Proposition: Assume $E$ has finite measure. Let the sequence of functions $\{f_n\}$ be uniformly integrable over $E$. If $\{f_n\} \rightarrow f$ pointwise a.e. on $E$, then $f$ is integrable over $E$.


Let $\delta_0 > 0$ respond to the $\epsilon=1$ challenge in the uniform integrability criteria for the sequence $\{f_n\}$.

Since $m(E)< \infty$, we may express $E$ as the disjoint union of a finite collection of measurable subsets $\{E_k\}^N_{k=1}$ such that $m(E_k)<\delta_0$ for $1\leq k \leq N$. For any $n$, by the monotonicity and additivity over domains property of the integral, $$\int_E |f_n| = \sum^N_{k=1} \int_{E_k} |f_n| < N.$$ We infer from Fatou's Lemma that $$\int_E |f| \leq \text{lim inf} \int_E |f_n| \leq N.$$ Thus $|f|$ is integrable over $E$.

My Question:

I don't understand why they have to break $E$ into disjoint subsets in order to use Fatou's Lemma. We know that $E$ is finite anyways, so why can't we just directly apply the lemma?

Thanks in advance.

  • $\begingroup$ @DavideGiraudo This is the definition our textbook gives "A family F of measurable functions on E is said to be uniformly integrable over E provided for each $\epsilon > 0$, there is a $\delta > 0$ such that for each $f \in F$, if $A \subseteq E$ is measurable and m(A) < $\delta$, then $\int_A |f| < \epsilon$. $\endgroup$
    – user58289
    Mar 2, 2013 at 20:11
  • 2
    $\begingroup$ They break $E$ into pieces to use uniform integrability, not Fatou. $\endgroup$
    – Julien
    Mar 2, 2013 at 20:18
  • 1
    $\begingroup$ Since m(E)<∞, we may express E as the disjoint union of a finite collection of measurable subsets {Ek}Nk=1 such that m(Ek)<δ0 for 1≤k≤N... How so? $\endgroup$
    – Did
    Mar 2, 2013 at 20:19
  • 3
    $\begingroup$ @Artus This definition does not even guarantee that $(f_n)$ is bounded in $L^1$, a property which seems desirable if the family must be said to be uniformly integrable... $\endgroup$
    – Did
    Mar 2, 2013 at 20:30
  • $\begingroup$ Any assumption on the measure space? $\endgroup$ Mar 12, 2013 at 15:07

1 Answer 1


In Bogachev, Measure theory, volume 1, theorem 1.12.9 asserts the following:

Let $(\Omega,\mathcal F,\mu)$ be a measure space with a finite non-negative measure. Then for each $\delta>0$, we can find an integer $N$ and a finite partition of $\Omega$, $\{S_1,\dots,S_N\}$ such that for each $i$, either $\mu(S_i)\leqslant \delta$ or $S_i$ is an atom of measure $>\delta$.

So all works nice when $\mu$ is atomless. But not when for example $X=\{a,b\}$ with $\mu\{a\}=\mu\{b\}=1$ (the mentioned assertion in the proof doesn't hold).


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