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?
Best Answer
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.