Covering Finitely Many Points
A finite set can be spread across a large part of the real line, but its Lebesgue outer measure is still zero. The reason is that outer measure is defined by an infimum over interval covers: for any prescribed positive length budget, we can cover each point with a sufficiently short interval and make the sum of the lengths fit within that budget. The locations of the points do not affect this construction.
We work with Lebesgue outer measure \(m^*\) on subsets of \(\mathbb{R}\), not with a measure restricted to a sigma-algebra. For \(E\subseteq\mathbb{R}\), \(m^*(E)\) is the infimum of the sums of the lengths of countable interval covers of \(E\), as established in the construction of Lebesgue outer measure. In the covers below, we use open intervals; the length of \((a,b)\), where \(a<b\), is \(b-a\). The outer measure is nonnegative by its definition. We use the earlier result that \(m^*\) is monotone: if \(A\subseteq B\), then \(m^*(A)\le m^*(B)\).
Proof. Since \(F\) has exactly \(n\) distinct points, write \(F=\{x_1,\ldots,x_n\}\), with no repetitions. Fix any \(\delta>0\), and for each \(i=1,\ldots,n\) define the open interval \[ I_i=\left(x_i-\frac{\delta}{2n},\,x_i+\frac{\delta}{2n}\right). \] The point \(x_i\) is the midpoint of \(I_i\), so \(x_i\in I_i\). Consequently, \(F\subseteq\bigcup_{i=1}^n I_i\). The length of each interval is \[ \left(x_i+\frac{\delta}{2n}\right)-\left(x_i-\frac{\delta}{2n}\right)=\frac{\delta}{n}. \] There are \(n\) intervals, so the total length is \(\sum_{i=1}^n\delta/n=\delta\). Thus the displayed intervals give the required cover. \(\square\)
Zero Outer Measure for Finite Sets
Proof. First suppose \(F\) contains \(n\ge1\) distinct points. The proposition shows that, for every \(\delta>0\), there is an interval cover of \(F\) with total length \(\delta\). By the definition of \(m^*(F)\) as the infimum over the total lengths of such covers, \(m^*(F)\le\delta\) for every \(\delta>0\). Also \(m^*(F)\ge0\). These inequalities force \(m^*(F)=0\): if \(m^*(F)>0\), choosing \(\delta=m^*(F)/2\) would give \(m^*(F)\le m^*(F)/2\), a contradiction.
If \(F=\varnothing\), the empty family of intervals covers \(F\) and has total length zero. Since outer measure is nonnegative, \(m^*(\varnothing)=0\). This proves the result in all cases. \(\square\)
The crucial quantifier is “for every \(\delta>0\).” A single small cover proves only an upper bound by that cover’s total length; it does not by itself prove that the infimum is zero. The construction works because the length budget can be chosen as small as desired.
Worked Example: Three Points Far Apart
Let \(F=\{-2,0,3\}\). Construct a cover with total length \(1/10\), then determine \(m^*(F)\).
There are \(n=3\) distinct points. Set \(\delta=1/10\). The proposition assigns each point an interval of length \[ \frac{\delta}{n}=\frac{1/10}{3}=\frac{1}{30}. \] An interval of length \(1/30\) centered at \(x\) has radius \(1/60\). Thus the cover is \[ \left(-2-\frac{1}{60},-2+\frac{1}{60}\right),\qquad \left(-\frac{1}{60},\frac{1}{60}\right),\qquad \left(3-\frac{1}{60},3+\frac{1}{60}\right). \] Each point is the midpoint of its interval, so all three points are covered. The sum of the lengths is \[ \frac{1}{30}+\frac{1}{30}+\frac{1}{30}=\frac{3}{30}=\frac{1}{10}. \] This cover proves \(m^*(F)\le1/10\), but the theorem requires covers with arbitrarily small total length, not just this one. For any \(\eta>0\), use intervals centered at \(-2,0,3\) of length \(\eta/3\) each. Their total length is \(\eta\), so \(m^*(F)\le\eta\) for every \(\eta>0\). Nonnegativity then gives \(m^*(F)=0\). Check: for the specified cover, the three lengths add to \(1/10\), as required; the conclusion of zero outer measure follows from the separate arbitrary-\(\eta\) argument.
Worked Example: Five Points in a Bounded Interval
Let \(G=\{0,\tfrac14,\tfrac12,\tfrac34,1\}\). Give an explicit cover of total length \(1/20\), then verify that the outer measure is zero.
