$\mu(E_{\epsilon})<+\infty$ such that if $F \cap E_{\epsilon} = \emptyset$, then $\left \| f \chi_{F} \right \|_{p} < \epsilon$.

Question

Let $f \in L_{p}(X,\Sigma,\mu)$, $1 \leq p < \infty$ and $\epsilon>0$. There exists a set $E_{\epsilon} \in \Sigma$ with $\mu(E_{\epsilon})<+\infty$ sucht that if $F \cap E_{\epsilon} = \emptyset$, then $\left \| f \chi_{F} \right \|_{p} < \epsilon$.

I tried to construct the set $E_{\epsilon}=\left \{ X \setminus A : A \in \Sigma, \left \| f \chi_{A} \right \|_{p} < \epsilon \right \}$, then it satisfies the above property, but wasn't able to show that this set has finite measure. Actually, $E_{\epsilon}$ defined as above may not even belong to the $\sigma$-algebra $\Sigma$ when there is an uncountable quantity of sets $A$.

This is the exercise 6.P in "Elements of Integration and Lebesgue Measure", R Bartle.

Answer 1

Let $E_n = \{x \in X : |f(x)| > n\}$. Then $\cap_{n \ge 1} E_n = \emptyset$, i.e. $1_{E_n} \to 0$ pointwise, so $\int |f|1_{E_n} \to 0$ in $L^p$, by the dominated convergence theorem. Therefore, there is some $n$ with $\int |f|^p1_{E_n} \le \epsilon^p$. Let $E_\epsilon = E_n^c = \{x \in X : |f(x)| \le n\}$. Then $\mu(E_\epsilon) < \infty$ (since $f \in L^p$), and for any $F \in \Sigma$ with $F\cap E_\epsilon = \emptyset$, $F \subseteq E_n$ and thus $||f\chi_F||_p = (\int |f|^p\chi_F)^{1/p} \le (\int |f|^p \chi_{E_n})^{1/p} \le \epsilon$, as desired.

Answer 2

Let $E_n=\{x\in X: |f(x)|\geq \frac{1}{n}\}$ and observe that: $$\infty > \int_{E_n} |f|^p d\mu \geq \frac{1}{n} \mu (E_n) $$ thus $\mu(E_n) < \infty$ for each $n\in\mathbb{N}$. We show that: $$\lim_{n\rightarrow\infty} \int_{E_n} |f|^p d\mu = \int |f|^p d\mu $$ We have that $|f|^p\chi_{E_n}\leq|f|^p$ for each $n\in\mathbb{N}$ and, also, $|f|^p\chi_{E_n}\rightarrow|f|^p$. Thus, by the dominated convergence theorem we get the above expression. Finally, given $\epsilon>0$, there exists $N\in\mathbb{N}$ such that: $$\epsilon^p + \int_{E_N} |f|^p d\mu > \int |f|^p d\mu $$ by taking $E_{\epsilon}=E_N$, if $F\cap E_{\epsilon}=\emptyset$, then: $$\epsilon^p + \int_{E_\epsilon} |f|^p d\mu \geq \int |f|^p d\mu \geq \int_{E_{\epsilon}\cup F} |f|^p d\mu = \int_F |f|^p d\mu + \int_{E_{\epsilon}} |f|^p d\mu \implies$$ $$\epsilon^p > \int_F |f|^p d\mu$$