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?


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.


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.