let $f \geq 0$ be measurable on $E \subset \mathbb{R^n}$. If for any $\ n \geq 1$ there is a subset $E_n$ of $E$ with $\ m(E - E_n) < \frac{1}{n}$ and if
$\lim\limits_{n \rightarrow 0}\int_{E_n} \! f \ \text{exists}$ then show that f is lebesgue integrable.
What does it mean to say that the limit of the integral exists? I am assuming the
$$\lim\limits_{n \rightarrow 0}\int_{E_n} \! f(x) \, \mathrm{d}x =\liminf\limits_{n \rightarrow 0}\int_{E_n} \! f(x) \ \mathrm{d}x = \limsup\limits_{n \rightarrow 0}\int_{E_n} \! f(x) \, \mathrm{d}x$$
but is it finite?
First, I thought I am going to use the additivity property of the lebesgue integral over $E_n$ and its complement in $E$
But since we were given the limit over $E_n$, I thought about constructing a sequence of sets that converge to $E$.
Let $E_n = \{x \in E: f_n(x) \geq \frac{1}{n} \}$
then $E_n$ is monotone increasing to $E$. Then I can just apply the monotone convergence theorem of Levi and get that
$$\int_E \! f(x) \, \mathrm{d}x = \lim\limits_{n \rightarrow 0}\int_{E_n} \! f(x) \, \mathrm{d}x$$
Since $RHS$ exists, so does $LHS$, and thus $f$ is lebesgue integrable.
Am I missing something? I don't see where we used the fact that $\ m(E - E_n) < \frac{1}{n}$.
Your argument fails because a sequence $(E_n)$ is given to you and you cannot choose it.
It is understood that $\lim\limits_{n \rightarrow \infty}\int_{E_n} \! f \ \text{exists and is finite}$. Otherwise, the result is false.
First consider the case where $ \sum \mu (E\setminus E_n) <\infty$. Here $\mu (\lim \sup (E\setminus E_n)=0$ and so $\int_E f =\int _{E \cap D} f$ where $D$ is the complement of $\lim \sup (E\setminus E_n)$ which is $\lim \inf (E\setminus E_n)^{c}$ . Apply Fatou's Lemma and use the fact that $\int_{E_n} f$ is bounded to finish the proof.
For the general case you only have to replace $(E_n)$ by a subsequence.