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Measure Theory 6 | Lebesgue Integral [dark version] thumbnail

Measure Theory 6 | Lebesgue Integral [dark version]

5 min read

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TL;DR

Lebesgue integration is built on a measure space (X, A, μ) where μ assigns sizes to measurable sets in A.

Briefing

Lebesgue integration gets its full footing by first defining the integral for nonnegative “step” (simple) functions on an abstract measure space, then extending that definition to general nonnegative measurable functions via a supremum over all step functions that sit below them. The key payoff is a well-defined notion of area under a measurable function—independent of how a simple function is represented—built only from the measure μ of sets, even when the underlying space X is completely abstract.

A measure space is described as a triple (X, A, μ): a set X, a sigma-algebra A of measurable subsets, and a measure μ mapping sets in A to values in [0, ∞]. The integration target is a measurable function f: X → ℝ, meaning every preimage f⁻¹(E) of a Borel set E lies in A. The starting point is the characteristic function 1_A of a measurable set A, whose integral is required to match the set’s “volume”: ∫ 1_A dμ = μ(A). From there, simple functions (also called step functions) are introduced as finite linear combinations of characteristic functions: f(x)=∑_{i=1}^n c_i 1_{A_i}, with measurable A_i.

For simple functions, the integral is defined by summing rectangle areas: ∫ f dμ = ∑_{i=1}^n c_i μ(A_i). A subtlety appears immediately when μ(A_i)=∞. If negative coefficients are allowed, expressions like (−2)·∞ become problematic. The solution is to restrict attention to nonnegative simple functions, where all coefficients c_i are ≥ 0. This leads to the set S⁺ of nonnegative simple functions, which behaves like a “half” vector space: addition is allowed, but scaling by negative numbers is not.

Even for simple functions, the representation is not unique—splitting a set A_i into smaller measurable pieces changes the formula without changing the function’s graph. The integral’s definition is designed to be representation-independent, so the total “area” remains the same no matter how the rectangles are partitioned.

With the nonnegative simple case in place, the Lebesgue integral for general nonnegative measurable functions is built using approximation from below. For a measurable f ≥ 0 that may take infinitely many values, one considers all simple functions h ∈ S⁺ such that h(x) ≤ f(x) everywhere. Each such h has a definite integral from the earlier step. Because h is always below f, these integrals provide lower bounds on the desired integral of f. The Lebesgue integral is then defined as the supremum of all these lower bounds:

∫ f dμ = sup{ ∫ h dμ : h ∈ S⁺, 0 ≤ h ≤ f }.

This supremum always exists in [0, ∞] and can be infinite. When the value is finite, f is called μ-integrable. The construction is fundamentally order-driven on the y-axis (the real line), not on X itself: the measure μ only tells how big sets are, while the “approximation by step functions” uses the ordering of real numbers to squeeze f from below. The result is a general Lebesgue integral defined purely from measure and measurability, setting up the later properties that make it powerful.

Cornell Notes

The Lebesgue integral is first defined for nonnegative simple (step) functions on an abstract measure space (X, A, μ). For a simple function f(x)=∑_{i=1}^n c_i 1_{A_i} with c_i ≥ 0, the integral is ∫ f dμ = ∑_{i=1}^n c_i μ(A_i), matching the “area of rectangles” picture. This definition is well defined even though simple-function representations are not unique. To integrate a general nonnegative measurable function f, the construction uses monotone approximation from below: consider all simple functions h ≤ f, compute their integrals, and define ∫ f dμ as the supremum of those values. If the supremum is finite, f is μ-integrable; otherwise the integral is ∞.

Why does the construction start with characteristic functions and what requirement fixes their integral?

Characteristic functions 1_A of measurable sets A are the atomic building blocks. The integral is fixed so that ∫ 1_A dμ equals μ(A), meaning the “area under the graph” of 1_A matches the measure (volume) of the set A. This anchors the whole integration scheme to the measure μ on the sigma-algebra A.

