# why $g_n$ is measurable in the proof of Fatou's Lemma

Fatou Lemma: Suppose $\{f_n\}$ is a sequence of measurable functions with $f_n \geq 0$. If $\lim_{n\rightarrow\infty}f_n(x)=f(x)$ for a.e. $x$, then $$\int f \leq \liminf_{n\rightarrow\infty}\int f_n$$

Proof: Suppose $0\leq g \leq f$, where $g$ is bounded and supported on a set $E$ of finite measure. If we set $g_n(x)=\min(g(x),f_n(x))$, then $g_n$ is measurable. etc..

I don't know why $g_n$ is measurable. because the author didn't assume $g$ is measurable. can anyone explain why? thanks very much

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The assumption that $g$ is measurable is implicit but very much necessary here. – Did Jun 14 '13 at 5:52
Just to say that there is proof that don't using that $g$ is measurable. For example: Let $g_n(x)=\inf_{k\ge n} f_k(x)=\inf\{f_n(x),f_{n+1}(x),\ldots\}.$ Then $g_n \le f_n$ for every $n \in \mathbb{N}$ and $g_n \uparrow \displaystyle\liminf_{n\to \infty}f_n=f$. Then using TMC we find $$\int_A f = \int_A \lim_{n \to \infty} g_n = \lim_{n\to \infty} \int_A g_n = \liminf_{n\to \infty} \int_A g_n \le \liminf_{n \to \infty} \int_A f_n$$ To prove that $g_n(x)=\inf_{k\ge n} f_k(x)$ is measurable note that $g_n^{-1}([c,+\infty])=\bigcap_{n=1}^{\infty} \{x \in X: f_n(x)\ge c\}$ – Cortizol Jun 14 '13 at 7:35

$g$ is assumed to be measurable, but the author didn't explicitly say so.