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For a finite interval $(a,b)$ with Lebesgue measure, restrictions of continuous functions on $[a,b]$ form a dense subspace of the Lp space $L^p((a,b))$ for $1\le p<\infty$. One proof first approximates by simple functions. The regularity of Lebesgue measure approximates each of their measurable sets in measure by a finite union of intervals. An interval's indicator function is approximated by continuous functions taking values in $[0,1]$ and differing from that indicator function only in endpoint strips of total length tending to zero. Their Lp norm error is at most the $1/p$ power of that length. The triangle inequality completes the approximation of each simple function.

First suppose $m(E)<\infty$. The interval-cover definition of Lebesgue outer measure, together with $m^{\ast }(E)=m(E)$, provides bounded open intervals $I_j$ covering $E$ such that ${\sum }_j|I_j|<m(E)+\epsilon$. If the definition uses intervals that are not open, enlarge each by an amount whose total is smaller than the available error. For the open set $U={\bigcup }_j{I}_j$, countable subadditivity and additivity of Lebesgue measure giveThis subtraction is valid because both measures are finite.

For arbitrary $E$, including $m(E)=\infty$, set $E_n=E\cap [n,n+1)$ for $n\in \mathbb{Z}$, and choose positive weights $w_n=2^{-|n|}/3$, whose sum is one. Each $E_n$ has finite Lebesgue measure, so the preceding argument gives an open set $U_n\supseteq E_n$ with $m(U_n\setminus E_n)<\epsilon w_n$. Then $U={\bigcup }_{n\in \mathbb{Z}}{U}_n$ is open, contains $E$, andThus the required form of regularity of Lebesgue measure does not require $E$ itself to have finite measure.

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