I need to show that if $|f_n|\leq g$, and $g$ is integrable, then $\{f_n\}$ is uniformly integrable, i.e., $\underset{a\rightarrow\infty}\lim\sup_n\int_{[|f_n|\geq a]} |f_n| d\mu=0$.

Here is how I thought about it: Since $|f_n|\leq g$, and $g$ is integrable, then each $f_n$ is integrable. Thus, for each $n$, $|f_n| I_{[|f_n|\geq a]}\leq |f_n|$ and $|f_n| I_{[|f_n|\geq a]}\rightarrow 0$ almost everywhere, thus, by Lebesgue dominated convergence theorem, we have $\underset{a\rightarrow\infty}\lim\int_{[|f_n|\geq a]} |f_n|d\mu=0$ for each $n$, and hence, $$0=\sup_n \lim_{a\rightarrow\infty}\int_{[|f_n|\geq a]}|f_n|d\mu=\lim_{a\rightarrow\infty}\sup_n\int_{[|f_n|\geq a]}|f_n|d\mu.$$

However, I am not so sure about the last step of interchanging the limit and supremum? Any thoughts?

  • 3
    $\begingroup$ Simply use $$\int_{[|f_n|\geq a]} |f_n| d\mu\le\int_{[g\geq a]} g d\mu.$$ $\endgroup$ – Did Mar 16 '15 at 0:48

It might work to consider that $\sup_n |f_n|$ itself is measurable and bounded by $g$, so it's integrable; and $$\lim_{a \rightarrow \infty} \int_{|f_k| \ge a} \sup_n |f_n| \, \mathrm{d}\mu = 0 \; \; \mathrm{for} \; \mathrm{every} \; k$$ by the argument you gave above. This should be enough.

  • $\begingroup$ And how should I interchange the sup and integration? $\endgroup$ – Meemo Mar 16 '15 at 0:20
  • $\begingroup$ @Meemo You only need the inequality $\sup \int |f_n| \le \int \sup |f_n|$, which is true because $\int |f_k| \le \int \sup |f_n|$ for each $k$. $\endgroup$ – user223729 Mar 16 '15 at 0:22

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