1.3. The Lebesgue integral 55 Exercise 1.3.13 (Area interpretation of integral). If f : Rd → [0, +∞] is measurable, show that Rd f(x) dx is equal to the d+1-dimensional Lebesgue measure of the region {(x, t) ∈ Rd × R : 0 ≤ t ≤ f(x)}. (This can be used as an alternate, and more geometrically intuitive, definition of the unsigned Lebesgue integral it is a more convenient formulation for establishing the basic convergence theorems, but not quite as convenient for establishing basic properties such as additivity.) (Hint: Use Exercise 1.2.22.) Exercise 1.3.14 (Uniqueness of the Lebesgue integral). Show that the Lebesgue integral f → Rd f(x) dx is the only map from measurable un- signed functions f : Rd → [0, +∞] to [0, +∞] that obeys the following prop- erties for measurable f, g : Rd → [0, +∞]: (i) (Compatibility with the simple integral) If f is simple, then we have Rd f(x) dx = Simp Rd f(x) dx. (ii) (Finite additivity) Rd f(x)+ g(x) dx = Rd f(x) dx + Rd g(x) dx. (iii) (Horizontal truncation) As n → ∞, Rd min(f(x),n) dx converges to Rd f(x) dx. (iv) (Vertical truncation) As n → ∞, Rd f(x)1|x|≤n dx converges to Rd f(x) dx. Exercise 1.3.15 (Translation invariance). Let f : Rd → [0, +∞] be mea- surable. Show that Rd f(x + y) dx = Rd f(x) dx for any y ∈ Rd. Exercise 1.3.16 (Linear change of variables). Let f : Rd → [0, +∞] be measurable, and let T : Rd → Rd be an invertible linear transformation. Show that Rd f(T −1(x)) dx = | det T| Rd f(x) dx, or equivalently, that Rd f(Tx) dx = 1 | det T | Rd f(x) dx. Exercise 1.3.17 (Compatibility with the Riemann integral). Let f : [a, b] → [0, +∞] be Riemann integrable. If we extend f to R by declaring f to equal zero outside of [a, b], show that R f(x) dx = b a f(x) dx. We record a basic inequality, known as Markov’s inequality, that asserts that the Lebesgue integral of an unsigned measurable function controls how often that function can be large: Lemma 1.3.15 (Markov’s inequality). Let f : Rd → [0, +∞] be measurable. Then for any 0 λ ∞, one has m({x ∈ Rd : f(x) ≥ λ}) ≤ 1 λ Rd f(x) dx. Proof. We have the trivial pointwise inequality λ1{x∈Rd:f(x)≥λ} ≤ f(x).

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