Open Covers with Arbitrarily Small Excess

Why Enlarge a Cover?

In the definition of Lebesgue outer measure on \(\mathbb{R}\), a set is covered by countably many intervals and the total length of the intervals is made as small as possible. It is often useful to have the covering intervals open as well. An interval that is closed or half-open cannot usually be replaced by its interior: that can omit one or both endpoints. The useful replacement is instead a slightly larger open interval. The key is to distribute the allowed increase in length across the intervals so that the total increase remains small, even for a countable cover.

Throughout, \(E\subseteq\mathbb{R}\), and \(m^*\) denotes Lebesgue outer measure on \(\mathbb{R}\). No measurability assumption on \(E\) is made. An interval cover of \(E\) is a finite or countable family of intervals \((I_n)\) such that \(E\subseteq\bigcup_n I_n\). Write \(\ell(I)\) for the length of an interval \(I\): it is the difference of its supremum and infimum when these are finite, \(+\infty\) for an unbounded interval, and \(0\) for the empty interval. In particular, a singleton has length \(0\). The total length of a countable cover is the nonnegative series \(\sum_n\ell(I_n)\).

Enlarging One Interval

Proof. Since \(I\) is nonempty and has finite length, its endpoints \(a=\inf I\) and \(b=\sup I\) are finite, and \(\ell(I)=b-a\). Define \(J=(a-\delta/2,b+\delta/2)\). For every \(x\in I\), the definitions of infimum and supremum give \(a\le x\le b\), so \(a-\delta/2<x<b+\delta/2\), and therefore \(x\in J\). The interval \(J\) is open, and its length is \((b+\delta/2)-(a-\delta/2)=b-a+\delta=\ell(I)+\delta\). This also covers the degenerate case \(a=b\), when \(I\) is a singleton and \(J\) has length \(\delta\). \(\square\)

The lemma gives an exact increase for one interval. For a countable cover, assigning the same positive increase to every interval would generally produce an infinite total increase. Instead, the individual increases must have a finite sum smaller than the error we are allowed.

The Countable Open-Cover Construction

Proof. For each index \(n\ge1\), assign the positive amount \(\delta_n=\varepsilon/2^{n+1}\). The geometric series has sum \(\sum_{n=1}^{\infty}\delta_n=\varepsilon/2\). Since \(S<\infty\), no \(I_n\) can be unbounded: an unbounded interval has infinite length and would make \(S=\infty\). For every nonempty \(I_n\), apply the lemma with \(\delta=\delta_n\), obtaining an open interval \(J_n\supseteq I_n\) with \(\ell(J_n)=\ell(I_n)+\delta_n\). If \(I_n\) is empty, omit it from the cover; it contributes neither points nor length. The resulting open intervals cover \(E\), because every point covered by an original nonempty interval remains in its enlargement. For the total length, nonnegative series add term by term, and hence

This proves the assertion. If the original cover is finite, the same construction applies to its indices; the sum of the assigned increases is then no larger than \(\varepsilon/2\). \(\square\)

Why Open Covers Compute the Same Outer Measure

Let \(m^*_{\mathrm{open}}(E)\) denote the infimum of the total lengths of open-interval covers of \(E\), with value \(+\infty\) if there is no such cover of finite total length. Because every open interval is an interval, the definition of \(m^*\) immediately gives \(m^*(E)\le m^*_{\mathrm{open}}(E)\). For the reverse inequality, if \(m^*(E)=+\infty\), then there cannot be an open cover of finite total length, since it would also be an interval cover; so both infima are infinite. If \(m^*(E)<\infty\), take any \(\eta>0\). The defining property of an infimum gives an interval cover with total length less than \(m^*(E)+\eta/2\). Apply the theorem with error \(\eta/2\); the resulting open cover has total length less than \(m^*(E)+\eta\). Since this holds for every \(\eta>0\), \(m^*_{\mathrm{open}}(E)\le m^*(E)\). Therefore the two infima agree. This does not assert that either infimum is attained by a cover.

The finite-total-length hypothesis is important to the conversion theorem. A cover containing an unbounded interval has infinite total length; assigning a finite enlargement to that interval cannot make its length finite. In contrast, a cover with finite total length consists only of bounded nonempty intervals and empty intervals, so the endpoint construction applies to every interval that needs to cover a point.

Common Errors and Their Corrections

Giving every interval the full error. For the countable family \(I_n=[0,3^{-n}]\), suppose one enlarges every interval by \(1/5\). Then \(\ell(J_n)=3^{-n}+1/5\), and the sum of these lengths is infinite because the partial sums include \(N\) copies of \(1/5\), giving a contribution \(N/5\) for every \(N\). This does not produce a finite-total-length cover. The correction is to choose increases \(\delta_n\) whose series is bounded by the total error, as in \(\delta_n=\varepsilon/2^{n+1}\).

Replacing a closed interval by its interior. The interval \((0,1)\) does not cover \([0,1]\): it omits both \(0\) and \(1\). The correction is to expand beyond the endpoints, for example replacing \([0,1]\) with \((-\delta/2,1+\delta/2)\), whose length is \(1+\delta\) and which contains the entire closed interval.

In both corrections, the goal is not to find an open cover with exactly the original total length. It is to retain every covered point while controlling the total increase. The countable allocation works because the sum of all assigned increases is itself controlled, not because any individual interval is necessarily enlarged by a negligible amount.