Suppose $f_1, f_2, \dotsc$ is a collection of measurable functions which satisfy $\sup_n \int |f_n|^{1 + \gamma}\ d\mu< \infty$ for some $\gamma > 0$, and $\mu$ is a finite measure. I am being asked to show that the $\{f_n\}$ are uniformly absolutely continuous. That is, for each $\epsilon > 0$ there exists a $\delta > 0$ such that for all $A$ with $\mu(A) < \delta$ $$\int_{A} |f_n|\ d\mu < \epsilon$$ for all $n \geq 1$.
I'm not sure how to use the fact that $\sup_n \int |f_n|^{1 + \gamma}\ d\mu< \infty$. If I assume only that $\sup_n \int |f_n|\ d\mu< \infty$, it seems I can prove something weaker:
Suppose $A_1, A_2, \cdots$ is any sequence of sets for which $\mu(A_n) \to 0$ as $n \to \infty$, and let $m \geq 1$. Since $f_m \cdot 1_{A_n} \to 0$ a.e. and $|f_m \cdot 1_{A_n}| \leq |f_m|$ and $\int |f_m| < \infty$, the dominated convergence theorem implies that $\int_{A_n} |f_m| \to 0$ as $n \to \infty$.
But I'm not sure where to go from here. I think this can be done by first proving that the $f_n$ are uniformly integrable, but I would like to find a direct proof if possible.
Best Answer
Let $p=1+\gamma$. Then $\frac1q=\frac{p-1}{p}=\frac{\gamma}{1+\gamma}$ so $q=\frac{1+\gamma}{\gamma}$. By Holder's inequality,
$\int_A |f_n|\,d\mu \leq \|f_n\|_p\|1\|_q=\left[\int_A |f_n|^{1+\gamma}\,d\mu\right]^{\frac{1}{1+\gamma}}\left[\mu(A)\right]^{\frac{\gamma}{1+\gamma}}\leq \left[\sup_n\int_X |f_n|^{1+\gamma}\,d\mu\right]^{\frac{1}{1+\gamma}}\left[\mu(A)\right]^{\frac{\gamma}{1+\gamma}} .$ It should be clear how to pick $\delta(\epsilon)$ so that the above is bounded above by $\epsilon$.