Suppose $f_n$ is a uniformly integrable sequence of functions defined on $[0, 1]$. How do I see that there is a subsequence $n_j$ where $\int_0^1 f_{n_j}g\,dx$ converges when $g$ is a real-valued bounded measurable function?
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It suffices to show the sequence of real numbers is bounded, i.e. \begin{align} \left| \int^1_0 f_n g\ dx \right|\leq M<\infty. \end{align} Since $\{f_n\}$ is a uniformly integrable family of functions, then there exists a $\delta>0$ such that \begin{align} \int_A |f_n|< 1 \end{align} whenever $m(A)< \delta$ for all $n \in \mathbb{N}$. In particular, if we sub-divide the interval $[0, 1]$ into intervals with length less than $\delta$, i.e. $[0, 1] = \bigcup^N_{k=1} I_k$ where $m(I_k)<\delta$ \begin{align} \int^1_0|f_n| \leq \sum^N_{k=1}\int_{I_k} |f_n| <N \end{align} for all $n \in \mathbb{N}$. Finally, using the fact that $g \in L^\infty[0, 1]$, we get \begin{align} \left|\int^1_0 f_n g\ dx \right| \leq ||g||_\infty \int^1_0 |f_n|\ dx < ||g||_\infty N \end{align} for all $n \in \mathbb{N}$.
Jacky Chong
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My interpretation is that the subsequence should be independent of $g.$ – zhw. Oct 02 '16 at 05:15
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@zhw I thought since $L^\infty[0, 1]= (L^1[0, 1])^\ast$, then to say there exists one subsequence $f_{n_k}$ such that $g(f_{n_k})$ converges for all continuous linear functional $g$ seems very strange. – Jacky Chong Oct 02 '16 at 05:30
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@zhw I could be very wrong. – Jacky Chong Oct 02 '16 at 05:32
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@JackyChong it is true and is called the Dunford Pettis theorem – algebroo Jul 17 '24 at 06:52