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An increasing sequence of integers has full density when its range has natural density one. Convergence in density of a sequence means that, for each fixed tolerance, the exceptional indices have natural density zero.

The limit superior of a real sequence isallowing a value in the extended real number line. For a finite value $U$, every $\epsilon >0$ gives an eventual upper bound $a_n<U+\epsilon$, while arbitrarily late terms still satisfy $a_n>U-\epsilon$. The second condition concerns recurrence at arbitrarily large indices, not the fraction of indices where a large value occurs.

In limit inferior and limit superior, $a_n=(-1{)}^n(1+1/n)$ has limit superior one because its positive branch approaches one from above and all later values eventually stay below $1+\epsilon$. Changing any finite number of terms leaves the limit superior unchanged: once the tail of a sequence begins beyond those positions, all its suprema are the same as before. The sparse peaks can determine a limit superior example shows that even a set of peak indices with zero natural density can decide the upper long-run edge.

For a set $S$ of nonnegative integers, its natural density is the limiting proportion of the first $N$ nonnegative integers that belong to $S$:provided this limit exists. Write $C_S(N)=|S\cap \{0,\dots ,N-1\}|$ for the counting function. The numerator is a finite cardinality; the denominator is exactly $N$, including zero. Every existing natural density lies between zero and one.

For the even nonnegative integers $E=\{0,2,4,\dots \}$, there are $m$ elements below $N=2m$ and $m+1$ below $N=2m+1$. Thus $C_E(N)=\lceil N/2\rceil$ using the ceiling function, and $d(E)=1/2$. For the square numbers $Q=\{0,1,4,9,\dots \}$, the exact count for $N\ge 1$ isThe floor function counts the allowed indices $1,\dots ,\lfloor \sqrt{N-1}\rfloor$, and the extra one counts zero. At $N=10$, the square numbers below ten are $0,1,4,9$, so the proportion is $4/10$; at $N=100$, it is $10/100$. The count grows like the square root of $N$, while the denominator grows like $N$.

The prime numbers also have natural density zero. If $P$ is their set, the Prime number theorem gives $C_P(N)\sim N/\log N$, hence $C_P(N)/N\to 0$. The Prime number theorem is a substantial result about the count, rather than a consequence of the fact that there are infinitely many prime numbers. ^[ref]

Why use natural density when cardinality already measures size? The even nonnegative integers, the square numbers, and the prime numbers all have the same countably infinite cardinality, yet their proportions among large initial integer intervals differ. Natural density distinguishes frequent membership from sparse membership. Equivalently, under a uniform distribution on a finite set of $N$ possible nonnegative integers, $C_S(N)/N$ is the probability of landing in $S$. A set without natural density shows why this limiting proportion need not exist.

For a probability measure-preserving system, strong mixing means that for every pair of measurable sets $A,B$,A sequence $a_n$ has convergence in density of a sequence to $a$ when for each $\epsilon >0$ the exceptional set $\{n:|a_n-a|\ge \epsilon \}$ has natural density zero. A weakly mixing measure-preserving transformation has convergence in density of $\mu (T^{-n}A\cap B)$ to $\mu (A)\mu (B)$ for every $A,B$. Equivalently, because these correlations are bounded,Indeed the mean of the absolute discrepancy is at least $\epsilon$ times the exceptional frequency, while it is at most $\epsilon$ plus a uniform bound times that frequency. A signed Cesaro convergence of a sequence without absolute values is insufficient to define weak mixing.

The standard three-cut Chacon map is obtained by cutting and stacking. Start on $[0,1)$ with normalized Lebesgue measure, an initial tower consisting of $[0,2/3)$, and a reservoir $[2/3,1)$ for spacers. At stage $j$, cut every level of the current tower into three equal subintervals, producing three subcolumns. Add one new interval of the same width above the middle subcolumn. Stack the first subcolumn at the bottom, then the middle subcolumn with its spacer, then the third at the top. Define the partial transformation by translation from each level to the next, leaving the current top unmapped. These assignments extend the earlier partial transformation. Repeat indefinitely.

