Covering Sparse Sequences with Summable Lengths

From a Sparse Sequence to a Length Budget

A sequence can have points spread across the line, or accumulate rapidly near a limit, while still admitting a simple cover: place a short interval around each term and make the total of the interval lengths finite. The lengths then give an upper bound for the outer measure of the set of terms. The choice of intervals matters: a useful construction must cover every point, and its length calculation must account for the full width of each interval, not just its radius.

Throughout, \(m^*\) denotes Lebesgue outer measure on \(\mathbb{R}\). We use its interval-cover definition: for \(E\subseteq\mathbb{R}\), \(m^*(E)\) is the infimum of \(\sum_j(b_j-a_j)\) over finite or countable covers of \(E\) by open intervals \((a_j,b_j)\). The empty set has outer measure zero. No measurability assumption on \(E\) is needed for this definition or for the bounds below.

Covering principle: If \(E\) is contained in a finite or countable collection of open intervals \(I_j=(a_j,b_j)\), then \(m^*(E)\leq\sum_j(b_j-a_j)\). In particular, an interval with center \(x\) and radius \(r>0\), namely \((x-r,x+r)\), has length \(2r\).

The first statement follows directly from the definition: the displayed intervals are one admissible cover, and an infimum is no larger than the value of any admissible cover. The factor of two in the radius statement comes from subtracting the endpoints: \((x+r)-(x-r)=2r\). We now use this simple principle to turn convergent length series into explicit bounds.

A General Summable-Cover Estimate

Theorem: Let \((x_n)_{n\geq1}\) be any sequence of real numbers, and let \(r_n>0\) be radii such that \(\sum_{n=1}^{\infty}2r_n<\infty\). For \(S=\{x_n:n\geq1\}\), define \(I_n=(x_n-r_n,x_n+r_n)\). Then \(S\subseteq\bigcup_{n=1}^{\infty}I_n\), and \(m^*(S)\leq\sum_{n=1}^{\infty}2r_n\). For each \(N\geq1\), the tail set \(S_N=\{x_n:n\geq N\}\) also satisfies \(m^*(S_N)\leq\sum_{n=N}^{\infty}2r_n\).

Proof. For every \(n\), \(r_n>0\), so \(x_n-r_n<x_n<x_n+r_n\); hence \(x_n\in I_n\). Every element of \(S\) equals some \(x_n\), so \(S\subseteq\bigcup_{n=1}^{\infty}I_n\). Each interval has length \((x_n+r_n)-(x_n-r_n)=2r_n\). The intervals therefore form an admissible cover with total length \(\sum_{n=1}^{\infty}2r_n\), and the definition of \(m^*\) gives the asserted bound. Applying the same argument to the intervals with indices \(n\geq N\), which cover \(S_N\), gives the tail bound. If some sequence terms repeat, the same cover remains valid: repeating an interval can only leave the stated upper bound unchanged or make it less efficient. \(\square\)

The tail estimate can be useful even when the bound for the whole sequence is not sharp. Since the length series converges, its tails tend to zero. Thus this particular construction gives covers of later terms with arbitrarily small total length. The conclusion is an upper bound from a chosen cover; it does not say that the infimum defining outer measure is attained by that cover.

Worked Covers for Two Sparse Sequences

Worked Example: Powers of Two with Geometrically Decreasing Radii

Let \(S=\{2^{-n}:n\geq1\}\). Around \(x_n=2^{-n}\), choose radius \(r_n=2^{-(n+3)}\) and define \(I_n=(2^{-n}-2^{-(n+3)},2^{-n}+2^{-(n+3)})\). Since the radius is positive, the center \(2^{-n}\) lies strictly inside \(I_n\), so these intervals cover \(S\).

The length of the \(n\)-th interval is twice its radius: \(2r_n=2\cdot2^{-(n+3)}=2^{-(n+2)}\). The total length is a geometric series. Its first term, at \(n=1\), is \(2^{-3}=1/8\), and the common ratio is \(1/2\), so

\sum_{n=1}^{\infty}|I_n| =\sum_{n=1}^{\infty}2^{-(n+2)} =\frac{1/8}{1-1/2} =\frac14.

