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Is there an example to see why the dominated convergence theorem fails when there is no integrable function dominates the sequence $f_n(x)$?
Also for Fatou's lemma, is there an example where the strict inequality holds?, i.e: $$\int_X f(x)\text{d}\mu < \liminf_{n\to\infty} \int_X f_n(x) \text{d}\mu$$

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2 Answers 2

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Let $X$ be the line with Lebesgue measure and let $f_n(x) = \chi_{[n,n+1]}(x)$. This is a counterexample for both questions.

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  • $\begingroup$ Thank you Umberto P. Could you please clarify it because I am not familiar with these theorems. $\endgroup$
    – Fabian
    Aug 14, 2015 at 14:08
  • $\begingroup$ What do you need clarified? $\endgroup$
    – Umberto P.
    Aug 14, 2015 at 14:13
  • $\begingroup$ For D.C.T. Why $f_n(x)$ is not dominated?? Does the sequence go to infinity? $\endgroup$
    – Fabian
    Aug 14, 2015 at 14:18
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    $\begingroup$ The sequence $f_n$ converges everywhere to $0$, so $f(x) = 0$. Note that if $f_n(x) \le g(x)$ for all $x \in \mathbb R$, then in particular $g(x) \ge 1$ for all $x \in [1,\infty)$ so that $g$ cannot be integrable. $\endgroup$
    – Umberto P.
    Aug 14, 2015 at 14:23
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    $\begingroup$ Either $[0,\infty)$ or $(-\infty,\infty)$ is an acceptable domain. $\endgroup$
    – Umberto P.
    Aug 14, 2015 at 15:14
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To complement Umberto P.'s example, here is another which shows that the inequality in Fatou's lemma can be strict: let $(X,\mu)$ be a probability space, and let $E \subset X$ be a measurable subset of positive (but not full) measure. Define $$ f_n(x) = \begin{cases} \chi_E(x), & \textrm{ $n$ odd, }\\ \chi_{E^c}(x) = 1-\chi_E(x), & \textrm{ $n$ even}. \end{cases} $$ It follows that $$ \int_X \left(\liminf_{n \to \infty} f_n \right) d\mu = 0 < \min\{ \mu(E), 1-\mu(E) \} = \liminf_{n \to \infty} \int_X f_n d\mu. $$

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