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Any help with the following problem is appreciated.

Given: a sequence of nonnegative functions $(g_n)$ which are U.I. (uniformly integrable) in $\mathcal{L}^1(0,1)$ with $\sup_n \Vert g_n \Vert_1 < \infty$.

Claim: there exist $ g \in \mathcal{L}^1(0,1)$ and a subsequence $(g_{n_k})$ such that $$\forall \, \text{ measurable }E \subset [0,1]\qquad \lim_{k \to \infty} \int_E g_{n_k} =\int_E g .$$

I am having difficulty proving the claim.

Davide Giraudo
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user62089
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2 Answers2

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Here's the sketch I would follow.

  1. Consider the measures $\nu_n$ defined by $\nu_n(E) = \int_E g_n\,dm$. This is a bounded sequence of positive measures, so by Helly's selection theorem (or Banach-Alaoglu or something analogous) we can extract a subsequence $\nu_{n_k}$ converging vaguely to some measure $\nu$, i.e. $\int h g_n\,dm \to \int h\,d\nu$ for every $h \in C_c((0,1))$.

  2. Use the uniform integrability to show that $\nu$ is absolutely continuous to $m$ (the characterization of uniform integrability in terms of uniform absolute continuity would be helpful) and hence by Radon-Nikodym is of the form $\nu(E) = \int g \,dm$ for some $g$.

  3. For a compact set $K$, approximate $1_K$ by continuous compactly supported functions to conclude that $\int_K g_n\,dm \to \int_K g\,dm$.

  4. Consider the set $\mathcal{L}$ of all measurable sets $E$ for which $\int_E g_n\,dm \to \int g\,dm$. Use a monotone class argument to show that $\mathcal{L}$ consists of all measurable sets.

If you use this for a homework problem, please credit me (and give the URL of this answer) in your submission.

Nate Eldredge
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  • Is this a correct approach:
    1. Prove the sequence, $G_n(x) = \int_0^x g_n(t) dt $ for $0 \leq x \leq 1$ is uniform absolute continuous.
    2. Apply uniform boundedness principle to prove $G_n$ is uniformly bounded.
    3. Use Diagonal argument to extract subsequence that converges for all rationals in $[0,1]$.
    4. (3) $\Rightarrow$ Convergence is for all $x \in [0,1]$ and infact uniform
     – user62089 Apr 08 '13 at 19:41
  • @pondy: Sure, in fact that's one way to prove Helly's selection theorem which I mention in step 1. – Nate Eldredge Apr 08 '13 at 21:43
  • The argument does prove existence and convergence to an absolute continuous function $\int g$. Can I then do the following $\vert \int_E (g_{n_k}-g) \vert \leq \int (g_{n_k} - g) \rightarrow 0$ for large enough $n_k$ for any measurable subset $E$ of $[0,1]$, to prove the claim – user62089 Apr 08 '13 at 22:19
  • I solved the problem using Arzela-Ascoli. I was dumb not to see that before. – user62089 Apr 09 '13 at 19:14
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  1. Let $\{E_j,j\in\Bbb N\}$ be an algebra generating the Borel $\sigma$-algebra of the unit interval. By boundedness, we can find a nested sequence of infinite subsets of the set of natural numbers $(I_j,j\in\Bbb N)$ such that $\left(\int_{E_j}h_nd\lambda,n\in I_j\right)$ is convergent.

  2. By a diagonal argument, we can find a subsequence $(h_{n_k},k\in\Bbb N)$ such that for each $j$, the sequence $\left(\int_{E_j}h_{n_k}d\lambda,k\in \Bbb N\right)$ is Cauchy.

  3. By an approximation argument and the assumption of uniform integrability, we can show that for each $E\subset [0,1]$ measurable, the sequence $\left(\int_Eh_{n_k}d\lambda,k\in\Bbb N\right)$ is convergent.

  4. Let $\mu_k(E):=\int_Eh_{n_k}d\lambda$. Then $\mu_k$ is a finite measure. By 3. and this thread, the map $\mu\colon E\mapsto \lim_{k\to +\infty}\mu_k(E)$ is a measure. By boundedness in $L^1$, this is a finite measure. Furthermore, $\mu$ is absolutely continuous with respect to Lebesgue measure. Radon-Nykodym theorem gives $g$.

Community
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Davide Giraudo
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  • Is this a correct approach: 1. Prove the sequence, $G_n(x)=\int_0^x g_n(t)dt$ for $0\le x \le 1$ is uniform absolute continuous. 2. Apply uniform boundedness principle to prove $(G_n)$ is uniformly bounded. 3. Use Diagonal argument to extract subsequence that converges for all rationals in $[0,1]$. 4. (3) $\Rightarrow$ Convergence is for all $x \in [0,1]$ and in fact uniform – user62089 Apr 08 '13 at 19:58
  • This will give you that the limit exists for a sub-sequence for any interval. It's a step to the result, but as it stands it's not enough. – Davide Giraudo Apr 08 '13 at 20:10
  • What is it that is missing? – user62089 Apr 08 '13 at 20:44
  • We need the property to be true for all the measurable subsets of $[0,1]$, not only intervals. And how do you show the existence of $g$? – Davide Giraudo Apr 08 '13 at 20:50
  • (3) $\Rightarrow$ the subsequence converges and that $\int g$ is also absolutely continuous. – user62089 Apr 08 '13 at 21:17