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Is it possible to extend the dominated convergence theorem without the restriction of the limit function be integrable?

In precise words, is the next statement true?

Let $(f_{n})$ be a succession of m-measurable functions in $X$, and let $g$ be an integrable function (respect to m-measure) in $X$, such that $|f_{n}|\leq g$ almost everywhere in $X$. If $\lim f_{n}=f$ almost everywhere in $X$ for a real function $f$ in $X$, not necessarily measurable then: 1) $f = h$ almost everywhere for an integrable function h and 2) $\lim \int_{X} f_{n}(x)dm = \int_{X}h(x)dm$ ?

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  • $\begingroup$ By $m$ do you mean the Lebesgue measure or a general measure? $\endgroup$ – copper.hat Oct 10 '14 at 6:45
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As $f_n \to f$ almost everywhere, we can choose $X_0$ such that $m(X_0)=m(X)$ and

$$\lim_{n \to \infty} f_n(x)=f(x)$$

for all $x \in X_0$. Then

$$\tilde{f}(x) := \begin{cases} f(x) & x \in X_0 \\ 0 & \text{otherwise} \end{cases}.$$

defines a measurable function satisfying $\tilde{f}=f$ almost everywhere. By assumption, $|f_n| \leq g$ almost everywhere and therefore we find

$$|f| = \lim_{n \to \infty} |f_n| \leq g$$

almost everywhere. Hence, $|\tilde{f}| \leq g$ almost everywhere. Consequently, we conclude that $\tilde{f}$ is integrable and we may apply the "standard" dominated convergence theorem to get the second statement.

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  • $\begingroup$ One should maybe still mention that $f$ is measurable as an (a.e.) limit of measurable functions. $\endgroup$ – PhoemueX Oct 10 '14 at 7:19
  • $\begingroup$ @PhoemueX Sure, you are right. $\endgroup$ – saz Oct 10 '14 at 11:30
  • $\begingroup$ The limit $f$ is not necessarily measurable without additional hypotheses (such as completeness). Let $E$ is a non-measurable set contained in a set of measure zero. Let $f=1_E$ and $f_n = 0$. Then $f_n \to f$ ae.. but $f$ is not measurable. $\endgroup$ – copper.hat Oct 10 '14 at 15:10
  • $\begingroup$ @copper.hat I see your point. I have rewritten my answer. $\endgroup$ – saz Oct 10 '14 at 15:42

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