0
$\begingroup$

Let $\{f_n\} \to f$ pointwise on $E$, then $\int_E f \leq \liminf \int_E f_n$.

The book claims that it suffices to show that if $h$ is any bounded measurable function of finite support for which $0\leq h \leq f$ on E, then $\int_E h \leq \liminf \int_E f_n$. Why is this the case?

Note: this is from page 82 of Royden's real analysis.

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

This is enough because, if $A_f$ is the set of bounded measurable functions with finite support and $0\leq h\leq f$, then $$\int_Ef=\sup_{h\in A_f}\int_Eh.$$

$\endgroup$
1
  • $\begingroup$ so if we define $h_n = min\{h, f_n\}$, why does $h_n -> h$? $\endgroup$
    – user1559897
    Commented May 12, 2016 at 2:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .