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    Measurability of $\inf_k f_k$ and $\sup_k f_k$, Fatou's lemma, and the monotone convergence theorem from Fatou

    P-OSPIM

    Area
    Real Analysis
    Topics
    Convergence of Integrals, Fatou, Measure Theory
    Status
    draft
    1. Let \(E \subseteq \RR^n\) be bounded. Prove that the following are equivalent:

      1. For any \(\epsilon>0\) there exist an open set \(G\) and a closed set \(F\) such that \(F \subseteq E \subseteq G\) and \(m(G\sm F) < \epsilon\).

      2. There exist a \(G_\delta\) set \(V\) and an \(F_\sigma\) set \(H\) such that \(m(V\sm H) = 0\).

    2. Let \(f_k\) be a sequence of extended real-valued Lebesgue measurable function.

      1. Prove that \(\inf_k f_k, \sup_k f_k\) are both Lebesgue measurable function.

        Hint: argue that \[ \ts{x \st \inf_k f_k(x) < a} = \Union_k \ts{x \st f_k(x) < a} .\]

      2. Carefully state Fatou’s Lemma and deduce the Monotone Converge Theorem from it.

    Condition 2 of part (a) needs \(H \subseteq E \subseteq V\); as printed, it holds for every \(E\) with \(V = H = \emptyset\).

    Solution

    1. Part (a), with \(H \subseteq E \subseteq V\) in condition 2.

    This is P-7ITXP. If (1) holds, take \(G_k\) and \(F_k\) for \(\eps = 1/k\) and put \(V = \bigcap_k G_k\), \(H = \bigcup_k F_k\); then \(m(V \setminus H) \le m(G_k \setminus F_k) < 1/k\). Conversely, intersect \(V\) with an open ball containing \(E\), write it as a decreasing intersection of open sets of finite measure and \(H\) as an increasing union of closed sets, and apply continuity of measure from above and below.

    2. For measurable \(f_k\) with values in \([-\infty, \infty]\), \(\sup_k f_k\) and \(\inf_k f_k\) are measurable.

    For every \(a \in \RR\), \(\theset{\sup_k f_k \le a} = \bigcap_k\theset{f_k \le a}\) and \(\theset{\inf_k f_k < a} = \bigcup_k\theset{f_k < a}\) are countable intersections and unions of measurable sets.

    3. Fatou’s lemma: for measurable \(f_n \ge 0\), \(\int \liminf_n f_n \le \liminf_n \int f_n\).

    4. If \(0 \le f_1 \le f_2 \le \cdots\) are measurable and \(f_n \to f\) pointwise, then \(\int f_n \to \int f\).

    \(\int f_n\) is nondecreasing, so \(\lim_n\int f_n\) exists in \([0,\infty]\). Since \(f_n \le f\), \(\lim_n \int f_n \le \int f\). Since \(f = \liminf_n f_n\), Fatou’s lemma gives \(\int f \le \liminf_n \int f_n = \lim_n \int f_n\).