From Outer Approximation to Inner Approximation
Outer regularity approximates a measurable set from the outside by open sets. To approximate from within, the useful move is to apply outer regularity not to the set itself, but to its complement inside a bounded closed interval. Taking a relative complement then produces a closed subset of the original set, and the measure of the part left out is controlled by the outer-approximation error.
Throughout, \(m\) is Lebesgue measure on \((\mathbb{R},\mathcal{L})\), where \(\mathcal{L}\) is the Lebesgue sigma-algebra. All sets considered below are Lebesgue measurable unless stated otherwise. We use outer regularity in the following standard form, established earlier: if \(A\in\mathcal{L}\) has finite measure and \(\varepsilon>0\), there is an open set \(G\supseteq A\) such that \(m(G\setminus A)<\varepsilon\). The difference formula follows from measurability and \(m(A)<\infty\).
Proof. Set \(A=I\setminus E\). Because \(I\) and \(E\) are Lebesgue measurable, \(A\) is Lebesgue measurable. Also \(A\subseteq I\), so \(m(A)\le b-a<\infty\). If \(E=\varnothing\), take \(F=\varnothing\), and the loss is zero. Otherwise, apply outer regularity to \(A\): for the given \(\varepsilon>0\), choose an open set \(G\supseteq A\) with \(m(G\setminus A)<\varepsilon\). Define \(F=I\setminus G\). Since \(I\) is closed and \(G\) is open, \(F=I\cap G^c\) is closed. If \(x\in F\), then \(x\in I\) and \(x\notin G\); as \(A\subseteq G\), this implies \(x\notin A=I\setminus E\), so \(x\in E\). Thus \(F\subseteq E\). Finally, if \(x\in E\setminus F\), then \(x\in E\subseteq I\) and \(x\in G\), while \(x\notin A\). Hence \(x\in G\setminus A\), proving \(E\setminus F\subseteq G\setminus A\). Monotonicity of measure gives \(m(E\setminus F)\le m(G\setminus A)<\varepsilon\). If \(a=b\), then \(I\) has measure zero and the same construction and inequalities remain valid. \(\square\)
The interval \(I\) is doing real work: it makes the complement \(A=I\setminus E\) have finite measure, and the set \(I\setminus G\) stays inside \(I\). No completeness assumption is needed beyond the fact that the sets in the argument are Lebesgue measurable.
Worked Example: Removing a Small Neighborhood of a Gap
Let \(I=[0,3]\) and \(E=[0,3]\setminus(1,2)\). Construct a closed \(F\subseteq E\) with \(m(E\setminus F)<1/10\).
The complement inside \(I\) is \(A=(1,2)\), with \(m(A)=1\). Choose \(\delta=1/40\) and let \(G=(1-\delta,2+\delta)\), an open superset of \(A\). Its excess over \(A\) consists of two intervals, each of length \(\delta\), so
Taking the complement relative to \(I\) gives \(F=I\setminus G=[0,39/40]\cup[81/40,3]\). These are closed intervals contained in \(E\). The part of \(E\) removed is \([39/40,1]\cup(2,81/40)\), up to endpoints, which do not affect Lebesgue measure. Its measure is
Check: \(m(E)=2\), while \(m(F)=39/40+(3-81/40)=39/40+39/40=39/20\). Thus \(m(E)-m(F)=2-39/20=1/20\), agreeing with the direct calculation.
How the Measure Error Is Recorded
Because \(F\subseteq E\) and both sets are measurable, the disjoint decomposition \(E=F\cup(E\setminus F)\) gives an exact relation between the measure lost and the measure of the approximating set. Finiteness matters here: it ensures that the subtraction below is between finite quantities.
Proof. The sets \(F\) and \(E\setminus F\) are disjoint and measurable, and their union is \(E\). Finite additivity therefore gives \(m(E)=m(F)+m(E\setminus F)\). The theorem makes the last term less than \(\varepsilon\), so \(m(F)>m(E)-\varepsilon\). Every measurable subset \(F\subseteq E\) has \(m(F)\le m(E)\) by monotonicity. Thus the supremum over closed subsets is at most \(m(E)\), and the approximations just constructed show it is at least \(m(E)\). \(\square\)
Worked Example: A Finite Union of Intervals
Let \(I=[0,5]\) and \(E=I\setminus((1,2)\cup(3,4))\). Find a closed subset \(F\subseteq E\) whose measure loss is less than \(1/5\), and verify the loss in two ways.
