Common Errors in Countable Cover Arguments

Three Checks for Countable Covers

A countable interval cover is a family of intervals whose union contains the entire target set. In arguments involving such covers, a small numerical error is not harmless unless the errors have a finite total, and a family that covers a dense subset need not cover the target set. A third issue is logical: an infimum of possible cover lengths need not be the length of any actual cover. These are separate checks, and passing one does not automatically pass the others.

We work on \(\mathbb{R}\). For any set \(E\subseteq\mathbb{R}\), its Lebesgue outer measure \(m^*(E)\) is the infimum of the sums of lengths of countable interval covers of \(E\). Outer measure is defined for every subset; \(E\) need not be measurable. We use the interval-cover convention from the earlier material, including degenerate intervals of length zero. A cover \((I_j)_{j\ge1}\) of \(E\) must satisfy \(E\subseteq\bigcup_{j\ge1} I_j\), and its total length is \(\sum_{j\ge1}|I_j|\), possibly \(+\infty\). By contrast, Lebesgue measure \(m\) is defined on the Lebesgue sigma-algebra \(\mathcal{L}\). We will use the established fact \(m^*([0,1])=1\), but no measurability assumption is needed for the cover checks below.

Infima Give Approximations, Not Necessarily Minimizers

The safe use of an infimum is to ask for a cover within a prescribed positive error. This is exactly what the definition guarantees when the infimum is finite.

Proof. Write \(\alpha=m^*(E)=\inf S_E\). Suppose, to the contrary, that no cover length is less than \(\alpha+\varepsilon\). Then every \(s\in S_E\) satisfies \(s\ge\alpha+\varepsilon\), so \(\alpha+\varepsilon\) is a lower bound for \(S_E\). This contradicts the fact that \(\alpha\) is the greatest lower bound, since \(\alpha+\varepsilon>\alpha\). Thus some \(s\in S_E\) satisfies \(s<\alpha+\varepsilon\), and the cover giving that length is the required cover. \(\square\)

The strict inequality matters: the proposition supplies a cover with length less than the requested threshold, not necessarily one with length equal to the infimum. If \(m^*(E)=\infty\), this finite approximation statement does not apply. Also, the proposition does not assert that the chosen cover is optimal. Earlier material on covering infima gives examples where the infimum is not attained.

Errors Must Have a Finite Total

Suppose a construction makes one choice for each positive integer \(n\), with an error allowance \(\varepsilon_n>0\). To conclude that the total error is at most \(\varepsilon\), it is not enough that every individual allowance is small or that \(\varepsilon_n\to0\). The required condition is \(\sum_{n=1}^{\infty}\varepsilon_n\le\varepsilon\).

Proof. For the geometric allocation, the partial sum is \(\sum_{n=1}^{N}\varepsilon 2^{-n}=\varepsilon(1-2^{-N})\). Since \(2^{-N}\to0\), the infinite sum is \(\varepsilon\). For the general assertion, add the cost bounds through index \(N\): \(\sum_{n=1}^{N}(c_n+\varepsilon_n)\le\sum_{n=1}^{N}c_n+\sum_{n=1}^{N}\varepsilon_n\le\sum_{n=1}^{\infty}c_n+\varepsilon\). The partial sums of the nonnegative costs increase to their infinite sum, so the same upper bound holds for the total cost. \(\square\)

In cover arguments, the choices may be intervals or whole interval covers. When each set \(E_n\) has finite outer measure, the approximation proposition lets us choose a cover of \(E_n\) with cost less than \(m^*(E_n)+\varepsilon_n\). If the errors are summable, the combined costs are controlled. The point of the lemma is the bookkeeping: each error must be chosen before summing, and the sum of all allowances—not just each allowance separately—must fit the budget.

A Cover Must Include Every Point

Length estimates only concern a family that actually covers the set in question. Covering many points, a dense subset, or all but a few endpoints is not enough. The definition requires the set inclusion \(E\subseteq\bigcup_n I_n\), so the check is pointwise: take an arbitrary \(x\in E\) and identify an interval that contains it.

Common Mistakes and a Practical Audit

A common wrong calculation is to set \(\varepsilon_n=\varepsilon/n\) and then write \(\sum_n\varepsilon_n=\varepsilon\). This incorrectly treats the divergent harmonic series as though it summed to \(1\). For example, with \(\varepsilon=1\), the first four allowances total \(1+1/2+1/3+1/4=25/12>1\), and the full series diverges. The correction is to choose a genuinely summable sequence, such as \(\varepsilon_n=\varepsilon 2^{-n}\), and verify its sum before using it.

Two other invalid shortcuts are “choose a cover whose length equals the infimum” and “these intervals cover the points I have been considering, so they cover \(E\).” The first needs an attainment result that the definition of infimum does not provide; the second needs an inclusion proof for every point of \(E\). Before accepting a countable-cover argument, check the following:

• Target: Is the set being covered stated precisely?
• Coverage: For each \(x\) in that set, which interval contains \(x\)? Are endpoints included?
• Infimum: Is the claim only an approximation within \(\varepsilon\), or is attainment separately justified?
• Budget: Are the individual errors positive and does their series have the claimed finite sum?
• Total: Are the interval lengths summed with the correct indexing and no omitted intervals?

These checks separate the geometric claim from the numerical estimate. A valid proof needs both: a family that covers the whole target and a correct bound on the total length of that family.