Exercise 1.5.11 (Uniform integrability on finite measure spaces)

Question

Let $(X,\mathcal{B},\mu)$ be a finite measure space, and let $\left(fight)_{n\in\mathbb{N}}$ be a sequence of measurable functions from $X$ to $\mathbb{C}$. Show that $\left(fight)_{n\in\mathbb{N}}$ is uniformly integrable if and only if $\sup_{n\in\mathbb{N}}\int_{\{x\in X: |f_n(x)|\geq M\}}|f_n|\ \mathrm{d}\mu 	o 0$ as $M	o\infty$.

Elu Thingol · Accepted Answer

$
ewcommand{\intx}[1]{\int_{X} #1 \mathop{}\!\mathrm{d}{\mu}}$
The ``only if'' direction is obvious and follows directly from the definition. For the ``if'' direction, assume that
$$\lim_{M	o\infty} \sup_{n\in\mathbb{N}}\int_{\{x\in X: |f_n(x)|\geq M\}}|f_n|\ \mathrm{d}\mu = 0.$$
\begin{enumerate}[label=(oman*)]
\item By additivity of the integral, we can write
$$ \sup_{n\in\mathbb{N}}\intx{|f_n|} =  \sup_{n\in\mathbb{N}}\intx{|f_n|\cdot 1_{\{x\in X: |f_n(x)|\leq M}} + \sup_{n\in\mathbb{N}}\intx{|f_n|\cdot 1_{\{x\in X: |f_n(x)|> M} }\leq \mu(X)\cdot M + \epsilon < \infty$$
where we have chosen $M\in \mathbb{R}$ to match the $\epsilon>0$ in the theorem assumption.
\item Follows directly.
\item Follows from the monotonicity of the integral
$$\lim_{\delta 	o 0} \sup_{n\in\mathbb{N}}\intx{|f_n|\cdot 1_{\{x\in X: |f_n(x)|\leq \delta\}}} \leq \lim_{\delta 	o 0} \mu(X) \cdot \delta = 0.$$
\end{enumerate}

Exercise 1.5.11 (Uniform integrability on finite measure spaces)

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Exercise 1.5.11 (Uniform integrability on finite measure spaces)

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