# Show that $\|f \chi_F\|_p < \epsilon$

Let $f \in L_p(X, \mathcal{A}, \mu)$, $1 \leq p < \infty$, and let $\epsilon > 0$. Show that there exists a set $E_\epsilon \in \mathcal{A}$ with $\mu(E_\epsilon) < \infty$ such that if $F \in \mathcal{A}$ and $F \cap E_\epsilon = \emptyset$, then $\|f \chi_F\|_p < \epsilon$.

I was wondering if I could do the following:

$$\int_X |f|^p = \int_{X-E_{\epsilon}}|f|^p + \int_{E_\epsilon}|f|^p \geq \int_{F}|f|^p + \int_{E_\epsilon}|f|^p$$

$$\int_{X}|f\chi_F|^p=\int_{F}|f|^p \leq \int_X |f|^p- \int_{E_\epsilon}|f|^p < \epsilon$$

Please just point out any fallacies within my logic as I have found this chapter's exercise to be quite difficult, so any help is appreciated.

• OK, so how do you get $\int_X |f|^p - \int_{E_{\epsilon}} |f|^p = \epsilon$ (actually you want it to be $<\epsilon$)? Commented May 6, 2012 at 7:03
• Oh yea typo there thanks but yea I mean that's what I want I would still want to show that. Commented May 6, 2012 at 7:08

Your estimates are fine since everything is finite, now you must exhibit the set $E_\epsilon$. For this consider $E_{\frac{1}{n}} = \{ x\in X : |f(x)|^p \geq \frac{1}{n} \}$, now prove that $g_n=|f\chi_{E_{\frac{1}{n}}}|^p$ gives a monotone increasing sequence, with $|g_n|\leq |f|^p$ and $g_n(x)\to |f(x)|^p$ everywhere in $X$
• Yes, I love this answer especially because I did a very similar problem in $L_1$ where I defined my $E_n$ almost analogous to yours. Thanks! Commented May 6, 2012 at 17:40