By W. W. Sawyer

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**Extra resources for A Path to Modern Mathematics**

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To this end we verify that ω{/ϋ^, χ) < ω(/, χ). Indeed, in every segment Δ the following inequalities hold sup/JiK^) < sup/(x), Δ Δ inf/JiXx) ^ inf/(x). Δ Δ Therefore, ω{/ϋ^. Δ) < ω(/. Δ). Contracting the segment Δ to the point X, we obtain the required inequality. This inequality also proves the inclusion f(ω)(/Χί, X ) > α ) c= £ ( ω ( / , χ ) > α). 6) Now it is sufficient to observe that the set Εχ{ω{/, χ) > α) = is discrete. ), is discrete and can be covered by a finite system of intervals { S j of total length <ε.

103). ·We shall show that Conditions (1'}-(3') actually correspond to Conditions (I}-(6) of the problem of integration, formulated for characteristic functions. , congruent sets have the same measure. · Condition (2) requires that the measure of a set contained in [a, b) be equal to the sum of the measures of its parts belonging to [a, c) and [c, b) (the point c may possibly be counted twice). We shall see that, formulated in this manner, Condition (2) is a corollary of Conditions (3) and (4). Condition (3) makes sense for the measure problem if and only iff and cp are characteristic functions of nonoverlapping sets, since only in 54 4 LEBESGUE'S MEASURE AND INTEGRATION this case is f + qJ a characteristic function (of the sum of these sets).

1 THE PROBLEM OF INTEGRATION 51 For this purpose it is sufficient to note, using Conditions (4) and (3) and Eq. 1), that O~l(q>-f)dx= " From the inequalities lq>dx+ l(-f)dx= lq>dx- l f dx . 1) we obtain Furthermore, from Condition (3) we obtain (n being an integer) b b a a f nfdx = n ff dx , and applying the last equality to fin, 1 b b 1 - f f dx = f - f dx. 3) for a rational k. Now let k be irrational and k l be rational. Then II kf dx - k1If dxl < Ilk - k11lfl dx< Ik - k11(suPlfl +,,) I dx, where" (0 < " < 1) is chosen such that Ik - k11 (suplfl + ,,) will be rational.

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