Given $(X,\mathcal{A}, \mu)$ a measurable space. Let $ p \in ]1,\infty[$, and $(f_{n})_{n}\subseteq L^{p}(X,\mu)$ so that $\sup_{n}||f_{n}||_{p}<\infty$ and $f_{n}\to 0$, $n \to \infty$ $\mu-$a.e.
i) Prove $ \forall \epsilon > 0$ and $n \in \mathbb N$, $X = \{f_{n} \leq \epsilon |g|\} \cup \{|f_{n}|\geq \epsilon^{2}\}\cup\{|g|\leq \epsilon\}$
ii) Show $\lim_{n\to \infty}\int_{X}|f_{n}|^{p-1}|g|d\mu = 0 = \lim_{n \to \infty}\int_{X}|g|^{p-1}|f_{n}|d\mu$
My ideas:
i) The inclusion $\{f_{n} \leq \epsilon |g|\} \cup \{|f_{n}|\geq \epsilon^{2}\}\cup\{|g|\leq \epsilon\} \subseteq X$ is trivial. I honestly do not know where to begin on $X \subseteq \{f_{n} \leq \epsilon |g|\} \cup \{|f_{n}|\geq \epsilon^{2}\}\cup\{|g|\leq \epsilon\}$.
ii) I believe that i) is then used to split up $g$, and then hopefully get $\lim_{\epsilon \to 0}\lim_{n\to\infty}(\int_{X}|f_{n}|^{p-1}|g|\chi_{\{|f_{n}|\leq\epsilon\}}d\mu+\int_{X}|f_{n}|^{p-1}|g|\chi_{\{|f_{n}|\geq\epsilon^2\}}d\mu+\int_{X}|f_{n}|^{p-1}|g|\chi_{\{|g|\leq\epsilon\}}d\mu)=0$
Any guidance is greatly appreciated.
Hypothesis on $g$ is missing. The result is true if $g \in L^{p}$. For i) just observe that if $|f_n| <\epsilon ^{2}$ and $|g|>\epsilon$ then $|f_n| <\epsilon ^{2} <\epsilon |g|$. Can you finish the proof of i) now?
I will prove the first part of 2) and the second part is very similar. So let us show that $\int_A |f_n|^{p-1} |g|d\mu $ can be made small (by suitable choice of $\epsilon$) if $A$ is one of the sets $A_1=\{|f_n| \leq \epsilon |g|\}, A_2=\{|f_n| \geq \epsilon ^{2}\}$ and $A_3=\{|g|\leq \epsilon\}$.
For $A=A_1$: use DCT. Since the integrand is dominated by $\epsilon |g|^{p}$ which is integrable, the integral tends to $0$.
For $A=A_2$: apply Holder's inequality (with conjugate exponents $\frac p {p-1}$ and $p$) to see that the integral is less than or equal to $\|f_n\|_p(\int_{A_2} |g|^{p}d\mu)$. Note that $\|f_n\|$ is bounded and $\mu(A_2) \to 0$ as $n \to \infty$.
Finally for $A=A_3$ use a similar argument and note that $\int |g|^{p} I_{\{|g| \leq \epsilon\}} \to 0$ as $\epsilon \to 0$.