54 1. Measure theory (x) (Reflection) If f + g is a simple function that is bounded with finite measure support (i.e. it is absolutely integrable), then we have Simp Rd f(x) + g(x) dx = Rd f(x) dx + Rd g(x) dx. Do the horizontal and vertical truncation properties hold if the lower Lebesgue integral is replaced with the upper Lebesgue integral? Now we restrict attention to measurable functions. Definition 1.3.13 (Unsigned Lebesgue integral). If f : Rd → [0, +∞] is measurable, we define the unsigned Lebesgue integral Rd f(x) dx of f to equal the lower unsigned Lebesgue integral Rd f(x) dx. (For non-measurable functions, we leave the unsigned Lebesgue integral undefined.) One nice feature of measurable functions is that the lower and upper Lebesgue integrals can match, if one also assumes some boundedness: Exercise 1.3.11. Let f : Rd → [0, +∞] be measurable, bounded, and van- ishing outside of a set of finite measure. Show that the lower and upper Lebesgue integrals of f agree. (Hint: Use Exercise 1.3.4.) There is a con- verse to this statement, but we will defer it to later notes. What happens if f is allowed to be unbounded, or is not supported inside a set of finite measure? This gives an important corollary: Corollary 1.3.14 (Finite additivity of the Lebesgue integral). Let f, g : Rd → [0, +∞] be measurable. Then Rd f(x) + g(x) dx = Rd f(x) dx + Rd g(x) dx. Proof. From the horizontal truncation property and a limiting argument, we may assume that f, g are bounded. From the vertical truncation property and another limiting argument, we may assume that f, g are supported inside a bounded set. From Exercise 1.3.11, we now see that the lower and upper Lebesgue integrals of f, g, and f + g agree. The claim now follows by combining the superadditivity of the lower Lebesgue integral with the subadditivity of the upper Lebesgue integral. In the next section we will improve this finite additivity property for the unsigned Lebesgue integral further, to countable additivity this property is also known as the monotone convergence theorem (Theorem 1.4.43). Exercise 1.3.12 (Upper Lebesgue integral and outer Lebesgue measure). Show that for any set E ⊂ Rd, Rd 1E(x) dx = m∗(E). Conclude that the upper and lower Lebesgue integrals are not necessarily additive if no measurability hypotheses are assumed.

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