Properties of Lebesgue Measure

How Lebesgue Measure Changes Under an Affine Map

Lebesgue measure has several useful structural properties beyond the approximations developed in the previous tutorial. A translation moves a set without changing its measure, as established by the Translation Invariance of Lebesgue Measure Theorem. Scaling behaves differently: multiplying every point by a nonzero constant multiplies the measure by the absolute value of that constant. We will first establish this fact for outer measure, then use it to study affine images of measurable sets.

Recall that \(m^*(S)\) denotes the Lebesgue outer measure of an arbitrary subset \(S\subseteq\mathbb{R}\), while \(\lambda(E)=m^*(E)\) when \(E\) is Lebesgue measurable. The outer-measure statement is useful because it does not require the set being transformed to be measurable.

Proof. Suppose first that \(m^*(S)<\infty\), and let \(\varepsilon>0\). By the definition of Lebesgue outer measure using countable open-interval covers, there are open intervals \(I_1,I_2,\ldots\) covering \(S\) such that

Since \(c\neq 0\), each \(cI_j=\{cx:x\in I_j\}\) is an open interval of length \(|c||I_j|\), and these intervals cover \(cS\). Therefore

Letting \(\varepsilon\) decrease to zero gives \(m^*(cS)\leq |c|m^*(S)\). If \(m^*(S)=\infty\), this inequality holds in the extended-real sense automatically. Apply the same inequality to \(cS\) with the nonzero constant \(1/c\). Since \((1/c)(cS)=S\), we obtain

which gives the reverse inequality \(m^*(cS)\geq |c|m^*(S)\). These inequalities also show that if either outer measure is infinite, so is the other. Together they prove the claimed equality. \(\square\)

The argument depends on how interval lengths change under scaling: the sign of \(c\) affects the direction of an interval but not its length. A translation likewise preserves interval lengths, which is why translation invariance and scaling fit naturally together.

Proof. First suppose \(c\neq 0\). By the Borel Representative Modulo a Null Set Theorem, there is a Borel set \(B\supseteq E\) such that \(\lambda(B\setminus E)=0\). The map \(T\) is a homeomorphism of \(\mathbb{R}\), so \(T(B)\) is Borel. By the Scaling of Lebesgue Outer Measure Theorem and translation invariance, \(T(B\setminus E)\) has outer measure zero. Every set of outer measure zero is Lebesgue measurable, as established in the construction of Lebesgue measure. Since \(T\) is one-to-one,

Thus \(T(E)\) is obtained by removing a measurable set from the Borel set \(T(B)\), so \(T(E)\) is Lebesgue measurable. Now apply the outer-measure scaling result and translation invariance:

If \(c=0\), then \(T(E)=\{t\}\) when \(E\) is nonempty, and \(T(E)=\varnothing\) when \(E\) is empty. Both sets are Borel and have measure zero. This proves the theorem in every case. \(\square\)

Worked Applications of Affine Scaling

Open Sets as Disjoint Unions of Intervals

Another useful property of Lebesgue measure is that the measure of an open set can be read from its interval components. The essential point is to decompose the open set into countably many disjoint open intervals. Countable additivity then turns this geometric description into a formula for its measure.

Proof. Suppose \(O\) is nonempty. For \(x,y\in O\), declare \(x\sim y\) if every point between \(x\) and \(y\), including the endpoints, belongs to \(O\). This relation is reflexive because \(x\in O\), symmetric by its definition, and transitive because the union of the intervals between \(x\) and \(y\), and between \(y\) and \(z\), contains the interval between \(x\) and \(z\). Thus it partitions \(O\) into disjoint equivalence classes.

Each class \(I\) is an interval: if \(x,y\in I\), every point between them belongs to \(O\) and is equivalent to \(x\). It is also open. To see this, fix \(x\in I\). Since \(O\) is open, there is \(\delta>0\) such that \((x-\delta,x+\delta)\subseteq O\). Every \(y\) in that neighborhood is equivalent to \(x\), so the neighborhood lies in \(I\). Therefore each class is an open interval, possibly unbounded. Distinct classes are disjoint, and their union is \(O\).

Every nonempty open interval contains a rational number. Choose one rational from each class. Because the classes are disjoint, the chosen rationals are distinct. Since the rationals are countable, there are at most countably many classes. Finally, each interval \(I_j\) is measurable and has measure equal to its length by the established theorem on the Lebesgue measure of bounded intervals, together with the corresponding unbounded-interval case. Countable additivity on the disjoint intervals gives

If the sum is infinite, the equality is understood in the extended real numbers. If \(O=\varnothing\), the decomposition has no intervals and both sides are zero. \(\square\)

Why These Properties Matter

The affine-image theorem is the basic measure calculation behind changes of scale in one dimension. A stretch by a factor of \(3\) triples lengths and measures; a reflection combined with a stretch by a factor of \(3\) does the same, because the factor is \(|c|\), not \(c\). A translation contributes no additional factor. These facts make it possible to transform measurable sets while keeping track of their size.

The open-set decomposition provides a complementary perspective. Open sets may have complicated descriptions involving many intervals, but their disjoint components reduce their measure to a sum of interval lengths. In particular, if the component lengths have a finite sum, the open set has finite measure; if their sum diverges, its measure is infinite. The disjointness requirement matters: overlapping intervals cannot simply have their lengths added without correcting for overlap.

These results concern Lebesgue measure, not ordinary length as a pointwise property of every set. The affine-image theorem applies to Lebesgue measurable sets, while the interval decomposition applies to open sets. One should not infer from the open-set formula alone that an arbitrary measurable set is a union of intervals. The approximation theorems from the previous tutorial are what allow general measurable sets to be related to open sets without claiming they have the same structure.