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 $h_n -> h$?

  • $\begingroup$ 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

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 $h$.


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