integration - If ${f_n}$ uniformly integral, then ${ f_n - f }$ uniformly integrable?

Let $\{f_n\}$ be uniformly integrable, on space $(\Omega, F, P)$. $f_n \rightarrow f$ in measure.

The notes I was reading then stated two facts i do not follow:

• By Fatou's $\int_\Omega |f| dP \le \sup_n \int_\Omega |f_n | dP.$

we should have "$\liminf |f_n|$" on the LHS. How is this implied?

• $\{f_n -f\}$ is uniformly integrable.

I tried to show $\sup_n \int_{|f_n-f|>N} |f_n-f| dP$ can be bounded but could not split the integral into easier pieces. How does this follow?

Answer

If $f_n \to f$ in measure, then there is a subsequence $\{f_{n_k}\}$ with $f_{n_k}(x) \to f(x)$ almost everywhere. Fatou's lemma gives you

$$\int_\Omega |f| \, dP = \int_\Omega \lim_{k \to \infty} |f_{n_k}| \, dP \le \lim_{k \to \infty} \int_\Omega |f_{n_k}| \, dP \le \sup_n \int_\Omega |f_n| \, dP.$$

Uniform integrability means two things:

• $\displaystyle \sup_n \int_\Omega |f_n| \, dP < \infty$, and




• $\forall \epsilon > 0 \ \exists \delta > 0$ $P(A) < \delta \implies \displaystyle \sup_n \int_A |f_n| \, dP < \epsilon$.

The remark about Fatou tells you that $\displaystyle \int_\Omega |f| \, dP < \infty$. In particular, for any $\epsilon > 0$ there exists $\delta > 0$ with the property that $P(A) < \delta$ implies $\displaystyle \int_A |f| \, dP < \epsilon.$

Now work with $\epsilon$ and $\delta$ and the fact that $$\int_A |f_n - f| \, dP \le \int_A |f_n| \, dP + \int_A |f| \, dP$$ to show $\{|f_n - f|\}$ is uniformly integrable.