Show that $\text{ sup}_n \int{|f_n|^pdm<C}$ implies $\int_{E}{|f_n|dm}<\epsilon \text{ for large enough } n$.

Question

My teacher claimed the following, without further explanation:

Let $\{f_n \}$ be a sequence of integrable functions on $\Bbb R$ with Lebesgue measure $m. \text{ Meaning, } \{f_n\}\subset L^1 \bigl( \Bbb R,m \bigl)$.

Moreover, let $p>1$ such that: $\text{ sup}_n \int{|f_n|^pdm<C}$ when $C>0$ is some constant.

Then, for every $\epsilon>0$ there exit $\delta>0$ such that for every Lebesgue measurable $E$ with $m(E)<\delta$, we have:

$\text{sup}_n\int{|f_n|dm}<\epsilon$.

Is it true? If so, why is it true? I just cannot connect the dots.

Thanks!

Answer 1

That comes for instance from Hölder's inequality:

$$\int_E |f| \ d \mu = \int_\Omega \mathbf{1}_E |f| \ d \mu \leq \|f\|_{\mathbb{L}^p (\Omega)} \|\mathbf{1}_E \|_{\mathbb{L}^{p^*} (\Omega)} = \|f\|_{\mathbb{L}^p (\Omega)} \mu(E)^{\frac{1}{p^*}} = \|f\|_{\mathbb{L}^p (\Omega)} \mu(E)^{1-\frac{1}{p}}.$$

So, if $\sup_n \int_\Omega |f_n|^p \ d \mu \leq C$ and $\mu(E) \leq \delta$, then

$$\sup_n \int_E |f_n| \ d \mu \leq C^{\frac{1}{p}} \delta^{1-\frac{1}{p}}.$$

Note that this holds for all $n$, and does not require the functions $f_n$ to be integrable.