If $h_j$ is the number of levels and $w_j$ their width, thenHence $h_j=(3^{j+1}-1)/2$, $w_j=2/3^{j+1}$, and the stage-$j$ tower has measure $1-3^{-j-1}$. The spacers consume total measure ${\sum }_{j\ge 0}{w}_{j+1}=1/3$, precisely the reservoir. The increasing partial maps give the Chacon map modulo null sets. The tower tops and unused reservoir have measures tending to zero, and the construction yields an invertible measure-preserving transformation. With $0$ marking an original level and $1$ a spacer, the tower words satisfy $W_0=0$ and $W_{j+1}=W_j{W}_{j}1W_j$, so the first new word is $0010$. This describes the spacer placement without needing a proof of well-definedness. The three-cut convention agrees with the classical constant spacer vector $(0,1,0)$ described in Ryzhikov's construction.

For the correlation assertion, take the L2 inner product to be $⟨u,v⟩=\int u\overline{v}d\mu$, linear in its first argument, and put $U=U_T$. The Koopman operator is an isometry on $L^2$, even when $T$ is not invertible. For $n\ge k$ the hint givesTo extend rigorously to all test functions, centre the observable: $v=f-a1$. Invariance of the integral gives $⟨U^{n}v,v⟩=⟨U^{n}f,f⟩-|a{|}^2\to 0$. LetFor each fixed $k$, $⟨U^{n}v,U^{k}v⟩\to 0$ by the same isometry identity. Therefore the limit is zero for every finite linear combination of these orbit vectors. By the Cauchy-Schwarz inequality and $\Vert U^{n}v{\Vert }_2=\Vert v{\Vert }_2$, approximation extends this conclusion to every $g\in M$: the approximation error in the correlation is bounded by $\Vert v{\Vert }_2\Vert g-g_0{\Vert }_2$ uniformly in $n$. For $g\in M^{\perp }$, the correlation is identically zero because $U^{n}v\in M$. The orthogonal decomposition by a closed subspace now gives the result for all $g\in L^2$. ThusThis is decay of autocorrelation implies weak convergence of an observable; no assumption that $T$ is an ergodic transformation, and no invertibility hypothesis was used, and $a=0$ is included.

Finally, strong mixing immediately implies the stated diagonal limit by taking $B=A$. Conversely, suppose that limit holds for every $A$. Apply the correlation result with $f=1_A$ and $g=1_B$. Its hypothesis is exactly $⟨U^n{1}_A,1_A⟩\to \mu (A{)}^2$, and its conclusion isHence strong mixing is equivalent to diagonal set-correlation criterion for mixing. Using the orbit-span proof avoids an invalid polarization argument that would recover only the sum of the two directed cross-correlations from diagonal correlations.

For an increasing sequence $(n_k)$ of nonnegative integers, full-density sequence means that its range $S$ has natural density one:For a sequence $(z_n)$ in a metric space, convergence in density of a sequence to $z$ means that, for every $\epsilon >0$,In particular, a bounded nonnegative scalar sequence converging to zero in density has averages tending to zero: the average is at most $\epsilon$ plus its bound times the proportion of indices above $\epsilon$.

Here is a Hilbert space form of the Van der Corput lemma, including the density version. Suppose $\Vert u_n\Vert \le M$, and putThenSince $0\le c_h\le M^2$, it suffices for $(c_h)$ to converge to zero in density. A frequently used special case is that the correlation average tends to zero for every fixed $h\ge 1$.

To prove the lemma, extend the finite list $u_0,\dots ,u_{N-1}$ by zero outside this range, and write $S_N={\sum }_{n=0}^{N-1}{u}_n$. For $1\le H\le N$,The Cauchy-Schwarz inequality for this sum, followed by expansion of the squared Hilbert space norm, givesThe coefficient $H-h$ counts the pairs of shifts separated by $h$. Dividing by $H^2{N}^2$, bounding real parts by absolute values, and first taking the limit superior in $N$ for fixed $H$ yieldsThe right side tends to zero as $H\to \infty$ under the stated hypothesis. This proves both the lemma and its density consequence.

