By Terence Tao

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**Extra info for An Introduction To Measure Theory (January 2011 Draft)**

**Example text**

N=1 n=1 On the other hand, from countable subadditivity one has ∞ ∞ En ) ≤ m( n=1 m(En ), n=1 and the claim follows. Next, we handle the case when the En are bounded but not necessarily compact. We use the ε/2n trick. Let ε > 0. 7, we know that each En is the union of a compact set Kn and a set of outer measure at most ε/2n . Thus m(En ) ≤ m(Kn ) + ε/2n and hence ∞ ∞ m(En ) ≤ ( n=1 m(Kn )) + ε. n=1 Finally, from the compact case of this lemma we already know that ∞ m( ∞ Kn ) = n=1 m(Kn ) n=1 while from monotonicity ∞ ∞ Kn ) ≤ m( m( n=1 n=1 En ).

Now we connect the Riemann integral to Jordan measure in two ways. 24 (Basic properties of the Riemann integral). Let [a, b] be an interval, and let f, g : [a, b] → R be Riemann integrable. Establish the following statements: (1) (Linearity) For any real number c, cf and f +g are Riemann b b b integrable, with a cf (x) dx = c · a f (x) dx and a f (x) + g(x) dx = b a f (x) dx + b a g(x) dx. e. f (x) ≤ g(x) for all b b x ∈ [a, b]) then a f (x) dx ≤ a g(x) dx. 8A function f : [a, b] → R is piecewise continuous if one can partition [a, b] into finitely many intervals, such that f is continuous on each interval.

N=1 Finally, from the compact case of this lemma we already know that ∞ m( ∞ Kn ) = n=1 m(Kn ) n=1 while from monotonicity ∞ ∞ Kn ) ≤ m( m( n=1 n=1 En ). 2. Lebesgue measure 37 Putting all this together we see that ∞ ∞ m(En ) ≤ m( n=1 En ) + ε n=1 for every ε > 0, while from countable subadditivity we have ∞ m( ∞ En ) ≤ n=1 m(En ). n=1 The claim follows. Finally, we handle the case when the En are not assumed to be bounded or closed. Here, the basic idea is to decompose each En as a countable disjoint union of bounded Lebesgue measurable sets.