3
$\begingroup$

Let $(X, \Sigma, \mu)$ be a probability space and let $\mathscr{F}$ be a norm-bounded subset of $L^{1}(\mu)$.

We say that $\mathscr{F}$ is equi-integrable if for every $\epsilon>0$ there is some $\delta>0$ such that for any $A \in \Sigma$ with $\mu(A) \leq \delta$ and for all $f \in \mathscr{F}$, $$ \int_{A}|f| d \mu \leq \epsilon $$

Alternatively, being equi-integrable is equivalent to

$$ \lim _{C \rightarrow \infty} \sup _{f \in \mathscr{F}}\int_{\{|f|>C\}}|f| d \mu=0 \quad \quad (1)$$

According to Theorem 4.5.6 in Measure Theory by Bogachev we have

If $f_{n}$ be a sequence in $L^{1}(\mu)$ and for each $A \in \Sigma$ the sequence $\int_{A} f_{n} d \mu$ has a finite limit, then $\left\{f_{n}\right\}$ is bounded in $L^{1}(\mu)$ and is equi-integrable.

I'm confused about the last statement. Consider the sequence $f_n=n I_{[0,\frac{1}{n}]}$. It is norm-bounded and it is not equi-integrable as it does not satisfy (1). However, $\lVert{f_n}\rVert_{L^1}=1$. This implies that $\{f_n\}$ seen as a subset of $(L^{\infty})^*$, is bounded in the norm of $(L^{\infty})^*$. By Banach-Alaoglu it has a convergent subsequence, that converges to a (potentially only finitely additive) measure $\nu\in (L^{\infty})^*$. Therefore, passing to the subsequence, for every $A\in \Sigma$, $$\lim_n\int_A f_n d\mu=\lim_n\int f_n I_A d\mu=\int I_A d\nu=\nu(A).$$ Therefore, the subsequence seems to satisfy the hypothesis in Theorem 4.5.6 in Bogachev.

What am I missing?

Thanks!

$\endgroup$
0
2
$\begingroup$

One defect in your argument is the following: $L^{\infty}$ is not separable and the unit ball of its dual is not metrizable in weak* topology. So you can only say that is a sub-net of $(f_n)$ which converges which is weaker than Bogachev's hypothesis. The statement by Bogachev is a special case of Vitali-Hahn-Saks Theorem. See p. 158 of Linear operators by Dunford and Schwartz, Vol. 1

$\endgroup$
1
  • $\begingroup$ Thanks! Your answer was very helpful. $\endgroup$
    – Condor5
    Jul 27 at 14:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.