# Thread:

1. ## Lebesgue Integral

Let $f \in L_p(X)$,1 $\le p < \infty$ and let $\epsilon > 0$. Show that there exists a set $E_\epsilon \subseteq X$ with $m(E_\epsilon) < \infty$ such that if $F \subseteq X$ and $F \cap E_\epsilon = \phi$,then $\int_F |f|^p dm < \epsilon^p$

I let $E_\epsilon = \{ x \in X : |f(x)| \ge \delta_\epsilon \}$,where $\delta_\epsilon > 0$ and $F = \{ x \in X : |f(x)| < \delta_\epsilon \}$.

$\int_F |f|^P dm < \int_F (\delta_\epsilon)^p dm = (\delta_\epsilon)^p m(F)$
I am not sure whether my construction of the sets are correct because I can not make any conclusion regarding $m(F)$.
Can anyone comment on this?

2. Originally Posted by problem
Let $f \in L_p(X)$,1 $\le p < \infty$ and let $\epsilon > 0$. Show that there exists a set $E_\epsilon \subseteq X$ with $m(E_\epsilon) < \infty$ such that if $F \subseteq X$ and $F \cap E_\epsilon = \phi$,then $\int_F |f|^p dm < \epsilon^p$

I let $E_\epsilon = \{ x \in X : |f(x)| \ge \delta_\epsilon \}$,where $\delta_\epsilon > 0$ and $F = \{ x \in X : |f(x)| < \delta_\epsilon \}$.

$\int_F |f|^P dm < \int_F (\delta_\epsilon)^p dm = (\delta_\epsilon)^p m(F)$
I am not sure whether my construction of the sets are correct because I can not make any conclusion regarding $m(F)$.
Can anyone comment on this?
I think you have to go right back to the definition of the Lebesgue integral to do this properly. For a positive function such as $|f|^p$, the usual way to define its integral (as described here, for example) is that it is the supremum of the integrals of nonnegative simple functions majorised by $|f|^p$. If the integral $\int_X|f|^pdm$ is finite (as it is if $f \in L_p(X)$) then each of these simple functions must have finite integral and therefore finite support. We can find a simple function s such that $s\leqslant|f|^p$ and $\int_Xs\,dm>\int_X|f|^pdm - \varepsilon^p$. Now take $E_\varepsilon$ to be the support of s. You should find that this does the required job.
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