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Let $(X,\Sigma,\mu)$ be a finite measure space. Let $\mathcal{F}$ be a family of measurable functions $f:X\to\mathbb{R}$. Prove that if $$\lim_{t\to\infty}\left(\sup_{f\in\mathcal{F}}\int_{\{x\in X:|f(x)|\ge t\}}|f|d\mu \right)=0,$$

then $$\sup_{f\in\mathcal{F}}\int_X|f|d\mu<\infty,$$

and for all $\epsilon >0$ there exists $\delta >0$ such that: $$A\in\Sigma,\mu(A)<\delta\Longrightarrow \sup_{f\in\mathcal{F}}\int_A|f|d\mu<\epsilon.$$

For the first part.

Let $t>0$ such that: $\displaystyle\sup_{f\in\mathcal{F}}\int_{\{x\in X:|f(x)|\ge t\}}|f|d\mu<1$.

Fix $f\in\mathcal{F}$. Then $$\displaystyle\int_X|f|d\mu=\int_{\{|f|\ge t\}}|f|d\mu+\int_{\{|f|<t\}}|f|d\mu\le 1+t\mu(X)<\infty.$$

And $1+t\mu(X)$ does not depend of $f$, so we get $\sup_{f\in\mathcal{F}}\int_X|f|d\mu<\infty$.

Is that correct?

I don't know how to do the second part. Could it be true that $v(A):=\sup_{f\in\mathcal{F}}\int_A|f|d\mu$ is a finite measure? I wanted to try something similar to that known result when $\mathcal{F}$ is just one function (some call it absolutely continuous of the measure $v$, I think).

Any hint? Thank you.

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Your first part is correct.

For the second part, try to bound $\int_A |f|\,d\mu$ similarly to how you bounded $\int_X |f|\,d\mu$ in the first part:

$$\int_A |f|\,d\mu=\int_{A\cap \{|f|<t\}}|f|\,d\mu+\int_{A\cap \{|f|\ge t\}}|f|\,d\mu \le t\mu(A)+\sup_{f\in \mathcal F}\int_{|f|\ge t}|f|\,d\mu$$ Then try to make the right side as small as possible, by choosing $t$ and $\delta$ appropriately.

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  • $\begingroup$ Thank you so much! I couldn't see that, it was a bit simple. $\endgroup$ – JonSK Jan 15 '16 at 1:46
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The condition on the $lim sup$ implies that for every $\epsilon>0$ there exists $t_0$ such that $t>t_0$ implies that $\int_{f\in {\cal F}\{x\in X f(x)\geq t\}}\mid f\mid <\epsilon/4$.

Let $A$ be a subset of $X$, suppose that $\mu(A)<{{\epsilon}\over{4t_0}}$. For every $f\in {\cal F}$, $\int_A\mid f\mid\leq \int_{A\cap\{x\in X f(x)\leq t_0\}}\mid f\mid+ \int_{A\cap\{x\in X f(x)\geq t_0\}}\mid f\mid$.

But $\int_{A\cap\{x\in X f(x)\leq t_0\}}\mid f\mid\leq \mu(A)t_0\leq \epsilon/2$ and $\int_{A\cap\{x\in X f(x)\geq t_0\}}\mid f\mid\leq\epsilon/4$. This implies the result.

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  • $\begingroup$ Thank you! Seems to me that you don't need any $t>t_0$, just one $t_0$ with such property I think. $\endgroup$ – JonSK Jan 15 '16 at 1:46

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