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 $\int_E h \leq \liminf \int_E f_n$.

so if we define $h_n = min\{h, f_n\}$, why does $h_n -> h$?

$\endgroup$
  • $\begingroup$ Fix $x \in E$. We have two cases to consider. Case 1: If $h(x) < f(x)$ then since $f_n(x) \to f(x)$ we have $h(x) < f_n(x)$ for sufficiently large $n$, and therefore $h_n(x) = h(x)$ for sufficiently large $n$. Case 2: If $h(x) = f(x)$ then for sufficiently large $n$, both $h(x)$ and $f_n(x)$ are near $h(x) = f(x)$, hence $h_n(x) = \min\{h(x), f_n(x)\}$ is near $h(x) = f(x)$. $\endgroup$ – Bungo May 12 '16 at 14:30
1
$\begingroup$

Let $x \in E$ and $\epsilon >0$. Then there exists $N \in \mathbb{N}$ such that for all $n \geq N$, $|f_n(x)-f(x)|<\epsilon$. Note that $h_n \leq h \leq f$. For a given $n \geq N$, there are two possibilities: either $h_n(x)=h(x)$ or $h_n(x)=f_n(x)$. In the first case, $|h_n(x)-h(x)|=0<\epsilon$. In the second, $f_n(x) =h_n(x) \leq h(x) \leq f(x)$, so $|h_n(x)-h(x)|\leq |f_n(x)-f(x)| < \epsilon$.

Intuitively, $h_n \to h$ because we are truncating $h$ by the $f_n$'s, which approach something larger than $h$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.