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?



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

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

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