$\lim_{n\to\infty} ||f_n-f||_p=0 $ prove that for every $\epsilon>0$ there exists a $\delta$ with $\mu(A)<\delta $-> $\int_A|f_n|^p d\mu <\epsilon$

Question

$(S,A,\mu)$ is a measure space

Let $p\in [1,\infty), f \in L^p(S), (f_n)_{n\in N} \subset L^p(S)$ and $\lim_{n\to\infty} ||f_n-f||_p=0 $

prove that for every $\epsilon>0$ there exists a $\delta>0$ with $\mu(A)<\delta \Rightarrow \int_A|f_n|^p d\mu <\epsilon$

............ My first ideas:

$0 \leftarrow_{n\to \infty} ||f_n-f||_p\ge||f_n||_p-||f||_p$

So i have: $\lim_{n\to \infty} ||f_n||_p =||f||_p $

Would be happy about hints where I add $\epsilon$ and $\delta$

If the $\delta$, $\varepsilon$ inequality should hold uniformly for all $n$ this looks like a form of uniform integrability. And there seems to be a proof here: https://math.stackexchange.com/questions/1430329/if-f-n-to-f-in-l1-show-that-for-every-epsilon-0-there-exists-delt?rq=1 — hgmath, Nov 23 '23 at 16:40
Does this answer your question? For every $\epsilon>0$ there exists $\delta>0$ such that $\int_A|f(x)|\mu(dx) < \epsilon$ whenever $\mu(A) < \delta$ — Mittens, Nov 23 '23 at 21:32

score 0 · Answer 1 · answered Nov 23 '23 at 16:57

By Minkowski's inequality you get that

$$ \Vert f_n\cdot\mathbf{1}_A \Vert_p \leq \Vert (f_n-f)\cdot\mathbf{1}_A \Vert_p + \Vert f\cdot \mathbf{1}_A \Vert_p. $$

Using that $\{ f_n^p \}_{n=1}^\infty\subseteq L^1(S)$ and $f^p\in L^1(S)$, you can essentially use this answer.

I am implicitly assuming that you know how to show that if $\Vert g\Vert_p<\infty$, then there is a $\delta_\epsilon>0$ for any $\epsilon>0$, such that $\int_A \vert g\vert^p d\mu<\epsilon$ whenever $\mu(A)<\delta_\epsilon$. You can show this using Hölder's inequality on $g$ and $\mathbf{1}_A$.

$\lim_{n\to\infty} ||f_n-f||_p=0 $ prove that for every $\epsilon>0$ there exists a $\delta$ with $\mu(A)<\delta $-> $\int_A|f_n|^p d\mu <\epsilon$

1 Answers1