# Is the sequence of functions $f_n=\chi_{[n,n+1]}$ uniformly integrable?

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?

• I think your answer is fine! – астон вілла олоф мэллбэрг Feb 16 '17 at 3:37
• Thank you! Will you have done it any different? – Cnine Feb 16 '17 at 3:39
• No. I would have done it the same way. – астон вілла олоф мэллбэрг Feb 16 '17 at 3:41
• 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. – Cnine Feb 16 '17 at 3:52
• Of course, that too would work. – астон вілла олоф мэллбэрг Feb 16 '17 at 3:53

## 1 Answer

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.