I'm trying to solve exercise 6R from Bartle's "Elements of Integration" that states:
Let $(f_n)$ be a Cauchy sequence in $L_p$. Then there exists a measurable set $E_\epsilon$ with $\mu(E_\epsilon)<\infty$ such that if $F$ is measurable and $F\cap E_\epsilon=\emptyset$ then: $$\int_F|f_n|^pd\mu<\epsilon^p $$ for each $n\in\mathbb{N}$
So far, I tried to exercise 6.P that states:
"if $f\in L_p$, then there exists a measurable set $E_\epsilon$ with $\mu(E_\epsilon)<\infty$ such that if $F$ is measurable and $F\cap E_\epsilon=\emptyset$ then: $\int_F|f|^pd\mu<\epsilon^p $ "
by taking the $E_{n \epsilon}$ for each $f_n$ (I'm also using the fact that I can take $E_{n \epsilon}=\{x\in X: |f_n(x)-f_m(x)|\geq \frac{1}{N_n}\}$ for some $N_n$ as I do on the second comment of $\mu(E_{\epsilon})<+\infty$ such that if $F \cap E_{\epsilon} = \emptyset$, then $\left \| f \chi_{F} \right \|_{p} < \epsilon$., for $f_m$ fixed, or replaced with $f$ such that $f_n\rightarrow f$), and I can find $\{E_{n\epsilon}\}$ such that $\mu(E_{n\epsilon})<\infty$ and if $F\cap E_{n\epsilon}$ then $\int_F|f|^pd\mu<\epsilon^p$.
My next step would be take $E_\epsilon=\bigcup_{n=1}^{\infty}E_{n\epsilon}$ so that if $F\cap E_\epsilon=\emptyset$ then $F\cap E_{n\epsilon}=\emptyset$ for each $n\in\mathbb{N}$ but I cannot find my way of proving that $\mu(E_\epsilon)<\infty$
If anyone could help, it would be much appreciated
Edit: I think I found a way:
If $f\in L_p$ such that $||f_n-f||_p\rightarrow 0$ by Minkowski:
$$\left(\int_E |f_n|^p d\mu \right)^\frac{1}{p} \leq \left(\int_E |f_n-f|^p d\mu \right)^\frac{1}{p} + \left(\int_E |f|^p d\mu\right)^\frac{1}{p} < \epsilon/2 + \left(\int_E |f|^p d\mu\right)^p $$
for $n\geq N$, for some $N\in\mathbb{N}$, for any $E$ measurable. By 6.P, we have that there exists measurable sets $E_0,E_1,...,E_{N-1}$, with $\mu(E_k)<\infty$ such that if $F\cap E_k$: $$\left(\int_F |f|^p d\mu\right)^\frac{1}{p} < \epsilon/2 \hspace{0.5cm} \textrm{for} \hspace{0.25cm} k=0 $$ $$\left(\int_F |f_k|^p d\mu\right)^\frac{1}{p} < \epsilon \hspace{0.5cm} \textrm{for} \hspace{0.25cm} k=1,...,N-1$$ Hence, taking $E_\epsilon=\cup_{k=0}^{N-1}$, if $F\cap E_\epsilon=\emptyset$ then: $$\left(\int_F |f_n|^p d\mu \right)^\frac{1}{p} < \epsilon \hspace{0.5cm} \textrm{for any} \hspace{0.2cm} n \in \mathbb{N}$$