I wish to prove or disprove that the sequence of functions $f_n=\chi_{[n,n+1]}$ is uniformly integrable?

At a glance my judgement is YES, it is uniformly integrable.

From the definition of Uniform integrability, that's

A sequence ${f_n}$ is called uniformly integrable if $\forall \epsilon >0 \exists \delta > 0 $ such that if $E \subset X$, $E$ measurable and $\mu (E)< \delta $ then $\forall n$ $\int_E |f_n| d\mu < \epsilon$.

So I let $E \subset R$ such that $\mu (E)<\delta$ then $\int_E|f_n|=\int|f_n|\chi_E \leq \mu (E)<\delta$.

So in this case $\epsilon =\delta$.

Does this make sense?

  • $\begingroup$ I think your answer is fine! $\endgroup$ – астон вілла олоф мэллбэрг Feb 16 '17 at 3:37
  • $\begingroup$ Thank you! Will you have done it any different? $\endgroup$ – Cnine Feb 16 '17 at 3:39
  • $\begingroup$ No. I would have done it the same way. $\endgroup$ – астон вілла олоф мэллбэрг Feb 16 '17 at 3:41
  • $\begingroup$ oh okay, I was also considering $E=[k,k+\frac{1}{k^2}]$ for some natural number, then $\mu (E)=\frac{1}{k^2}<\delta$ THEN $\int_E |f_n|=\int_{[n,n+1]}|f_n|\chi_{E} \leq \mu (E)< \delta$. so $\epsilon=\delta$, and its the same idea though. $\endgroup$ – Cnine Feb 16 '17 at 3:52
  • $\begingroup$ Of course, that too would work. $\endgroup$ – астон вілла олоф мэллбэрг Feb 16 '17 at 3:53

The approach is correct with this definition of uniform integrability.

A concern about this one is that a uniform integrable family with this definition may not contain an integrable function. For example, the family consisting of the constant function equal to $1$ would be uniformly integrable.

A common definition is that a family $\mathcal F$ of function on a measure space $\left(X,\mathcal A,\mu\right)$ is uniformly integrable if for each positive $\varepsilon$, there exists an integrable function $g$ such that $$ \sup_{f\in\mathcal F}\int_{\left\{\left\lvert f\right\rvert\gt g\right\}}\left\lvert f\right\rvert\mathrm d\lambda\lt \varepsilon. $$ With this definition, the family $(f_n)$ defined by $f_n=\chi_{[n,n+1]}$ is not uniformly integrable. Indeed, let $\varepsilon=1/2$. If a function $g$ is such that $$\sup_{n\geqslant 1}\int_{\left\{\left\lvert f_n\right\rvert\gt g\right\}} f_n\mathrm d\lambda\lt 1/2$$ then $\lambda\left(\{g\lt 1\}\cap [n,n+1]\right)\lt 1/2$ hence $\lambda\left(\{g\geqslant 1\}\cap [n,n+1]\right)\geqslant 1/2$ for all $n$ hence $g$ cannot be 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.