<Home>Qualifying Exams><Analysis>Fall 2010 - Problem 1

Fall 2010 - Problem 1

construction, dominated convergence theorem, Fatou's lemma

For this problem, consider just Lebesgue measurable functions f ⁣:[0,1]→R\func{f}{\br{0,1}}{\R} together with the Lebesgue measure.

  1. State Fatou's lemma (no proof required).
  2. State and prove the Dominated Convergence Theorem.
  3. Give an example where fn(x)→0f_n\p{x} \to 0 a.e., but ∫fn(x) dx→1\int f_n\p{x} \,\diff{x} \to 1.
Solution.

  1. Let fn,f ⁣:[0,1]→R\func{f_n, f}{\br{0,1}}{\R} be non-negative Lebesgue measurable functions for n≥1n \geq 1. Then

    ∫[0,1]lim inf⁡n→∞ fn dx≤lim inf⁡n→∞ ∫[0,1]fn dx.\int_{\br{0,1}} \liminf_{n\to\infty}\,f_n \,\diff{x} \leq \liminf_{n\to\infty}\,\int_{\br{0,1}} f_n \,\diff{x}.

  2. Let fn,f,g ⁣:[0,1]→R\func{f_n, f, g}{\br{0,1}}{\R} be Lebesgue measurable functions such that g∈L1g \in L^1, ∣fn∣≤g\abs{f_n} \leq g for all n≥1n \geq 1, and fn→ff_n \to f a.e.. Then ∥fn−f∥1→0\norm{f_n - f}_1 \to 0.

    Notice that fn→ff_n \to f a.e. implies ∣f∣≤g  ⟹  2g−∣fn−f∣≥0\abs{f} \leq g \implies 2g - \abs{f_n - f} \geq 0 and ∣fn−f∣→0\abs{f_n - f} \to 0 a.e., so by Fatou's lemma,

    ∫[0,1]2g dx=∫[0,1]lim inf⁡n→∞  (2g−∣fn−f∣) dx≤lim inf⁡n→∞ ∫[0,1]2g−∣fn−f∣ dx=lim inf⁡n→∞ ∫[0,1]2g dx+lim inf⁡n→∞ (−∫[0,1]∣fn−f∣ dx)=∫[0,1]2g dx−lim sup⁡n→∞ ∫[0,1]∣fn−f∣ dx.\begin{aligned} \int_{\br{0,1}} 2g \,\diff{x} &= \int_{\br{0,1}} \liminf_{n\to\infty}\,\,\p{2g - \abs{f_n - f}} \,\diff{x} \\ &\leq \liminf_{n\to\infty}\,\int_{\br{0,1}} 2g - \abs{f_n - f} \,\diff{x} \\ &= \liminf_{n\to\infty}\,\int_{\br{0,1}} 2g \,\diff{x} + \liminf_{n\to\infty}\,\p{-\int_{\br{0,1}} \abs{f_n - f} \,\diff{x}} \\ &= \int_{\br{0,1}} 2g \,\diff{x} - \limsup_{n\to\infty}\,\int_{\br{0,1}} \abs{f_n - f} \,\diff{x}. \end{aligned}

    Because g∈L1g \in L^1, ∫[0,1]2g dx\int_{\br{0,1}} 2g \,\diff{x} is finite, and so we may rearrange to get

    lim sup⁡n→∞ ∫[0,1]∣fn−f∣ dx≤0  ⟹  lim⁡n→∞∫[0,1]∣fn−f∣ dx=0.\limsup_{n\to\infty}\,\int_{\br{0,1}} \abs{f_n - f} \,\diff{x} \leq 0 \implies \lim_{n\to\infty} \int_{\br{0,1}} \abs{f_n - f} \,\diff{x} = 0.