What exactly makes a function “simple” (step) in this setting, and how does that lead to a rectangle-sum formula?

A simple function has the form f(x)=∑_{i=1}^n c_i 1_{A_i}, where A_i are measurable sets and c_i are real constants. Since 1_{A_i} is constant (either 0 or 1) on each measurable region, f takes only finitely many values. That finite-valued structure lets the integral be computed as a sum of rectangle areas: each term contributes c_i·μ(A_i).

Why restrict to nonnegative simple functions S⁺, and what goes wrong with negative coefficients when μ(A_i)=∞?

If μ(A_i)=∞ and a coefficient c_i is negative, the product c_i·μ(A_i) becomes ambiguous (e.g., (−2)·∞), so the integral would not be well defined. Restricting to c_i ≥ 0 ensures the only infinity contributions are of the form (+constant)·∞, which can be consistently treated as ∞ in the sum.

How can the integral of a simple function be well defined when its representation is not unique?

A simple function’s representation can change without changing the function itself—for instance, splitting a measurable set A_i into two parts A_i' and A_i'' changes the sum’s terms but not the function’s values. The integral is designed so that the total contribution remains the same because μ is additive over measurable partitions, so the rectangle areas recombine to the same total.

How does the Lebesgue integral for a general nonnegative measurable function f get defined using simple functions?

For general f ≥ 0, one looks at all simple functions h ∈ S⁺ that satisfy h ≤ f pointwise. Each such h has a defined integral from the simple-function stage. Since h lies below f, these integrals act as lower bounds. The integral of f is defined as the supremum of all these lower bounds: ∫ f dμ = sup{∫ h dμ : h ∈ S⁺, h ≤ f}.

What does “μ-integrable” mean in this framework?

A nonnegative measurable function f is called μ-integrable when its Lebesgue integral is finite, i.e., ∫ f dμ < ∞. If the supremum defining the integral equals ∞, then f is not μ-integrable (though the integral still exists as an extended real value).

Review Questions

  1. In what way does the definition of ∫ f dμ for nonnegative measurable f rely on approximating f from below, and why is a supremum the natural choice?
  2. How does restricting to S⁺ (nonnegative simple functions) resolve the ambiguity involving products like (−2)·∞?
  3. Explain why the integral of a simple function should not depend on how the measurable sets A_i are partitioned in its representation.

Key Points

  1. 1

    Lebesgue integration is built on a measure space (X, A, μ) where μ assigns sizes to measurable sets in A.

  2. 2

    Characteristic functions satisfy ∫ 1_A dμ = μ(A), tying integration directly to the measure.

  3. 3

    Simple (step) functions are finite sums f=∑ c_i 1_{A_i}, enabling an integral defined as a rectangle-sum: ∫ f dμ = ∑ c_i μ(A_i).

  4. 4

    To avoid undefined expressions involving ∞, the construction restricts to nonnegative coefficients c_i ≥ 0, forming S⁺.

  5. 5

    The integral for simple functions is representation-independent because splitting measurable sets recombines rectangle areas via additivity of μ.

  6. 6

    For general nonnegative measurable f, the integral is defined as the supremum of ∫ h dμ over all simple h ∈ S⁺ with 0 ≤ h ≤ f.

  7. 7

    A function is μ-integrable exactly when its Lebesgue integral is finite (not equal to ∞).

Highlights

The integral for simple functions becomes a literal “sum of rectangle areas”: ∫(∑ c_i 1_{A_i}) dμ = ∑ c_i μ(A_i).
Nonnegativity is not cosmetic: it prevents ambiguous products like (−2)·∞ when μ(A_i)=∞.
The general Lebesgue integral is defined by squeezing from below: ∫ f dμ is the supremum of integrals of all simple h ≤ f.
Even though simple-function representations can change, the integral stays the same because μ handles measurable partitions consistently.

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