1.1. Prologue: The problem of measure 13 is independent of the choice of partition used to demonstrate the piecewise constant nature of f. We will denote this quantity by p.c. b a f(x) dx, and refer to it as the piecewise constant integral of f on [a, b]. Exercise 1.1.21 (Basic properties of the piecewise constant integral). Let [a, b] be an interval, and let f, g : [a, b] R be piecewise constant functions. Establish the following statements: (1) (Linearity) For any real number c, cf and f + g are piecewise con- stant, with p.c. b a cf(x) dx = c p.c. b a f(x) dx and p.c. b a f(x) + g(x) dx = p.c. b a f(x) dx + p.c. b a g(x) dx. (2) (Monotonicity) If f g pointwise (i.e. f(x) g(x) for all x [a, b]), then p.c. b a f(x) dx p.c. b a g(x) dx. (3) (Indicator) If E is an elementary subset of [a, b], then the in- dicator function 1E : [a, b] R (defined by setting 1E(x) := 1 when x E and 1E(x) := 0 otherwise) is piecewise constant, and p.c. b a 1E(x) dx = m(E). Definition 1.1.6 (Darboux integral). Let [a, b] be an interval, and f : [a, b]→ R be a bounded function. The lower Darboux integral bf(x) a dx of f on [a, b] is defined as b a f(x) dx := sup g≤f, piecewise constant p.c. b a g(x) dx, where g ranges over all piecewise constant functions that are pointwise bounded above by f. (The hypothesis that f is bounded ensures that the supremum is over a non-empty set.) Similarly, we define the upper Darboux integral b a f(x) dx of f on [a, b] by the formula b a f(x) dx := inf h≥f, piecewise constant p.c. b a h(x) dx. Clearly, b a f(x) dx b a f(x) dx. If these two quantities are equal, we say that f is Darboux integrable, and refer to this quantity as the Darboux integral of f on [a, b]. Note that the upper and lower Darboux integrals are related by the reflection identity b a f(x) dx = b a f(x) dx. Exercise 1.1.22. Let [a, b] be an interval, and f : [a, b] R be a bounded function. Show that f is Riemann integrable if and only if it is Darboux
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