# On equi-integrability (also called uniform integrability)

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!

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