# Sequence of functions uniformly integrable but not bounded

Find a sequence $\{ f_n : E \to \mathbb{R}, m(E) < \infty\}$ of pointwise convergent functions that is uniformly integrable and not bounded by a single integrable function (i.e. there is no integrable function $g$ such that $|f_n| \leq g$ for all $n$). Show your Claim. Show that, for each example that one can find in the above problem, the limit function is integrable.

I was thinking $f_n =\frac{1}{n} \chi_{[0,n]}$ but I am not sure, anyone can help? Thank.

• Switch $n$ and $1/n$ and you get a nice example on $E=[0,1]$. – Giuseppe Negro Nov 9 '15 at 19:38
• @GiuseppeNegro this will not work as such $f_n$ will not be uniformly integrable. But slightly modification will do. – Zhanxiong Nov 9 '15 at 19:47

Check $E = (0, 1]$, $f_n(x) = \frac{n}{\log n} I_{(0, n^{-1})}(x), n = 1, 2, \ldots$. And $m$ is the Lebesgue measure on $E$.
You are on the right track, but an even simpler, more degenerate case can happen, an example that is a nice counterexample to a lot of things in analysis: set $f_n=\chi_{[n,n+1]}$. Then $\{f_n\}$ is obviously uniformly integrable, converges pointwise a.e. to $0$, and the "best" function that dominates is $g\equiv1$, which is of course not integrable.
• This example has problems. The usual definition of "uniformly integrable sequence" includes the following condition: for any $\epsilon>0$ there must exist a set $G_\epsilon$ of finite measure and such that $\left\lvert \int_{G_\epsilon^c} f_n\right\rvert \le \epsilon$ uniformly in $n$. Your example does not comply. EDIT: Moreover, the OP requires $m(E)<\infty$, $E$ being the domain of $f_n$. – Giuseppe Negro Nov 9 '15 at 19:34
• That condition is needed for Vitali's convergence theorem. en.wikipedia.org/wiki/Vitali_convergence_theorem Note that the sequence you provided does NOT converge in $L^1$ sense. – Giuseppe Negro Nov 9 '15 at 19:35
• Pick $\epsilon>0$ and set $G_\epsilon$ any set with $|G_\epsilon|\leq\epsilon$. Then $\int_{G_\epsilon}f_n\leq|G_\epsilon|\leq\epsilon$ – charlestoncrabb Nov 9 '15 at 19:36
• Sorry it was the complementary. Edited. Here $G^c$ means $\mathbb{R} \setminus G$. – Giuseppe Negro Nov 9 '15 at 19:37
• Is this definition not the same? en.wikipedia.org/wiki/Uniform_integrability, also Vitali's theorem requires $\mu(X)<\infty$? – charlestoncrabb Nov 9 '15 at 19:40