\varepsilon^4 n^2}\] Then, exactly as before: \(\lambda\big(\bigcup_{k \geq m} M_b(k,\varepsilon)\big) \leq \frac{D}{\varepsilon^4}\sum_{k \geq m}\frac{1}{k^2} \to 0\), so the set \(M_b(\varepsilon)\) of \(\alpha\) that are \(\varepsilon\)-bad for infinitely many \(n\) is null.
Finally \(\bigcup_{k}M_b(1/k)\) is null (Lemma 4.2), and so is the union over the \(r\) digits \(b\). \(\qquad \blacksquare\)
If \(r \geq 2\), almost all numbers are normal in base \(r\).
Proof. Fix \(m \in \mathbb{N}\) and \(n \in \mathbb{N}^{+}\), and let \(N\) be the set of reals that are not simply normal in base \(r^n\). Theorem 4.1 gives \(\lambda(N \cap [0,1)) = 0\). The integer part does not affect digits after the radix point, so \(N \cap [j, j+1) = j + (N \cap [0,1))\) is also null, and by Lemma 4.2 so is \(N \cap [0, r^m)\). Multiplying by \(r^{-m}\) shrinks lengths by a constant factor, so the set of \(\alpha \in [0,1)\) with \(r^m\alpha \in N\) is null.
The pairs \((m, n)\) are countable, so Lemma 4.2 finishes the job. \(\qquad \blacksquare\)
Almost all numbers are absolutely normal.
Proof. For each base \(r \in \{2, 3, 4, \ldots\}\) the non-normal numbers form a null set (Cor. 4.3). There are countably many bases, so by Lemma 4.2 their union is null. \(\qquad \blacksquare\)
| Borel (1909) | Hausdorff (1914) / Weitz | |
|---|---|---|
| Viewpoint | Probability (i.i.d. digits) | Counting interval lengths |
| Key estimate | Stirling \(\Rightarrow\) Gaussian tail | Fourth moment \(\Rightarrow \frac{D}{n^2}\) |
| Summability | \(\sum e^{-cn^{2\varepsilon}} < \infty\) | \(\sum \frac{1}{n^2} < \infty\) |
| Conclusion step | Borel–Cantelli | Same idea, done by hand |
| Bases covered | Any \(q \geq 2\) | \(2\) (Hausdorff), any \(r \geq 2\) (Weitz) |
Both roads lead to the same place: the “bad” sets are so small that they can be added up (countably many times) without ever reaching positive measure. So almost every real number is normal, yet explicit examples (e.g. \(\pi\), \(\sqrt{2}\)) are still unproven!
Borel, É. Les probabilités dénombrables et leurs applications arithmétiques. Suppl. Rend. Circ. mat. Palermo 27, 247–271 (1909).
Champernowne, D. G. The construction of decimals normal in the scale of ten. J. London Math. Soc. 8, 254–260 (1933).
Hausdorff, F. Grundzüge der Mengenlehre. Veit & Comp., Leipzig (1914).
Weitz, E. Hausdorff’s forgotten proof that almost all numbers are normal. arXiv:2102.00493v5 [math.NT] (2021).
Questions?