measure theory - If $mu(A)

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|

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.

Answer

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|Then try to make the right side as small as possible, by choosing $t$ and $\delta$ appropriately.