= Solution
For each $k$, choose $n_k$ with $\lambda(A\setminus E_{n_k}^{(k)})\le\varepsilon2^{-k-1}$ and set $B=\bigcap_{k\ge1}E_{n_k}^{(k)}$. <Countable subadditivity> gives $\lambda(A\setminus B)\le\varepsilon/2$. For any tolerance $\delta>0$, choose $k$ with $1/k<\delta$. For all $m\ge n_k$ and every $x\in B$, $|f_m(x)-f(x)|\le1/k$, proving <uniform convergence> on $B$.
If the <functions> are Borel measurable, $B$ itself is Borel and we can take $A_\varepsilon=B$. If “measurable” is interpreted in the completed Lebesgue sense, use <regularity of Lebesgue measure> to choose a Borel subset $A_\varepsilon\subseteq B$ 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 $f_n=\mathbf1_{[n,\infty)}$. 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 \b[finite measure is essential to the conclusion].
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