Here \(A=I\setminus E=(1,2)\cup(3,4)\). Set \(\delta=1/40\) and choose the open set
It contains \(A\). The added portions have four lengths \(\delta\), so \(m(G\setminus A)=4\delta=4/40=1/10<1/5\). The relative complement is
Each displayed interval lies in \(E=[0,1]\cup[2,3]\cup[4,5]\), so \(F\) is a closed subset of \(E\). Directly, the removed parts have total length \(4\delta=1/10\), hence \(m(E\setminus F)=1/10\). As a second check, \(m(E)=3\), and the three component lengths of \(F\) are \(39/40\), \(38/40\), and \(39/40\). Therefore \(m(F)=116/40=29/10\), and \(m(E)-m(F)=3-29/10=1/10\).
A Dense Null Complement
Inner approximation does not require the complement of \(E\) to contain an interval. For example, it can be dense in \(I\). The outer approximation is then built from a countable union of small open intervals, one around each point of the complement. Countable subadditivity controls the total measure of that union.
Worked Example: A Closed Set Mostly Inside the Irrationals
Let \(I=[0,1]\), \(E=[0,1]\setminus\mathbb{Q}\), and \(\varepsilon=1/2\). Construct a closed \(F\subseteq E\) and bound the measure lost.
The rationals in \([0,1]\) form a countable null set, so \(m(E)=1\). Enumerate them as \(q_1,q_2,\ldots\). For each \(n\ge1\), take the open interval \(J_n=(q_n-r_n,q_n+r_n)\), where \(r_n=1/2^{n+4}\), and set \(G=\bigcup_{n=1}^{\infty}J_n\). This open set contains every rational in \([0,1]\). By countable subadditivity,
Set \(F=[0,1]\setminus G\). This is closed, and it contains no rational in \([0,1]\), so \(F\subseteq E\). The loss \(E\setminus F=E\cap G\) has measure at most \(m(G)\le1/8<1/2\). In particular, \(m(F)=m(E)-m(E\setminus F)\ge7/8\). Check: the interval lengths sum to \(1/8\), so the bound is strictly below the requested error \(1/2\); the construction does not require the open intervals to be disjoint.
A Common Complement Mistake
A tempting but incorrect approach is to apply outer regularity directly to \(E\), obtain an open \(G\supseteq E\), and then call \(G^c\) an inner approximation of \(E\). In fact \(G^c\subseteq E^c\), so it points in the wrong direction. Even when one instead takes a complement relative to \(I\), the result \(I\setminus G\) is generally disjoint from \(E\), not a subset that approximates \(E\).
For a concrete check, take \(I=[0,1]\) and \(E=[0,1/2]\). An open superset such as \(G=(-1/10,3/5)\) contains \(E\). But \(I\setminus G=[3/5,1]\), which is not a subset of \(E\). The correction is to apply outer regularity to \(A=I\setminus E=(1/2,1]\), not to \(E\). For example, \(U=(2/5,11/10)\) is open, contains \(A\), and has \(m(U\setminus A)=1/10+1/10=1/5\). Now \(F=I\setminus U=[0,2/5]\) is closed and contained in \(E\); the loss is \(m(E\setminus F)=1/2-2/5=1/10\), which is bounded by the outer error \(1/5\).
The method is a basic way to transfer an outside approximation into an inside one: choose a bounded interval containing the set, approximate the relative complement from outside, and take the complement within that interval. The resulting set is closed, and the inclusion \(E\setminus F\subseteq G\setminus A\) is the precise reason the measure error is controlled.
Check Your Understanding
For each question, identify the set to which outer regularity should be applied before taking a relative complement.
- If \(E\subseteq[0,4]\) is Lebesgue measurable, what set should be approximated from outside to obtain a closed subset of \(E\)?
- For \(I=[0,2]\) and \(E=I\setminus(1/2,3/2)\), construct a closed inner approximation with error less than \(1/10\) by enlarging the removed interval.
- Why does \(E\setminus F\subseteq G\setminus(I\setminus E)\) hold for \(F=I\setminus G\)?
- Give a concrete example showing why the complement of an open superset of \(E\) need not be a subset of \(E\).
- Explain why the finite measure of \(I\setminus E\) is available when \(E\subseteq[a,b]\).