Therefore \(m^*(S)\leq1/4\). More generally, the cover using indices \(n\geq N\) has total length \(\sum_{n=N}^{\infty}2^{-(n+2)}=2^{-(N+1)}\), so \(m^*(\{2^{-n}:n\geq N\})\leq2^{-(N+1)}\).

Check: For \(n=1\), the center is \(2^{-1}=1/2\) and the radius is \(2^{-(1+3)}=1/16\). Thus the first interval is \((1/2-1/16,1/2+1/16)=(7/16,9/16)\). Its length is \(9/16-7/16=2/16=1/8\), exactly the \(n=1\) term of the length series, and it contains \(1/2\).

Worked Example: Reciprocal Squares with a Telescoping Bound

Let \(T=\{1/n^2:n\geq1\}\). Choose radius \(r_n=1/(4n^2)\) about \(x_n=1/n^2\). The resulting interval is \(J_n=(3/(4n^2),5/(4n^2))\): subtracting and adding \(1/(4n^2)\) to \(1/n^2\) gives these endpoints. The center is strictly between them, and the length is \(5/(4n^2)-3/(4n^2)=1/(2n^2)\).

To bound the total length, first consider the numerical series. Its first term is \(1\). For every \(n\geq2\), \(1/n^2\leq1/[n(n-1)]\), since \(n(n-1)\leq n^2\). Also, \(1/[n(n-1)]=1/(n-1)-1/n\). For every \(M\geq2\), it follows that

\sum_{n=1}^{M}\frac1{n^2} \leq 1+\sum_{n=2}^{M}\left(\frac1{n-1}-\frac1n\right) =1+\left(1-\frac1M\right) \leq2.

The partial sums are increasing and bounded above by \(2\), so the nonnegative series converges and its sum is at most \(2\). Consequently, the total length of the intervals \(J_n\) is at most \(1\), and \(m^*(T)\leq1\).

There is also a useful bound for the tail. If \(N\geq2\) and \(M\geq N\), the same telescoping comparison gives \(\sum_{n=N}^{M}1/n^2\leq1/(N-1)-1/M\leq1/(N-1)\). Passing to the limit in these partial sums yields \(\sum_{n=N}^{\infty}1/n^2\leq1/(N-1)\). Hence the intervals with indices \(n\geq N\) cover the corresponding tail of \(T\) with total length at most \(1/[2(N-1)]\).

Check: At \(n=1\), the interval is \((3/4,5/4)\); its length is \(5/4-3/4=1/2\), which agrees with \(1/(2\cdot1^2)=1/2\). The general estimate gives \(m^*(T)\leq1\), not an equality. A cover’s total length is an upper bound, and this construction is not intended to identify the infimum.

Allocating a Prescribed Length Budget

The same idea works for sequences that do not have a special pattern. If a desired total length is \(\eta>0\), assign a positive length to each point so that all assignments sum to at most \(\eta\). For example, the lengths \(\eta 2^{-n}\), \(n\geq1\), have total \(\eta\): the first term is \(\eta/2\), and the geometric ratio is \(1/2\), so their sum is \((\eta/2)/(1-1/2)=\eta\). An interval with a prescribed length \(\ell_n>0\) and center \(x_n\) is obtained by taking radius \(\ell_n/2\).

Length-budget corollary: For any sequence \((x_n)_{n\geq1}\) in \(\mathbb{R}\) and any \(\eta>0\), there are open intervals \(K_n\) containing \(x_n\) such that \(\sum_{n=1}^{\infty}|K_n|=\eta\). In particular, \(m^*(\{x_n:n\geq1\})\leq\eta\) for every \(\eta>0\), and therefore \(m^*(\{x_n:n\geq1\})=0\).

