Questions: Lebesgue Integral for Non-Negative Functions

Question 1

For a non-negative measurable function f that is not simple, how is ∫f dμ defined?

• A. As the limit of Riemann sums using equally-spaced partitions of the domain
• B. As the supremum of ∫φ dμ over all simple functions φ satisfying 0 ≤ φ ≤ f
• C. As the infimum of ∫ψ dμ over all simple functions ψ satisfying ψ ≥ f
• D. As the limit of ∫fₙ dμ where fₙ are the truncations fₙ = min(f, n)

Answer: (B)

The Lebesgue integral for non-negative f is defined by approximating from *below*: take all simple functions dominated by f, integrate each one (which is well-defined from the prior definition), and take the supremum of those values. This is philosophically distinct from Riemann sums (which partition the domain) — Lebesgue integration partitions the *range*. The infimum from above (option C) gives a different object; the truncation approach (option D) would work but is not the primary definition — it's a consequence via the Monotone Convergence Theorem.

Question 2

Let f(x) = 1/√x on (0,1] and f(0) = 0, with Lebesgue measure μ. What is ∫f dμ?

• A. Undefined — f is not bounded and therefore not Lebesgue integrable
• B. 0 — the singularity at x = 0 has measure zero and contributes nothing
• C. +∞ — the Lebesgue integral is defined but equals infinity
• D. 2 — the improper Riemann integral converges, and Lebesgue agrees

Answer: (D)

∫₀¹ x^{-1/2} dx = 2 as an improper Riemann integral, and the Lebesgue integral agrees with the improper Riemann integral whenever the latter exists and the function is non-negative. The Lebesgue integral is defined (by supremum of simple functions) and equals 2. Option A is wrong because the Lebesgue integral handles unbounded functions cleanly — it is defined for any non-negative measurable function, possibly equaling +∞. Option C would be correct for a function like 1/x (which has infinite integral on (0,1]) but not 1/√x.

Question 3

The Lebesgue integral for non-negative functions is undefined when the function is unbounded, since no simple function can approximate an infinite value.

• T. True
• F. False

Answer: False

The Lebesgue integral is *always* defined for non-negative measurable functions, whether bounded or not. The supremum of a set of non-negative numbers is either a finite non-negative number or +∞ — never undefined. If f is unbounded and the supremum is +∞, then ∫f dμ = +∞, which is a valid well-defined value in [0, +∞]. Indefiniteness (true 'undefined') only arises when you try to define the integral for a function with large positive and negative parts, where you might get ∞ − ∞. Non-negative functions avoid this problem entirely.

Question 4

If f ≤ g everywhere (both non-negative measurable), then every simple function dominated by f is also dominated by g, which implies ∫f dμ ≤ ∫g dμ.

• T. True
• F. False

Answer: True

This is the monotonicity property, and its proof follows directly from the supremum definition. If φ is simple and 0 ≤ φ ≤ f ≤ g, then φ is also a candidate in the supremum defining ∫g dμ. So every value in the set {∫φ : 0 ≤ φ ≤ f, φ simple} also appears in {∫φ : 0 ≤ φ ≤ g, φ simple}. The second set is at least as large, so its supremum is at least as large. Monotonicity is the key property that powers the Monotone Convergence Theorem: if fₙ ↑ f pointwise, the integrals ∫fₙ form a non-decreasing sequence bounded above by ∫f.

Question 5

Why is the Lebesgue integral for non-negative functions defined using a supremum over simple functions below f, and what goes wrong if you try to apply the same approach to functions that can take negative values?

Model answer: For non-negative functions, approximating from below is unambiguous: simple functions below f contribute less area than f, and the supremum captures the exact total area. The definition is always well-defined because the supremum of non-negative numbers is either finite or +∞, never undefined. For functions with negative values, the 'approximate from below' approach creates the problem ∞ − ∞: if f has a large positive part and a large negative part, the supremum of simple functions below f could be +∞ (capturing the positive part) and the 'correction' for the negative part would also be −∞, leaving the integral as ∞ − ∞, which is undefined. The solution is to decompose f into its positive part f⁺ = max(f, 0) and negative part f⁻ = max(−f, 0), integrate each using the non-negative definition, and compute ∫f = ∫f⁺ − ∫f⁻, which is well-defined as long as at least one of ∫f⁺, ∫f⁻ is finite.

This two-step structure — define for non-negative functions first, then extend by decomposition — is how measure theory builds integration rigorously. The non-negative case establishes the floor; the general case inherits its properties. The Monotone Convergence Theorem, Fatou's Lemma, and the Dominated Convergence Theorem are all proved first for non-negative functions and then extended.