We have the following example : for all $n\in \mathbb{N}$ and $x\in [0,1[$ we define : $f_n(x)=x^n$.
We know that $\forall x \in [0,1[$, $f_n$ is pointwise convergent to $f\equiv 0$. However its not uniformly convergent.
Now, I was wondering if we can apply Egorov's theorem to find a measurable set big enough where $f_n$ is uniformly convergent ?
Thanks in advance !
You don't have to be outrageously creative here. For $0<\delta <1$, we find $$|f_{n}(x)| \leq \delta^{n} \to 0, \quad x \in [0,\delta].$$ I also wanted to point out that there is an issue with your argumentation: on $[0,1)$, the limiting function of the sequence $(f_n)$ is indeed continuous (while it isn't on $[0,1]$). However, the convergence is still not uniform as $\sup_{x\in[0,1)} |f_n(x)| = \sup_{x\in[0,1]} |f_n(x)|$.