If $\left\{f_n\right\}$ are uniformly integrable and $f_n\overset{a.e.}{\rightarrow}f$ ($f$ measurable), is $f$ integrable? Can "uniformly integrable" be weakened to "integrable"?
Is the limit of uniformly integrable functions integrable?
3 Answers
Yes
$\int \left|f\right|\overset{\mathrm{a}}{\leq}\liminf\int |f_n|\overset{\mathrm{b}}{<}\infty$
b. Uniform integrability $\implies$ $\sup \int |f_n|<\infty$ (e.g. Klenke, Theorem 6.24i)
No, e.g. $f_n:=\mathbb{1}_{\left[-n,n\right]}$ (borrowed from Per Manne's comment below)
-
0@KopaLeo: I've corrected the proof. Hope it makes more sense now. – 2017-02-02
Put $f_n(x) = n(n+1)I_{(1/n+1, 1/n)}$, $x\in[0,1]$. Then $f_n\to 0$ in $[0,1]$ and the sequence of the $f_n$ is $L^1$-bounded. However, $f_n\not\rightarrow 0$ in $L^1$ as $n\to\infty$, since $\|f_n\|_1 = 1$ for all $n$.
-
2But this is an example of some integrable functions $f_n$ such that $f_n\to f$ almost everywhere and $f$ is integrable. Hence not an answer to the question asked. – 2012-08-09
Evan's accepted answer is almost correct and useful in most cases. However, it's missing a bit of nuance that is actually rather important when discussing other theorems like the Vitali Convergence Theorem.
Regarding your first question: The pointwise limit of uniformly integrable functions is only guarenteed to be integrable if their domain $E$ has finite measure. I am using $m(A)$ for any set $A$ to refer to Lebesgue measure.
As a counterexample in the case where $m(E) = \infty$, consider Per Manne's example. Set $E = [0, \infty)$ and $f_n = \chi_{[0, n]}$, the characteristic (indicator) function for $[0, n]$. The family $\{f_n\}$ is uniformly integrable: given $\epsilon > 0$, set $\delta = \epsilon$. Then for a measureable set $A$, $m(A) < \delta \implies \int_A |f_n| \le \int_A 1 < \delta = \epsilon$
However, $f_n \to f \equiv 1$ pointwise on $E$, and $\int_E |f| = \infty$ so $f$ is not integrable.
In general, another condition called tightness is used to justify the integrability of a pointwise limit of functions on any domain. Specifically, a family of measurable functions $F$ on a domain $E$ is tight over $E$ iff $\forall \, \epsilon > 0$, there exists $E_0 \subset E$ such that $\int_{E \setminus E_0} |f| < \epsilon \, \, \forall \, f \in F $
The proof of this is very similar to what Evan Aad described earlier, with an extra trick: I've sketched it below.
- Fix $\epsilon > 0$. Given $\{f_n\}$ measureable, uniformly integrable, and tight on $E$ where $\{f_n\} \to f$ pointwise on $E$, there exists some measurable set $E_0$ with finite measure s.t. $ \int_{E \setminus E_0} |f_n| < \epsilon \text{ and by Fatou's Lemma } \int_{E \setminus E_0} |f| < \epsilon$
- We have now reduced the problem to showing $f$ is integrable over $E_0$, as
$\int_E |f| = \int_{E \setminus E_0} |f| + \int_{E_0} |f|$ As Evan pointed out, uniform integrability implies $\sup \int_{E_0} |f_n| < \infty$ as $m(E_0) < \infty$, and then Fatou's Lemma implies $\int_{E_0} |f| < \int_{E_0} |f_n| < \infty$, completing the proof.
As a remark, if $m(E) < \infty$, then any family of functions is obviously tight; this is why the pointwise limit of uniformly integrable functions is guarenteed to be integrable if their domain has finite measure.
In response to your second question: As has been mentioned above, the pointwise limit of integrable functions is not necessarily integrable on any domain. An example of this with a domain of finite measure is $f_n = n$ on $E = [0,1]$. Then ${f_n} \to f \equiv \infty$ pointwise and clearly $f$ is not integrable even though $\int_E |f_n| = n$ for any $n$.
Citation: I am drawing from Sections 4.6 and 5.1 in Royden and Fitzpatrick, Real Analysis.