Proof of Lusin's Theorem by Simple Approximation

From Simple Approximation to Continuity

A bounded measurable function can be uniformly approximated by finite-valued measurable functions. Each such approximation is constant on finitely many measurable level sets, but that fact alone does not make it continuous: the level sets can meet along boundaries where the function jumps. The useful extra step is to replace each level set by a compact subset, losing only a small amount of measure. On the compact pieces, the finitely many values are separated, and continuity follows. Intersecting the resulting compact sets over all approximation stages gives a set where the original function is continuous.

We work in \((\mathbb{R},\mathcal{L},m)\), where \(m\) is Lebesgue measure. The ambient measure space is not finite; we assume the measurable set \(E\in\mathcal{L}\) has \(m(E)<\infty\). The function \(f:E\to\mathbb{R}\) is real-valued, bounded, and measurable with respect to the relative Lebesgue sigma-algebra on \(E\). Continuity on a subset of \(E\) means continuity for its relative topology.

The measure-theoretic input is the regularity of Lebesgue measure: if \(A\in\mathcal{L}\) has finite measure and \(\eta>0\), there is a compact \(C\subseteq A\) with \(m(A\setminus C)<\eta\). We use this standard regularity theorem without reproving it here. Since every level set used below is a measurable subset of \(E\), it has finite measure and the theorem applies.

Proof by Simple Approximation

Proof. The empty-set case is covered in the theorem statement, so suppose \(E\ne\varnothing\). Choose \(M<\infty\) such that \(|f(x)|\le M\) for every \(x\in E\). For each integer \(n\ge1\), define the finite-valued function \(\varphi_n(x)=2^{-n}\lfloor 2^n f(x)\rfloor\). Because \(f\) is measurable and the floor function has measurable level sets, \(\varphi_n\) is measurable. Boundedness of \(f\) implies that \(\varphi_n\) takes only finitely many values. The defining property of the floor gives, at every \(x\in E\),

In particular, \(\varphi_n\to f\) uniformly on \(E\). Write the nonempty level sets of \(\varphi_n\) as \(A_{n,1},\ldots,A_{n,J_n}\). Since these sets partition the nonempty set \(E\), \(J_n\) is a positive finite integer. They are measurable, pairwise disjoint, and each has finite measure.

For this \(n\), apply regularity to each \(A_{n,j}\), choosing a compact set \(C_{n,j}\subseteq A_{n,j}\) such that \(m(A_{n,j}\setminus C_{n,j})<\varepsilon/(2^{n+1}J_n)\). Set \(F_n=\bigcup_{j=1}^{J_n}C_{n,j}\). This is a finite union of compact sets, hence compact, and \(F_n\subseteq E\). Additivity over the disjoint level sets and the chosen bounds give

The restriction \(\varphi_n|_{F_n}\) is continuous. Indeed, it is constant on each \(C_{n,j}\). Any two distinct sets among this finite collection are disjoint compact subsets of \(\mathbb{R}\), so their distance is positive. If there is more than one nonempty compact piece, the minimum of the finitely many pairwise distances is positive. Consequently, each piece is open as well as closed relative to \(F_n\), and a function constant on each piece is continuous on \(F_n\). If there is only one piece, the restriction is constant and therefore continuous.

Now put \(K=\bigcap_{n=1}^{\infty}F_n\). The set \(K\) is closed in the compact set \(F_1\), since every \(F_n\) is closed in \(\mathbb{R}\); hence \(K\) is compact. Also \(K\subseteq E\). By countable subadditivity,

For every \(n\), \(K\subseteq F_n\), so \(\varphi_n|_K\) is continuous. The estimate \(0\le f-\varphi_n<2^{-n}\) holds on \(K\), and therefore the continuous functions \(\varphi_n|_K\) converge uniformly to \(f|_K\). To verify the continuity conclusion directly, let \(x\in K\) and \(\eta>0\). Choose \(n\) with \(2^{-n}<\eta/3\). Continuity of \(\varphi_n|_K\) at \(x\) gives a relative neighborhood of \(x\) in \(K\) on which \(|\varphi_n(y)-\varphi_n(x)|<\eta/3\). For every \(y\) in that neighborhood,

Thus \(f|_K\) is continuous. The construction also covers the possibility \(K=\varnothing\): the measure estimate still holds, and the restriction to the empty set is continuous. This proves the theorem. \(\square\)

Worked Examples of the Construction

A Common Pitfall: Levelwise Constancy Is Not Enough

Consider the measurable simple function \(s\) on \([0,1]\) defined by \(s(x)=0\) for \(0\le x\le1/2\) and \(s(x)=1\) for \(1/2<x\le1\). A tempting but incorrect argument says that \(s\) is continuous because it is constant on each of its two level sets. For a concrete test, let \(x_j=1/2-1/j\) for integers \(j\ge3\). Then \(x_j\to1/2\), but \(s(x_j)=0\) for every \(j\), whereas \(s(1/2)=0\); from the left there is no failure. From the right, let \(y_j=1/2+1/j\), also for \(j\ge3\). Then \(y_j\to1/2\), \(s(y_j)=1\), and \(s(1/2)=0\), so \(s\) is discontinuous at \(1/2\). Constancy on each level set does not control how different level sets meet.

The correction is to use compact subsets that stay away from the boundary between the level sets. For instance, \(K=[0,1/2-1/10]\cup[1/2+1/10,1]\) is compact; the restriction of \(s\) to each separated component is constant and hence continuous. The removed interval has measure \(1/5\), computed from its length \(1/10+1/10=1/5\). The general proof repeats this compact selection for every simple approximation, assigns a summable measure budget to all stages, and then uses uniform convergence on the intersection. It does not claim that the original function is continuous on all of \(E\), or even on an arbitrary compact subset of \(E\).

The method matters because it turns measurable information into continuity after discarding a set of controlled measure. The key ingredients are distinct: uniform quantization controls the values of the function; regularity supplies compact pieces; summability controls the total measure lost; and the uniform-limit argument transfers continuity from the approximants to \(f\). No assumption that \(E\) itself is compact or bounded is needed, only measurability and \(m(E)<\infty\).