Let f be a measurable function where ∫f⁺ dμ = +∞ and ∫f⁻ dμ = +∞. What is ∫f dμ?
A∫f dμ = 0 — the positive and negative infinite parts cancel each other out
B∫f dμ = +∞ — the positive part dominates when both parts are infinite
C∫f dμ is undefined — ∞ − ∞ has no well-defined value in this context
D∫f dμ = −∞ — the negative part subtracts from the positive, pulling the result to −∞
When both ∫f⁺ and ∫f⁻ are infinite, ∫f = ∫f⁺ − ∫f⁻ = ∞ − ∞, which is genuinely undefined — not zero, not infinite. Unlike a limit approaching ∞ − ∞ that might have a specific value, two separately infinite integrals have no well-defined net result. This is the exact problem the decomposition is designed to avoid: the definition of ∫f dμ = ∫f⁺ − ∫f⁻ only applies when at least one of the two integrals is finite.
Question 2 Multiple Choice
A measurable function f satisfies ∫f⁺ dμ = 5 and ∫f⁻ dμ = +∞. Which statement correctly describes ∫f dμ and whether f is integrable?
Af is integrable and ∫f dμ = 5, since only the finite part contributes
Bf is not integrable (∫|f| = ∞), but ∫f dμ = −∞ is still a well-defined extended real value
C∫f dμ is undefined because one integral is infinite — any infinite integral prevents the Lebesgue integral from being defined
Df is integrable and ∫f dμ = −∞, since the negative part dominates
Since ∫f⁺ = 5 (finite) and ∫f⁻ = +∞, the subtraction ∫f⁺ − ∫f⁻ = 5 − ∞ = −∞ is a well-defined extended real value (not the undefined ∞ − ∞). So ∫f dμ = −∞ exists. However, f is NOT integrable in the L¹ sense because ∫|f| = ∫f⁺ + ∫f⁻ = 5 + ∞ = +∞. Integrability (L¹) requires ∫|f| < ∞ — both parts must be individually finite. The integral can be defined as an extended real value without f being in L¹.
Question 3 True / False
If ∫|f| dμ < ∞ (f is in L¹), then both ∫f⁺ dμ and ∫f⁻ dμ must individually be finite.
TTrue
FFalse
Answer: True
Since |f| = f⁺ + f⁻, we have ∫|f| = ∫f⁺ + ∫f⁻. If this sum is finite, each non-negative term must be finite individually — a finite sum of non-negative quantities requires each summand to be finite. This is why L¹ integrability is the clean, useful condition: it guarantees ∫f = ∫f⁺ − ∫f⁻ is a well-defined, finite real number, enabling linearity, the dominated convergence theorem, and other essential tools.
Question 4 True / False
The decomposition f = f⁺ − f⁻ is necessary because the Lebesgue integral cannot handle negative function values.
TTrue
FFalse
Answer: False
The decomposition is not needed because the Lebesgue integral 'can't handle' negatives — it's needed to prevent the undefined form ∞ − ∞. If both f⁺ and f⁻ have finite integrals, signed values pose no difficulty whatsoever; ∫f = ∫f⁺ − ∫f⁻ is a straightforward finite subtraction. The problem arises only when both parts have infinite integrals, making the subtraction undefined. The decomposition is a mechanism to isolate and manage this specific ∞ − ∞ problem, not a restriction on negativity.
Question 5 Short Answer
Why does ∞ − ∞ pose a specific problem for defining the integral of a signed function, and how does the f⁺/f⁻ decomposition resolve it?
Think about your answer, then reveal below.
Model answer: The expression ∞ − ∞ is genuinely undefined — unlike a limit that approaches a specific value, two independently infinite quantities that 'cancel' cannot be assigned a consistent real value. For a signed function with infinite positive area above the x-axis and infinite negative area below, there is no consistent way to assign a net integral. The decomposition avoids this by integrating f⁺ and f⁻ separately using the non-negative Lebesgue integral (which is always well-defined in [0, +∞]). The subtraction ∫f⁺ − ∫f⁻ is only performed when at least one term is finite, ensuring the result is either a finite number or ±∞ — never the undefined ∞ − ∞.
This careful conditional definition is what distinguishes rigorous Lebesgue integration from a naive 'add up positive and negative areas' approach. The condition 'at least one of ∫f⁺ or ∫f⁻ is finite' is precisely the guard that prevents the undefined case.