How do we prove the following set is measurable?

Question

I was reading the proof of Egorov Theorem in the Real Analysis Book of Elias M Stein

Suppose $\{f_k\}$ is a sequence of measurable functions defined on the measurable set $E$ with $m(E)< \infty$ and assuming that $f_k \rightarrow f $ a.e on $E$

Then the set ${E_k}^n=\{x\in E| \hspace{2 mm} |{f_j}(x)-f(x)|<1/n \hspace{1 mm} \forall j>k\} $

The proof assumes that is function is measurable. I have been trying to use the following definition to prove it, but I am stuck

A set $A$ is said to be lebesgue measurable if $\forall \epsilon>0 \hspace{1 mm} \exists$ an open set $O$ such that $A\subset O $ such that $m_*(O-A)<\epsilon$ where $m_*$ is the exterior measure. I even tried thinking in terms of the equivalent definition of measurability using closed set but could not think of a way to prove it. Can you guys help me figure it out?

Answer 1

Here is one definition of a function being measurable. We need that $\{x \mid f(x) > \alpha \} \in \Sigma$ for all $\alpha \in \mathbb{R}$. This implies that the preimage of any interval in $\mathbb{R}$ is measurable. (Do you understand why?)

At the beginning of the proof for this theorem, you can assume without loss of generality that $\lim \limits_{n \to \infty} f_{n}$ exists and equals $f(x)$ for all $x \in E$.

Then, since each $f_{n}$ is measurable, the limit of the sequence, $f$, is measurable.

Finally, if $g$ is measurable so is $-g$. And if $f$, $g$ are measurable, so is $f + g$. Also, if $g$ is measurable, so is $|g|$. Using these properties, it's easy to see that $\{ x \in E \mid |f_{j} - f| < \frac{1}{n} \forall j \geq n \} = \bigcap \limits_{j = n}^{\infty} \{ x \in E \mid |f_{j} - f | < \frac{1}{n} \}$ is measurable, because it is the countable intersection of measurable sets.

Answer 2

The important fact is that all $f_k$ are measurable. Hence, so is $f$. This implies that $|f_j - f|$ is also measurable.

By definition, this means that the inverse image of every interval (alternatively of every Borel subset of $\Bbb{R}$) is measurable.

Hence,

$$ E_k^n = \bigcap_{j>k} |f_j -f|^{-1}([0,1/n)) $$

is measurable as a countable intersection of measurable sets (recall that the class of measurable sets is a $\sigma$-algebra).

EDIT: General comment: Just as you would normally not try to show that e.g.

$$ x \mapsto \sin(e^{x^2} - \sin(x)) $$

is continuous by computing for every $\varepsilon > 0$ some $\delta > 0$ such that ..., it is often best to not used the definition of measurable sets, but to use closure properties of measurable sets only (countable intersections, unions, complements of measurable sets are measurable, measurable functions form a vector space, inverse images of Borel sets under measurable functions are measurable, limits of measurable functions are measurable, ...).