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Let $\{f_n\}$ be a sequence of integrable functions on $X$ that is uniformly integrable. Suppose that $f_n\to f$ pointwise a.e. on $X$ and $f$ is measurable. Assume the measure space has the property that for each $\varepsilon \gt 0$, $X$ is the union of a finite collection of measurable sets, each of measure at most $\varepsilon$.

Can I get some hints to prove that $f$ is integrable over $X$?


since $f_n$ is uniformly integrable, there is a $\delta \gt0$ such that for each $n$ $$\int_E |f_n|~d\mu \lt 1,$$ where $E$ is a measurable subset of $X$. I can use Fatou's Lemma to show that on $E$ $$\int_E |f|~d\mu\lt 1.$$

Let $X = \bigcup_{n=1}^k E_n$ where $\mu(E_n) \le \varepsilon$ for each $n$. Then I can say $\mu(X) \le k\varepsilon$. I don't know how this is going help.

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up vote 5 down vote accepted

Choose a $\delta>0$ so that for each positive integer $n$ $$\tag{1}\int_E |f_n|<1,\quad\text{whenever}\quad\mu(E)<\delta.$$

Write $X=\bigcup\limits_{k=1}^M E_k$ where $\mu (E_k)<\delta$ for each $n=1$, $2$, $\ldots\,$, $M$. Then, taking advantage of $(1)$ in the last inequality below, it follows that for each positive integer $n$ we have $$\tag{2} \int_X |f_n|=\int_{\bigcup_{k=1}^M E_k} |f_n|\ \le\ \sum_{k=1}^M \int_{E_k} |f_n|\ \le\ \sum_{k=1}^M\, 1\ =\ M. $$ Since $f_n$ converges to $f$ pointwise on $X$, it then follows from $(2)$ and Fatou's Lemma that $$ \int_X |f|=\int_X \liminf_{n\rightarrow\infty} |f_n| \le \liminf_{n\rightarrow\infty}\int_X |f_n|\le M. $$ Thus, $f$ is integrable on $X$.

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Please, can you explain the first displayed equation. thanks. – Kuku Mar 30 '12 at 1:38
@kuku If the $E_i$ were disjoint, you'd have $\int_X |g|=\int_{\cup E_k}|g|=\int_{E_1} |g|+\int_{E_2}|g|+\cdots+\int_{E_M} |g|$. More generally, the first inequality holds for $X$ expressed as a, not necessarily disjoint, union of the sets $E_1$, $E_2$, $\ldots\,$, $E_M$. – David Mitra Mar 30 '12 at 1:42
Thanks very much. I got it now. – Kuku Mar 30 '12 at 1:48
@kuku You're welcome. Glad to help. – David Mitra Mar 30 '12 at 2:00

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