Function between two measurable functions is measurable Let $f$ be a function on $\mathbb{R}^n$. Assume that for any $\epsilon>0$, there exist measurable functions $g, h \in L^1(\mathbb{R}^n)$ such that $g(x) \leq f(x) \leq h(x)$ for all $x \in \mathbb{R}^n$ and $$
\int_{\mathbb{R}^n} (h(x)-g(x))\, dx < \epsilon
$$
Prove that $f$ is measurable on $\mathbb{R}^n$ and $f \in L^1(\mathbb{R}^n)$. 
My first thought was to show that $f(x)$ is a pointwise limit of measurable functions; but since all information we have about $f(x)$ is in $L^1$, it would be hard to consider pointwise limits. 
My problems are: 
(1) I am not sure how to prove $f$ is measurable. To establish measurability of $f(x)$, I need to go through the definition that $\{x: f(x)<c\}$ is measurable for all $c$. Now $\{x: f(x)<c\} = \{x: \exists h \, \text{measurable s.t.}\, f(x)\leq h(x)<c\}$, and the trouble is I am not sure how to express the latter as a countable union of measurable sets. 
(2) I was wondering if the following proof for $f \in L^1$ is correct: Assuming $f$ is measurable, we want to show that $f \in L^1(\mathbb{R}^n)$. For any $n\in \mathbb{N}$, there is $h_n, g_n \in L^1$ with $h_n(x)\leq f(x) \leq g_n(x)$ for all $x$ and $\int |f(x)-h_n(x)|\, dx = \int f(x)-h_n(x)\, dx \leq \int g_n(x)-h_n(x)\, dx < 1/n$. Hence $||f-h_n||_{L^1} < 1/n$, and thus $f \in L^1$ by completeness of $L^1$. 
In summary, I appreciate any help/hint on measurability of $f(x)$, and check on if the proof in (2) is correct. Thank you!  

(1) Measurability of $f(x)$. Thanks to the hint of @Robert Israel
Let $g_n, h_n \in L^1$ be such that $h_n(x) \leq f(x) \leq g_n(x)$ for all $x$ and $\int g_n(x)-h_n(x)\, dx < 1/n$. Let $h(x) = \limsup h_n(x)$ and $g(x) = \liminf g_n(x)$, so $h(x) \leq f(x) \leq g(x)$ for all $x$, and $h, g$ are measurable. Moreover, by Fatou's Lemma, $\int g(x)-h(x)\, dx \leq \liminf \int g_n(x)-h_n(x)\, dx =0$, so $g(x)=h(x)$ a.e. Thus $g(x)=h(x)=f(x)$ a.e., so $f$ is measurable. 
(2) $f \in L^1$. The original proof is correct, and an easier version is pointed out by @Bungo. 
