Prove the following without using Egoroff’s theorem. Let \((f_k)\) be measurable functions on a Lebesgue-measurable set \(E\subset\mathbb R^d\), with \(f_k\to f\) almost everywhere on \(E\). Suppose \(g\in L^1(E)\) and \(|f_k|\leq g\) for every \(k\). For each \(\varepsilon>0\), prove that there is a closed set \(A_\varepsilon\subseteq E\) with \(m(E\setminus A_\varepsilon)<\varepsilon\) such that \(f_k\to f\) uniformly on \(A_\varepsilon\).
Solution
1. Control the tail suprema in \(L^1\).
Choose a measurable null set \(Z\subset E\) outside which the given convergence holds and \(g\) is finite. On \(E_0=E\setminus Z\), passage to the limit gives \(|f|\leq g\).
For \(N\ge1\), define \[ h_N(x):=\begin{cases} \sup_{k\ge N}|f_k(x)-f(x)|,&x\in E_0,\\ 0,&x\in Z. \end{cases} \] Then each \(h_N\) is measurable, \[ 0\le h_{N+1}\le h_N\le 2g. \] The displayed domination holds almost everywhere, and \(h_N\downarrow0\) everywhere by its definition on \(Z\). Since \(g\in L^1(E)\), dominated convergence gives [@Fol13] \[ \int_E h_N\,dm\longrightarrow0. \]
2. Remove a set of arbitrarily small measure on which the tail bounds are bad.
Fix \(\varepsilon>0\). For each \(j\ge1\), choose \(N_j\) increasing so that \[ \int_E h_{N_j}\,dm<\frac{\varepsilon}{2^{j+2}j}. \] Set \[ B_j:=\{x\in E:h_{N_j}(x)>1/j\}. \] By Chebyshev’s inequality, \[ m(B_j) \le j\int_E h_{N_j}\,dm <\frac{\varepsilon}{2^{j+2}}. \] Hence, with \[ B:=Z\cup\bigcup_{j=1}^\infty B_j, \] we have \[ m(B)<\frac\varepsilon2. \] On \(E\setminus B\), for every \(j\) and every \(k\ge N_j\), \[ |f_k(x)-f(x)|\le h_{N_j}(x)\le\frac1j. \] Thus \(f_k\to f\) uniformly on \(E\setminus B\).
3. Replace the measurable good set by a closed good set.
Put \(F=E\setminus B\) and \(H=\mathbb R^d\setminus F\). For each positive integer \(j\), the measurable set \(H_j=H\cap[-j,j]^d\) has finite measure. Outer regularity gives an open \(O_j\supset H_j\) with \(m(O_j\setminus H_j)<\varepsilon/2^{j+1}\) [@Fol13]. Then \(O=\bigcup_jO_j\) is open and contains \(H\), and \[ m(O\setminus H)\leq\sum_{j=1}^\infty m(O_j\setminus H_j) <\varepsilon/2. \] Consequently \(A_\varepsilon=\mathbb R^d\setminus O\) is closed, lies in \(F\), and satisfies \(m(F\setminus A_\varepsilon)<\varepsilon/2\). This construction does not assume \(m(E)<\infty\). Therefore \[ m(E\setminus A_\varepsilon) \le m(B)+m\bigl((E\setminus B)\setminus A_\varepsilon\bigr) <\varepsilon. \] Since uniform convergence on a set passes to every subset, \(f_k\to f\) uniformly on \(A_\varepsilon\).
Hence there is a closed set \(A_\varepsilon\subset E\) with \[ m(E\setminus A_\varepsilon)<\varepsilon \] such that \(f_k\to f\) uniformly on \(A_\varepsilon\), as required.
Solution
1. At each accuracy level, the bad tail sets have finite measure decreasing to zero.
Choose a measurable null set \(Z\subset E\) outside which \(f_k(x)\to f(x)\) and \(g\) is finite. Put \(E_0=E\setminus Z\). On \(E_0\), the bound on \(f_k\) implies \(|f|\leq g\).
For positive integers \(j,N\), define \[ B_{N,j} :=\bigcup_{k\ge N} \left\{x\in E_0:|f_k(x)-f(x)|\ge \frac1j\right\}. \] For fixed \(j\), the sets \(B_{N,j}\) decrease as \(N\to\infty\). Pointwise convergence gives \[ \bigcap_{N=1}^\infty B_{N,j}=\varnothing. \] Also, because \(|f_k|,|f|\le g\), \[ B_{1,j}\subseteq\left\{x\in E:g(x)\ge\frac1{2j}\right\}. \] Chebyshev’s inequality gives \[ m(B_{1,j}) \le 2j\int_E g<\infty. \] Hence continuity from above of measure yields [@Fol13] \[ m(B_{N,j})\longrightarrow0 \qquad(N\to\infty). \]
2. Removing a set of small measure gives uniform convergence.
Choose \(N_j\) so that \[ m(B_{N_j,j})<\frac{\epsilon}{2^{j+2}}. \] Set \[ B:=Z\cup\bigcup_{j=1}^\infty B_{N_j,j}. \] Then \[ m(B)<\frac\epsilon2. \]
On \(E\setminus B\), convergence is uniform. Indeed, given \(\eta>0\), choose \(j\) with \(1/j<\eta\). If \(k\ge N_j\) and \(x\in E\setminus B\), then \(x\notin B_{N_j,j}\), so \[ |f_k(x)-f(x)|<\frac1j<\eta. \]
3. The good set contains a closed subset with arbitrarily small measure loss.
Let \(\mathbb R^d\) be the ambient Euclidean space, and put \(F=E\setminus B\) and \(H=\mathbb R^d\setminus F\). For each positive integer \(j\), use finite-measure outer regularity to choose an open \(O_j\supset H\cap[-j,j]^d\) whose excess measure is less than \(\epsilon/2^{j+1}\) [@Fol13]. The open set \(O=\bigcup_jO_j\) contains \(H\), and \[ m(O\setminus H) \leq\sum_j m\bigl(O_j\setminus(H\cap[-j,j]^d)\bigr) <\epsilon/2. \] Its complement \(A_\epsilon=\mathbb R^d\setminus O\) is closed, contained in \(F\), and satisfies \(m(F\setminus A_\epsilon)<\epsilon/2\). This holds whether or not \(E\) has finite measure. Then \[ m(E\setminus A_\epsilon) \le m(B)+m\bigl((E\setminus B)\setminus A_\epsilon\bigr) <\epsilon, \] and uniform convergence on \(E\setminus B\) implies uniform convergence on \(A_\epsilon\).