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For each $k$, choose $n_k$ with $\lambda (A\setminus E_{n_k}^{(k)})\le \epsilon 2^{-k-1}$ and set $B={\bigcap }_{k\ge 1}{E}_{n_k}^{(k)}$. Countable subadditivity gives $\lambda (A\setminus B)\le \epsilon /2$. For any tolerance $\delta >0$, choose $k$ with $1/k<\delta$. For all $m\ge n_k$ and every $x\in B$, $|f_m(x)-f(x)|\le 1/k$, proving uniform convergence on $B$.

If the functions are Borel measurable, $B$ itself is Borel and we can take $A_{\epsilon }=B$. If “measurable” is interpreted in the completed Lebesgue sense, use regularity of Lebesgue measure to choose a Borel subset $A_{\epsilon }\subseteq B$ differing from $B$ by a null set; the same measure bound and uniform convergence hold. Thus the requested Borel conclusion is valid under either convention. This proves the finite-measure Egorov theorem.

For an infinite-measure counterexample take $A=[0,\infty )$ and $f_n=1_{[n,\infty )}$. It converges pointwise to zero. Any subset of $A$ with complement of finite measure intersects every infinite tail $[n,\infty )$, so the supremum of $f_n$ there is one for every $n$. Hence finite measure is essential to the conclusion.

Yes. The regularity of Lebesgue measure says that for every Lebesgue measurable set $A\subseteq [0,1]$ and every $ϵ>0$, there is an open set $O\supseteq A$ withWrite $O\cap [0,1]$ as a countable disjoint union of intervals. Since its total measure is at most one, a finite subunion $B$ can be chosen withThen $B\in \mathcal{B}$ and $m(A△B)<ϵ$. Hence every Lebesgue measurable set lies in $\mathcal{C}$, while $\mathcal{C}\subseteq {\mathcal{L}}_{[0,1]}$ by definition. Therefore