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Using the folowing definition of uniform integrability(u.i.), I want to show that a set $\Phi$ of u.i. $\mathbb{R}^d$-valued functions is bounded in $L^1$ meaning: $\sup_{f \in \Phi} \int_{\Omega}\lVert f \rVert d\mu<\infty $ .

A nonempty set $\Phi$ of $\mathbb{R}^d$-valued measurable functions on $(\Omega,\mathcal{F})$ is called u.i. if for every $\epsilon >0$ there exists a $\mu-$ integrable function $\omega_\epsilon:\Omega \mapsto \mathbb{R}_+$ such that:

$\sup_{f \in \Phi} \int_{\{\lVert f \rVert >\omega_\epsilon\}}\lVert f \rVert d\mu<\epsilon $

As I think my approach is totally wrong i kindly ask for help. Here my thoughts: $\sup_{f \in \Phi} \int_{\Omega}\lVert f \rVert d\mu =\sup_{f \in \Phi} \int_{\lim_{n\to\infty}\cup_{k=1}^n\{\lVert f \rVert >\omega_k\}}\lVert f \rVert d\mu \\ = \sup_{f \in \Phi} \lim_{n\to\infty} \int_{\cup_{k=1}^n\{\lVert f \rVert >\omega_k\}}\lVert f \rVert d\mu \\ \leq \lim_{n\to\infty} \sup_{f \in \Phi} \int_{\cup_{k=1}^n\{\lVert f \rVert >\omega_k\}}\lVert f \rVert d\mu \\ \leq \lim_{n\to\infty} \sum_{k=1}^n\sup_{f \in \Phi} \int_{\{\lVert f \rVert >\omega_k\}}\lVert f \rVert d\mu \\ \leq \sum_{k=1}^\infty k = \infty $

Where for the second equality I did pull out the Limes using dominated convergence (I hope doing this I do not use whate i want wo show)

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  • $\begingroup$ Big-and-small: $\|f\|\le \omega_\epsilon+\|f\|1_{\{\|f\|>\omega_\epsilon\}}$, and so $\int_\Omega\|f\|\,d\mu\le \int_\Omega\omega_\epsilon+\epsilon$ for each $f\in\Phi$. $\endgroup$ – John Dawkins Oct 23 '16 at 16:00
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    $\begingroup$ Using $$\|f\|\leqslant\|f\|\cdot\mathbf 1_{\|f\|>\omega_1}+\omega_1$$ one gets, for every $f$ in $\Phi$, $$\int_\Omega\|f\|d\mu\leqslant\int_{\|f\|>\omega_1}\|f\|d\mu+\int_\Omega\omega_1d\mu\leqslant1+\int_\Omega\omega_1d\mu<\infty$$ $\endgroup$ – Did Oct 23 '16 at 16:01
  • $\begingroup$ Thank you Did and John Dawkins. You did show that $\Phi \subset L^1$ but is that equal to: $\sup_{f \in \Phi} \int_{\Omega}\lVert f \rVert d\mu<\infty $? $\endgroup$ – Adsertor Justitia Oct 23 '16 at 16:09
  • $\begingroup$ We each provided a finite upper bound for $\int_\Omega\|f\|\,d\mu$ valid uniformly for $f\in\Phi$. $\endgroup$ – John Dawkins Oct 23 '16 at 16:42
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Integrating the pointwise inequality $$\lVert f\rVert= \lVert f\rVert\mathbf 1\left\{ \lVert f\rVert\gt w_1\right\}+\lVert f\rVert\mathbf 1\left\{ \lVert f\rVert\leqslant w_1\right\}\leqslant \lVert f\rVert\mathbf 1\left\{ \lVert f\rVert\gt w_1\right\}+w_1,$$ we get that for each $f\in\Phi$, $$ \int_\Omega\lVert f\rVert\mathrm d\mu\leqslant \int_\Omega\lVert f\rVert\mathbf 1\left\{ \lVert f\rVert\gt w_1\right\}\mathrm d\mu +\int_\Omega w_1\mathrm d\mu\leqslant 1+\int_\Omega w_1\mathrm d\mu. $$ Since the latter bound is independent of $f$, we get that $$\sup_{f\in\Phi}\int_\Omega\lVert f\rVert\mathrm d\mu\leqslant1+\int_\Omega w_1\mathrm d\mu .$$

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