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To be specific, if we have a set of $\{f_n\}$ measurable functions, uniform integrable:

For any $\epsilon>0$, we can find a $\delta$, such that for any set E with $\mu\{E\}<\delta$, $|\int f_nd\mu|<\epsilon$.

I want to know how we can control $\int|f_n|d\mu$ from that. There is a similar question asked here. However that's just for the $f_n$ real case, where we can seperate $f$ to $f^+$ and $f^-$ on two measurable sets. I wonder how can we deal with the complex case?

My idea is as following:

$\int|f|d\mu=\int\sqrt{(Ref)^2+(Imf)^2}d\mu\le \sqrt{\int((Ref)^2+(Imf)^2)d\mu)}\sqrt{\mu(E)}$

While $\mu(E)$ is a finite constant, $\int((Ref)^2+(Imf)^2)d\mu=\int(Ref)^2d\mu+\int(Imf)^2d\mu$.

However from the uniform integrability we can only deduce that $\int(Ref)d\mu$ and $\int(Imf)d\mu$ are controlled, while for a finite measurable set $E$ we cannot get from $L^1$ to $L^2$ ($L^2 \subset L^1$) and this is where I'm stuck at.

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Vadupleix
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1 Answers1

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Write $f=u+iv$ with $u,v$ real valued. The reduction of complex case to the real is based on $$ \left|\int_E u\,d\mu \right| = \left|\operatorname{Re}\int_E f\,d\mu \right|\le \left|\int_E f\,d\mu \right| $$ and similarly $$ \left|\int_E v\,d\mu \right| = \left|\operatorname{Im}\int_E f\,d\mu \right|\le \left|\int_E f\,d\mu \right| $$ So, if $\{f_n\}$ is uniformly integrable, the corresponding families $\{u_n\}$ and $\{v_n\}$ are uniformly integrable. This gives you uniform $L^1$ bounds for $u_n$ and $v_n$. Finally, the triangle inequality yields $$\int_E |f|\,d\mu = \int_E (|u+iv|)\,d\mu\le \int_E |u|\,d\mu+\int_E |v|\,d\mu$$