For the polynomial application, fix a nonzero integer $m$ and let $u_n=e^{2\pi imP(n)}\in \mathbb{C}$. For each fixed $h\ge 1$, the correlation phase is linear:Thus, with $q_h=e^{2\pi i(2mah)}$,The coefficient $2mah$ is irrational, so $q_h\ne 1$. This geometric series is bounded in modulus by $2/(N|1-q_h|)$ and tends to zero. All $c_h$ vanish, so the Van der Corput lemma gives

The Weyl criterion now shows that the fractional parts form an equidistributed sequence. For completeness, its sufficiency here follows directly: the displayed limits give the correct average for every trigonometric polynomial, including the constant term. The Stone-Weierstrass theorem extends this to every continuous function by uniform approximation. Approximating an interval's indicator function above and below by continuous functions gives its length as the limiting frequency. HenceThis proves the equidistribution of a quadratic polynomial with irrational leading coefficient using the required correlation argument.

Include the positive integers in $[1,2),[4,8),[16,32),[64,128),\dots$, and exclude the intervening integer intervals and zero. Equivalently, include $[2^{2k},2^{2k+1})\cap \mathbb{Z}$ for $k\ge 0$. Each included block is large enough to raise the running proportion; the following excluded block lowers it again.

With $C_A(N)=|A\cap \{0,\dots ,N-1\}|$, the exact boundary counts follow from a geometric series:The first sum is empty at $k=0$. Dividing by the corresponding boundary, rather than using one denominator for both, givesFor example, $C_A(16)=1+4=5$, whereas $C_A(32)=1+4+16=21$, yielding proportions $5/16$ and $21/32$. For $4^k\le N\le 2\cdot 4^k$, the exact count is $(4^k-1)/3+N-4^k$; for $2\cdot 4^k\le N\le 4^{k+1}$ it stays $(4^{k+1}-1)/3$. Hence the proportion increases through each included block and decreases through each excluded block. These boundary subsequences give the actual limit inferior and limit superior:Since these differ, $A$ has no natural density. Its upper asymptotic density still records its largest limiting proportion.

Define a sequence by $a_n=1$ when $n$ is a power of two, including $1=2^0$, and $a_n=0$ otherwise. Every tail of a sequence contains both a later power of two and a later nonpower. Therefore $\mathrm{liminf}{a}_n=0$ and $\mathrm{limsup}{a}_n=1$, and the sequence does not converge. Its peak indices become increasingly sparse, yet never stop appearing.

Through index $N\ge 1$, the number of peaks is $1+\lfloor {\log }_{2}N\rfloor$, so their fraction isThe set of powers of two has natural density zero. The same fraction is the Cesàro mean of the first $N$ terms, which tends to zero even though the limit superior stays one. At $N=1024$, eleven peaks give mean $11/1024\approx 0.0107422$. An average and a long-run upper edge answer different questions about the same sequence.

A subsequential limit of a sequence $(a_n)$ is a value $L$ approached by some subsequence $(a_{n_k})$ with strictly increasing indices $n_k$. For $a_n=(-1{)}^n(1+1/n)$, the subsequences at even integers and odd integers give subsequential limits one and minus one. A finite value appearing among only finitely many initial terms does not become a subsequential limit merely by being a member of the image of a function represented by the sequence.

For a bounded sequence of real numbers, its subsequential limits form a nonempty bounded set that is also a closed set. The Bolzano-Weierstrass theorem guarantees nonemptiness, and its extreme values are the limit inferior and limit superior. The sequence in sparse peaks can determine a limit superior has the two subsequential limits zero and one, despite almost all indices in the natural density sense carrying zero.

For a subset $A$ of the positive integers, its upper asymptotic density isThe limit superior keeps the largest limiting proportion even when the running proportions oscillate. Because they lie in $[0,1]$, this limit superior always exists in $[0,1]$. If natural density exists, upper asymptotic density equals it.

The convention here uses $1,\dots ,N$, while natural density above uses $0,\dots ,N-1$. Replacing one interval by the other changes a count by at most one, so the proportions differ by at most $1/N$ and have the same limit superior. For the set without natural density above, the count in $\{1,\dots ,16\}$ is six rather than five, because $16$ is included and zero is absent. Nevertheless its upper asymptotic density is exactly $2/3$. The even positive integers have upper asymptotic density $1/2$ and the prime numbers have upper asymptotic density zero.

The useful feature is that a large proportion along arbitrarily large initial integer intervals already supplies structure. The Szemerédi theorem says that positive upper asymptotic density forces arithmetic progressions of every finite length, with positive common difference, even without an existing natural density. Szemerédi's 1975 paper resolved this general finite-length problem; it makes density a bridge from counting to additive structure. ^[ref]