Proof. Set \(K_n=(x_n-\eta 2^{-(n+1)},x_n+\eta 2^{-(n+1)})\). Its radius is \(\eta2^{-(n+1)}>0\), so it contains \(x_n\), and its length is \(2\eta2^{-(n+1)}=\eta2^{-n}\). The sum of these lengths is \(\eta\sum_{n=1}^{\infty}2^{-n}=\eta\). By the covering principle, \(m^*(\{x_n:n\geq1\})\leq\eta\). This holds for every \(\eta>0\). Since outer measure is nonnegative, it must be zero: if it were \(c>0\), choosing \(\eta=c/2\) would contradict \(c\leq\eta\). \(\square\)

Worked Example: An Unbounded Sequence with a Fixed Budget

Consider the set \(U=\{1,2,3,\ldots\}\) and choose the budget \(\eta=1/2\). The construction gives \(K_n=(n-2^{-(n+2)},n+2^{-(n+2)})\), because the radius is \(\eta2^{-(n+1)}=(1/2)2^{-(n+1)}=2^{-(n+2)}\). Each \(K_n\) contains the integer \(n\), and its length is \(2\cdot2^{-(n+2)}=2^{-(n+1)}\). Thus

\sum_{n=1}^{\infty}|K_n| =\sum_{n=1}^{\infty}2^{-(n+1)} =\frac{1/4}{1-1/2} =\frac12, \qquad m^*(U)\leq\frac12.

Check: For \(n=1\), the radius is \(2^{-(1+2)}=1/8\), so \(K_1=(1-1/8,1+1/8)=(7/8,9/8)\). Its length is \(9/8-7/8=2/8=1/4\), agreeing with \(2^{-(1+1)}=1/4\). The intervals cover an unbounded set because every positive integer is the center of one of them; no single bounded interval is being used to cover all of \(U\).

The value \(1/2\) is a bound supplied by this particular budget, not a claim that the outer measure equals \(1/2\). Choosing any positive budget \(\eta\) instead produces a cover with total length \(\eta\), which is the relevant feature of the construction.

A Radius Error to Avoid

A common mistake is to add the radii and call that sum the total interval length. In the powers-of-two construction, the radius at index \(n\) is \(2^{-(n+3)}\). Incorrectly using that radius as the interval length would produce the claimed total \(\sum_{n=1}^{\infty}2^{-(n+3)}=(1/16)/(1-1/2)=1/8\). But an interval with this radius has length \(2\cdot2^{-(n+3)}=2^{-(n+2)}\), so the actual total is \(1/4\), not \(1/8\). At the first index, this difference is visible directly: the radius is \(1/16\), while the interval \((7/16,9/16)\) has length \(1/8\).

There is a second point to keep in view: an upper bound from an explicit cover need not be optimal. The covering principle says that the outer measure is no greater than the total length used. It does not say that the intervals are disjoint, that their union has that same measure, or that no shorter cover exists. The method is valuable because it gives a concrete, checkable length budget and, when the lengths have summable tails, increasingly small covers for the later terms.

Check Your Understanding

For each question, verify both that the proposed intervals contain the intended points and that their full lengths have the claimed sum.

For \(x_n=3^{-n}\) and radius \(r_n=3^{-(n+2)}\), compute the interval length series and its sum. What bound does it give for the outer measure of \(\{3^{-n}:n\geq1\}\)?
Suppose intervals centered at \(x_n\) have radii \(r_n=1/(5n^2)\). Give a convergent-series upper bound for the outer measure of \(\{x_n:n\geq1\}\), using the telescoping comparison in the reciprocal-squares example.
For the powers-of-two cover in the first worked example, calculate the total length of the intervals with indices \(n\geq4\).
Given a sequence \((x_n)\) and a budget \(\eta>0\), what radius should be assigned at index \(n\) if the desired interval length is \(\eta2^{-n}\)?
Explain why the bound \(m^*(E)\leq L\) from one explicit cover does not imply \(m^*(E)=L\).