# An extension of Dominated Convergence Theorem

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) $\lim \int_{X} f_{n}(x)dm = \int_{X}h(x)dm$ ?

• By $m$ do you mean the Lebesgue measure or a general measure? – copper.hat Oct 10 '14 at 6:45

## 1 Answer

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.

• One should maybe still mention that $f$ is measurable as an (a.e.) limit of measurable functions. – PhoemueX Oct 10 '14 at 7:19
• @PhoemueX Sure, you are right. – saz Oct 10 '14 at 11:30
• 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. – copper.hat Oct 10 '14 at 15:10
• @copper.hat I see your point. I have rewritten my answer. – saz Oct 10 '14 at 15:42