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A sequence {$f_n$} of measurable functions is called uniformly integrable if

$$\lim_{M \to \infty} \sup_{n} \int_{[|f| >= M]} |f_n|\ \mathrm{d}\mu = 0$$

Show that if $|f_n| \leq g$ for all $n$ and $g$ is integrable, then $\{f_n\}$ is uniformly integrable.

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What did you try? –  Did Feb 27 '13 at 21:41
    
@did Are you asking yourself? –  no identity Feb 27 '13 at 21:44

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up vote 2 down vote accepted

Hint: $|f_n|\chi_{\{|f_n|\geqslant M\}}\leqslant g\chi_{\{g\geqslant M\}}$, so the problem reduces to show that $\lim_{M\to \infty}\int g\chi_{\{g\geqslant M\}}=0$. Use monotone convergence theorem.

The first claim of the hint follows from $\{|f_n|\geqslant M\}\subset \{g\geqslant M\}$. We can show, approximating $g$ by a simple functions, that for each $\varepsilon >0$, we can find $\delta>0$ such that for all measurable $A$ satisfying $\mu(A)<\delta$, then $\int_Agd\mu<\varepsilon$. To conclude, we have that $\mu\{g\geqslant M\}\leqslant \frac 1M\int gd\mu$.

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monotone convergence theorem or dominant convergence theorem? –  Cortizol Feb 27 '13 at 21:56
    
Monotone convergence theorem is enough. –  Davide Giraudo Feb 27 '13 at 22:03
    
OP says "Show that if $|f_n| \leqslant g$ for all $n$ and $g$ is integrable", so I thought is DCT (measure-theory is relative new for me, so sorry to objections :)) –  Cortizol Feb 27 '13 at 22:08
    
DCT is also valid here (but we can use monotone convergence theorem with the sequence $g_n:=g\chi_{\{g<n\}}$). –  Davide Giraudo Feb 27 '13 at 22:09
    
@Davide could you post a solution to this problem so I have time to study it for my exam which is tomorrow? Thanks if you get this in time. –  tyur43 Feb 28 '13 at 21:57

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