# Can anyone help me understand one step on the proof of Fatou's lemma?

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

The book (Royden) 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 $$h_n \to h$$?

• 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 $h(x) = f(x)$.
– user169852
Commented May 12, 2016 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 $h$.