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Multidimensional Integration 2 | The n-dimensional Lebesgue Measure [dark version] thumbnail

Multidimensional Integration 2 | The n-dimensional Lebesgue Measure [dark version]

4 min read

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TL;DR

Lebesgue measure in R^n is constructed via product measures, starting from the one-dimensional Lebesgue measure Λ on measurable sets in R.

Briefing

Lebesgue measure in n-dimensional space is built by a single idea: start with measurable sets on the real line, then define the n-dimensional measure on R^n by taking product sets and extending them to a full sigma-algebra. The payoff is a rigorous notion of “length,” “area,” and “volume” in higher dimensions—along with a corresponding n-dimensional Lebesgue integral—without inventing new definitions for each dimension.

In one dimension, the construction begins with a sigma-algebra of Lebesgue-measurable subsets of R, denoted L. Every set A in this sigma-algebra has a well-defined Lebesgue measure Λ(A), taking values in [0, ∞]. This measure generalizes the familiar notion of length on the number line.

To move to two dimensions, the key step is to use the Cartesian product structure of R^2 = R × R. Given two Lebesgue-measurable sets A1 and A2 in R, their product A1 × A2 forms a “generalized rectangle” in R^2. For these product sets, the two-dimensional Lebesgue measure is defined by the product rule: Λ_2(A1 × A2) = Λ(A1) · Λ(A2). From there, the measure is extended to a maximal sigma-algebra of subsets of R^2, called L(R^2), so that many more sets than just rectangles become measurable.

This extension matters because most subsets of R^2 are not simple rectangles, yet the resulting two-dimensional Lebesgue measure still assigns them an “area” in a way consistent with the measure axioms. The resulting sigma-algebra includes all Borel sets (all open and closed sets), and it is complete: any subset of a measure-zero set is also measurable. The measure is normalized so that the unit square has area 1, and it is translation invariant—shifting a measurable set in the plane does not change its measured area.

With Lebesgue measure in place, defining integrals becomes straightforward. The n-dimensional Lebesgue integral is the usual Lebesgue integral taken with respect to the n-dimensional Lebesgue measure. In practice, this means writing an integral over R^n with measure Λ_n, often denoted simply as ∫ f(x) d x when the dimension is clear.

The same product-measure construction repeats in higher dimensions. For R^3, one takes U ⊂ R^2 and A3 ⊂ R and defines volume via Λ_3(U × A3) = Λ_2(U) · Λ(A3). Iterating this process yields the n-dimensional Lebesgue measure Λ_n on R^n, defined on the sigma-algebra L(R^n). It retains the same core properties: completeness, inclusion of Borel sets, normalization on the unit n-cube (measure 1), and translation invariance. The integral in n dimensions follows immediately from measure theory, setting up the next challenge: computing these integrals explicitly when functions and sets are sufficiently “nice.”

Cornell Notes

n-dimensional Lebesgue measure on R^n is constructed using product measures. Start with the Lebesgue measurable sets L on R and their measure Λ(A). For R^2, measurable rectangles A1 × A2 get area Λ_2(A1 × A2)=Λ(A1)Λ(A2), and this rule is extended to a sigma-algebra L(R^2) that contains all Borel sets and is complete. The resulting measure is normalized (unit square has measure 1) and translation invariant. Once Λ_n is defined on R^n, the n-dimensional Lebesgue integral is obtained directly as the Lebesgue integral with respect to Λ_n, so no separate “new” integral definition is needed for each dimension.

How does the construction of Lebesgue measure in higher dimensions avoid redefining everything from scratch?

It uses the product-measure framework from measure theory. One defines the measure first on Cartesian products of one-dimensional measurable sets (rectangles in R^2, boxes in R^3, and so on) using a product rule for measures. Then the measure is extended to a maximal sigma-algebra of measurable subsets in R^n. Because the Lebesgue integral is defined for any measure, the n-dimensional integral is automatically the Lebesgue integral with respect to Λ_n.

What is the exact rule for the two-dimensional Lebesgue measure on product sets?

For Lebesgue-measurable sets A1, A2 ⊂ R, the two-dimensional measure of the Cartesian product is defined by Λ_2(A1 × A2)=Λ(A1)·Λ(A2). This matches the familiar area computation for rectangles when A1 and A2 are intervals, and it becomes the starting point for extending the measure to more complicated subsets of R^2.

Why does the sigma-algebra need to be enlarged beyond rectangles?

Rectangles (Cartesian products of measurable sets) form only a limited collection of subsets of R^2. To assign areas consistently to arbitrary subsets, the construction extends the product sigma-algebra to a larger sigma-algebra L(R^2) so that many more sets become measurable. This extension is what makes the measure useful in practice, since most subsets are not rectangles.

Which properties of Lebesgue measure are preserved in the n-dimensional setting?

The n-dimensional Lebesgue measure Λ_n is complete (subsets of measure-zero sets are measurable), includes all Borel sets, is normalized so the unit n-cube has measure 1, and is translation invariant (shifting a measurable set in R^n does not change its measure). These mirror the key properties already established in one and two dimensions.

How is the n-dimensional Lebesgue integral notationally handled in the construction?

The integral is written as an integral over R^n with respect to Λ_n. When the dimension is clear, the measure symbol is often omitted and the notation becomes ∫ f(x) dx, where x is a point in R^n. If needed, components can be written explicitly (e.g., x=(x1,x2) in R^2), and the differential can be written with dimension-specific notation to avoid confusion.

Review Questions

  1. What product rule defines Λ_2(A1×A2) for measurable sets A1 and A2 in R?
  2. Which sigma-algebra properties (Borel inclusion and completeness) does L(R^n) satisfy, and why do they matter?
  3. How does translation invariance affect the measured value of a set under shifts in R^n?

Key Points

  1. 1

    Lebesgue measure in R^n is constructed via product measures, starting from the one-dimensional Lebesgue measure Λ on measurable sets in R.

  2. 2

    For measurable product sets in R^2, area is defined by Λ_2(A1×A2)=Λ(A1)·Λ(A2), matching the rectangle-length product when A1 and A2 are intervals.

  3. 3

    The measure is extended from product sets to a full sigma-algebra L(R^n) so that far more subsets than rectangles become measurable.

  4. 4

    The n-dimensional Lebesgue measure is complete, includes all Borel sets, is normalized so the unit n-cube has measure 1, and is translation invariant.

  5. 5

    Once Λ_n is defined, the n-dimensional Lebesgue integral is obtained directly as the Lebesgue integral with respect to Λ_n, using standard measure-theoretic definitions.

  6. 6

    Higher-dimensional volume (e.g., in R^3) follows the same pattern: Λ_3(U×A3)=Λ_2(U)·Λ(A3), and the process iterates to any n.

Highlights

The n-dimensional Lebesgue measure Λ_n is built by iterating product-measure constructions, not by inventing new measurement rules for each dimension.
On product sets A1×A2 in R^2, the measure satisfies Λ_2(A1×A2)=Λ(A1)Λ(A2), providing the foundation for “area” in higher dimensions.
The resulting sigma-algebra L(R^n) is complete and contains all Borel sets, ensuring measurability for a wide class of sets.
Normalization and translation invariance carry over: the unit n-cube has measure 1, and shifting a set in R^n does not change its measure.
The n-dimensional Lebesgue integral comes for free once Λ_n exists, since Lebesgue integration works for any measure.