維正 發問時間: 科學數學 · 1 0 年前

收斂in測度

Let X be a finite measure space. Let {f_n} and {g_n} be sequences of a.e. real, measurable functions that converge in measure to f and g, respectively. Then:

(1) {f_n g} converges in measure to fg.

(2) {f_n g_n} converges in measure to fg. [Hint: Consider first the case f=0, g=0.]

1 個解答

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  • L
    Lv 7
    1 0 年前
    最佳解答

    1)Since f_n -> f in measure, so for any f_(n_k) there exists f_(n_k(m)) -> f a.e.,hence (f_(n_k(m)))g -> fg a.e..Hence (f_n)g -> fg in measure2)Since f_n -> f in measure, so for any f_(n_k) there exists f_(n_k(m)) -> f a.e.,for this n_k, since n_k(m) is a subsquence of n_k and g_n -> g in measure, hence there is n_k(m_p) so that g_(n_k(m_p)) -> g a.e.hence f_(n_k(m_p)) g_(n_k(m_p)) -> fg a.e., which implies that(f_n)(g_n) -> fg in measure

    2006-12-01 22:40:06 補充:

    這裡面用到了一個定理

    f_n 依測度收斂到 f

    若且惟若

    對於 f_n 的任一子列 f_(n_k), 都存在更進一步的子列 f_(n_k(m)) 使得 f_(n_k(m)) -> f a.e.

    2006-12-01 22:50:44 補充:

    全空間的測度要有限這個定理才會對 ORZ

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