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**Extra resources for American Mathematical Monthly, volume 117, June July 2010**

**Sample text**

It is assumed that G 0 := {x ∈ R N : (0, x) ∈ G} = ∅ and u 0 ∈ G 0 . 1), we set h := 0 and := (J × R N ) ∩ G. Note that 0 = G 0 and, moreover, since h = 0, I+ = I for all I ∈ I. Trivially, if I ∈ I is such that W (I ) = ∅, then I ⊂ J and W (I ) = WG (I ). 1), we mean a function u ∈ W (I ), with I ∈ I, t such that u(0) = u 0 and u (t) = 0 k(t, s)g(s, u(s)) ds for almost every t ∈ I . 1). Set q := p/( p − 1). We impose the following hypothesis on g. (IDE) For every (t, z) ∈ G, there exist a relatively open interval I ⊂ R+ containing t, an open ball B ⊂ R N containing z, and a nonnegative function l ∈ L q (I ) such that I × B ⊂ G, the function I → R M , s → g(s, x) is measurable for all x ∈ B and is in L q (I ) for some x ∈ B, and moreover, g(s, x) − g(s, y) ≤ l(s) x − y ∀ s ∈ I, ∀ x, y ∈ B.

Xn ∈ Bi . 2) j =1 is measurable, where I J j denotes the characteristic function of J j . 2) and hence is measurable. Since i ∈ {1, . . , m} is m arbitrary and K = ∪i=1 K i , the function K → R M , s → g(s, v(s)) is measurable. (2) Let (t, z) ∈ G. 1) holds, and, moreover, the function I → R M , s → g(s, y) is in L q (I ) for some y ∈ B. Therefore, g(s, x) ≤ g(s, x) − g(s, y) + g(s, y) ≤ l(s) x − y + g(s, y) ≤ l(s) sup x − y + g(s, y) =: b(s) ∀ s ∈ I, ∀ x ∈ B. x∈B Since l ∈ L q (I ) and g(·, y) ∈ L q (I ), it follows that b ∈ L q (I ).

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