Lusin Approximation for Unbounded Measurable Functions

From Bounded Values to Lusin Approximation

Lusin’s theorem says, in one of its standard forms, that a measurable function behaves continuously after restricting its domain to a suitable compact set, with only a small amount of measure discarded. For an unbounded function, there is an additional issue: a theorem for bounded functions cannot be applied to the whole domain without first controlling where the function takes large values.

Throughout, \(E\) is a Lebesgue-measurable subset of \(\mathbb{R}\) with \(m(E)<\infty\), and \(f:E\to\mathbb{R}\) is finite-valued and Lebesgue measurable. “Finite-valued” means \(f(x)\in\mathbb{R}\) for every \(x\in E\), not \(+\infty\) or \(-\infty\). Continuity on a subset such as \(K\) always means continuity in the relative topology on that subset. The measure \(m\) is Lebesgue measure; the theorem below asserts a compact set and continuity everywhere on that set, not merely almost everywhere.

The result for bounded functions is the bounded Lusin theorem: if \(A\subseteq\mathbb{R}\) is Lebesgue measurable with finite measure and \(g:A\to\mathbb{R}\) is bounded and measurable, then for every \(\delta>0\) there is a compact \(K\subseteq A\) with \(m(A\setminus K)<\delta\) such that \(g|_K\) is continuous. We use that earlier result in the proof. The extension to unbounded \(f\) rests on choosing the bounded-value subset carefully.

Proof by Restricting to a Large Bounded-Value Set

Proof. For each positive integer \(n\), define \(T_n=\{x\in E:|f(x)|>n\}\). These sets are measurable, decrease as \(n\) increases, and have empty intersection: if \(x\) belonged to every \(T_n\), the finite number \(|f(x)|\) would exceed every positive integer. Since \(m(T_1)\le m(E)<\infty\), their measures tend to zero. To verify this last step directly, the sets \(D_n=T_n\setminus T_{n+1}\) are disjoint and \(\bigcup_{k\ge n}D_k=T_n\), because \(\bigcap_kT_k=\varnothing\). Thus \(m(T_n)=\sum_{k\ge n}m(D_k)\). The series has finite sum, bounded by \(m(E)\), so its tails tend to zero.

Choose a positive integer \(n\) such that \(m(T_n)<\varepsilon/2\), and set \(A_n=E\setminus T_n=\{x\in E:|f(x)|\le n\}\). This is a measurable set of finite measure, and \(f|_{A_n}\) is bounded and measurable. Apply the bounded Lusin theorem to \(A_n\) with \(\delta=\varepsilon/2\). It gives a compact \(K\subseteq A_n\) for which \(m(A_n\setminus K)<\varepsilon/2\) and \(f|_K\) is continuous. Since \(K\subseteq A_n\), the two pieces \(T_n\) and \(A_n\setminus K\) are disjoint and form \(E\setminus K\). Therefore

This proves both the required measure bound and continuity on \(K\). If \(E\) is empty, take \(K=\varnothing\); if \(E\) has measure zero, the same bounded Lusin application also gives the result. No assumption that \(f\) is bounded on all of \(E\) has been used. \(\square\)

Worked Applications

A Consequence and a Common Truncation Error

Proof. The theorem gives a compact \(K\) on which \(f|_K\) is continuous. For each \(x\in K\), continuity provides a relative neighborhood \(U_x\) in \(K\) on which \(|f(y)|<|f(x)|+1\). These neighborhoods cover \(K\). Compactness gives a finite subcover \(U_{x_1},\ldots,U_{x_r}\), so on all of \(K\), \(|f(y)|\le \max_{1\le j\le r}(|f(x_j)|+1)\), a finite bound. If \(K=\varnothing\), boundedness is vacuous. The measure estimate is the one supplied by the theorem. \(\square\)

Thus an unbounded measurable function on a finite-measure set is continuous and bounded on a compact subset after discarding an arbitrarily small measure. This is a statement about restriction to \(K\), not a claim that \(f\) is continuous on the original domain or that \(f\) is integrable. Continuity and boundedness on \(K\) alone do not establish integrability on all of \(E\).

A common mistake is to truncate \(f\) on the whole domain and then conclude that the original function is continuous on the compact set supplied for the truncation. The truncation can hide precisely the part of \(f\) that is unbounded or discontinuous. For a concrete example, on \(E=[0,1]\) set \(f(0)=3\) and \(f(x)=3+1/x+\chi_{\mathbb{Q}}(x)\) for \(x>0\). This function is measurable, finite at every point, unbounded near zero, and discontinuous. With threshold \(2\), the clipping \(g(x)=\min(f(x),2)\) is identically \(2\), since \(f(x)\ge3\) throughout \(E\). Applying bounded Lusin to \(g\) can retain all of \(E\), where \(g\) is continuous. But this says nothing about continuity of \(f\): in fact \(f\ne g\) everywhere, and the omitted tail \(\{|f|>2\}\) has measure \(1\). The correct proof first chooses a threshold whose tail has small measure, and then applies bounded Lusin on the set where the original function already equals its bounded restriction.