The set has five distinct points. For total length \(\delta=1/20\), choose five intervals, each of length \[ \frac{\delta}{5}=\frac{1/20}{5}=\frac{1}{100}. \] Their radius is \(1/200\). In order, the intervals are \[ \left(-\frac{1}{200},\frac{1}{200}\right),\quad \left(\frac14-\frac{1}{200},\frac14+\frac{1}{200}\right),\quad \left(\frac12-\frac{1}{200},\frac12+\frac{1}{200}\right),\quad \left(\frac34-\frac{1}{200},\frac34+\frac{1}{200}\right),\quad \left(1-\frac{1}{200},1+\frac{1}{200}\right). \] Each listed point is the midpoint of its interval. The total length is \[ 5\cdot\frac{1}{100}=\frac{5}{100}=\frac{1}{20}. \] For any \(\eta>0\), instead give each of the five points an interval of length \(\eta/5\). This produces a cover of total length \(5(\eta/5)=\eta\), so \(m^*(G)\le\eta\) for every \(\eta>0\). Since \(m^*(G)\ge0\), it follows that \(m^*(G)=0\). Check: the specified intervals have total length exactly \(1/20\), and the argument for zero uses a cover for every positive \(\eta\).
Worked Example: Repeated Entries Do Not Add Points
Consider the finite set \(H=\{2,2,-1,\tfrac73,-1\}\), written here with repetitions. Give a cover of total length \(3/8\), and explain how the repetitions affect the calculation.
A set records distinct elements, so \(H=\{-1,2,\tfrac73\}\) and has \(n=3\) points, not five. Take \(\delta=3/8\). The length assigned to each point is \[ \frac{\delta}{n}=\frac{3/8}{3}=\frac18, \] so each interval has radius \(1/16\). The three intervals are \[ \left(-1-\frac{1}{16},-1+\frac{1}{16}\right),\quad \left(2-\frac{1}{16},2+\frac{1}{16}\right),\quad \left(\frac73-\frac{1}{16},\frac73+\frac{1}{16}\right). \] Each of the three distinct points lies at the midpoint of one interval. The total length is \[ \frac18+\frac18+\frac18=\frac38. \] For an arbitrary \(\eta>0\), assigning length \(\eta/3\) to each of the three distinct points gives total length \(3(\eta/3)=\eta\). Thus \(m^*(H)\le\eta\) for every \(\eta>0\), and nonnegativity yields \(m^*(H)=0\). Check: counting the five entries would treat duplicate points as though they were new elements; the actual cover uses three intervals, one for each distinct point.
A Subset Consequence
Proof. A subset of a finite set is finite. By the theorem, \(m^*(F)=0\). Monotonicity of outer measure, applied to \(A\subseteq F\), gives \(m^*(A)\le m^*(F)=0\). Since \(m^*(A)\ge0\), we obtain \(m^*(A)=0\). \(\square\)
This consequence also illustrates the scope of the argument: it applies to subsets of a finite set because they remain finite. The proof does not claim that an arbitrary infinite set has zero outer measure. A cover must account for every point, and a construction for finitely many points does not, on its own, provide a finite length budget for infinitely many intervals.
Common Pitfall: Spending the Whole Budget at Every Point
Suppose \(F\) has four distinct points and the desired total length is less than \(1/10\). A wrong calculation is to assign an interval of length \(1/20\) to each point and then claim the total is \(1/20\), because that is the length “chosen for the cover.” In fact, the total is \[ 4\cdot\frac{1}{20}=\frac{4}{20}=\frac15, \] which is greater than \(1/10\). The correction is to divide the total budget among the four points. For example, intervals of length \(1/40\) each have total length \[ 4\cdot\frac{1}{40}=\frac{4}{40}=\frac{1}{10}. \] For a strict bound below \(1/10\), choose length \(1/80\) at each point; the total is \(4/80=1/20<1/10\). More generally, to achieve total length \(\delta>0\), use length \(\delta/n\) at each of \(n\) distinct points.
Another tempting shortcut is to cover a point by the degenerate interval \([x,x]\), call its length zero, and conclude that no positive-length intervals are needed. This relies on a degenerate interval rather than the open intervals used in the covering construction. The robust argument is to use genuine open intervals around the points: each has positive length, while their total can be made smaller than any prescribed positive number.
Check Your Understanding
Use the interval-cover definition and keep track of the total length, rather than only the length of one interval.
- For a set of seven distinct real numbers and a prescribed \(\delta>0\), what length should each of seven centered intervals have so that their total length is \(\delta\)?
- Construct intervals covering \(\{-3,1\}\) with total length \(1/4\). State each interval’s radius and verify the sum of their lengths.
- Why does a cover of a finite set with total length \(1/10\) not, by itself, prove that its outer measure is zero?
- How many intervals are needed in the construction for \(\{4,4,0,0\}\), and what length should each have for a total length budget \(\delta\)?
- Use monotonicity and the finite-set theorem to determine the outer measure of \(\{-1,0\}\)'s subset \(\{0\}\).