Teach Yourself Swedish

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Proof. e. In fact, let X n → X , or μ(X n n→∞ μ(X n n→∞ n→∞ X ) = 0. However, |X n+ν − X | ≥ X ) = 0 is equivalent to μ ν k = 1, 2, . . ⎛ But 1 μ |X n − X | ≥ k ≤ μ⎝ |X n+ν ν≥0 1 k → 0, n→∞ ⎞ 1 ⎠ − X| ≥ . k Thus μ |X n − X | ≥ 1 k → 0, n→∞ k = 1, 2, . . , or equivalently, μ Xn → X . n→∞ Remark 2. This need not be true if μ is not finite. Also, the inverse need not be true even if μ is finite. These points can be illustrated by examples (see Exercises 2(i) and 2(ii)). 2 Convergence in Measure is Equivalent to Mutual Convergence in Measure It is first shown that convergence in measure ensures the existence of a subsequence that converges almost everywhere.

We say that ϕ is an extension of ϕ, and ϕ is a restriction of ϕ , if C ⊂ C and ϕ = ϕ on C. 2 Outer Measures Definition 4. A set function μ◦ : P( ) → ¯ is said to be an outer measure, if (i) μ◦ ( ) = 0. , A ⊂ B implies μ◦ (A) ≤ μ◦ (B). , μ◦ ( ∞ n=1 An ) ≤ n=1 μ (An ). Remark 5. (i) μ◦ (A) ≥ 0 for all A, since ⊆ A implies 0 = μ◦ ( ) ≤ μ◦ (A) by (i) and (ii). (ii) It follows that μ◦ is finitely subadditive, since μ◦ ( nj=1 A j ) = μ◦ ( ∞ j=1 B j ), where B j = A j , j = 1, . . , n, B j = , j ≥ n + 1.

Is an outer measure. If μ is σ -finite on F, then μ∗ is σ -finite on P( ). If μ is finite on F, then μ∗ is finite on P( ). Proof. (i) Let A ∈ F. Then A ⊆ A so that μ∗ (A) ≤ μ(A) by the definition of μ∗ . Thus, it suffices to show that μ∗ (A) ≥ μ(A). Let A j ∈ F, j = 1, 2, . , A ⊆ ∞ j=1 A j . At this point we notice that j=1 A j need A ) need not be defined at all. So we work not belong in F and hence μ( ∞ j j=1 ∞ A ) = (A ∩ A ), while A ∩ A as follows: A = A ∩ ( ∞ j j ∈ F, since j=1 j j=1 ∞ ∞ A, A j ∈ F, j = 1, 2, .

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