Continuity on Almost All of a Finite-Measure Set
A Lebesgue-measurable function need not be continuous anywhere on its entire domain. Lusin’s theorem gives a useful alternative: on a finite-measure subset of the real line, the function is continuous after discarding a set of arbitrarily small measure. The remaining set can be chosen compact, which adds strong topological control to the measure estimate.
Throughout, \(m\) denotes Lebesgue measure on \(\mathbb{R}\), and \(\mathcal{L}\) its sigma-algebra. Let \(A\in\mathcal{L}\) satisfy \(m(A)<\infty\). A real-valued function \(f:A\to\mathbb{R}\) is Lebesgue measurable here if the inverse image of every Borel set is measurable for the relative sigma-algebra \(\{E\cap A:E\in\mathcal{L}\}\). “Finite-valued” means \(f(x)\in\mathbb{R}\) for every \(x\in A\), rather than allowing \(+\infty\) or \(-\infty\). Continuity on a subset \(K\subseteq A\) means continuity of the restricted function \(f|_K\) in the relative topology on \(K\); it does not mean continuity on all of \(A\).
The theorem is a statement about the measure of the discarded part of the domain, not about the size of the values of \(f\). It requires neither that \(f\) be bounded nor that it be integrable. The proof by simple approximation is developed in the subsequent treatment of Lusin’s theorem; the examples below illustrate the statement and its precise interpretation.
Reading the Compact-Set Conclusion
The compact set \(K\) is chosen after \(\varepsilon\) is specified. The inequality is strict: the omitted part must have measure less than \(\varepsilon\). There is no requirement that \(K\) be an interval or that \(f\) be continuous at points outside \(K\). Nor does the theorem say that \(f\) is close in value to a continuous function at every point of \(A\). It says that all points where continuity has not been secured can be placed in a measurable exceptional set of small measure.
Worked Example: An Unbounded Function on a Bounded Domain
Take \(A=(0,1)\) with Lebesgue measure and \(f(x)=1/x\). This function is finite-valued and continuous at every point of \(A\), but it is unbounded on \(A\). For \(\varepsilon=1/4\), choose \(K=[1/10,9/10]\). It is compact and contained in \(A\), and \(f|_K\) is continuous.
The omitted set is $$ A\setminus K=(0,1/10)\cup(9/10,1). $$ The two intervals are disjoint, so their measures add: $$ m(A\setminus K)=\frac{1}{10}+\frac{1}{10}=\frac{1}{5}<\frac{1}{4}. $$ On \(K\), the values run from \(1/(9/10)=10/9\) to \(1/(1/10)=10\); in particular, the restriction is bounded. Check: \(m(K)=9/10-1/10=4/5\), and \(m(A)-m(K)=1-4/5=1/5\), agreeing with the direct calculation of the omitted measure.
Worked Example: The Indicator of the Rationals
Let \(A=[0,1]\), and define \(f(x)=\chi_{\mathbb{Q}\cap[0,1]}(x)\). This is a Lebesgue-measurable, finite-valued function. It is discontinuous at every point of \(A\), but Lusin’s theorem still gives a large compact set on which its restriction is continuous. Here is an explicit construction for \(\varepsilon=1/4\).
Enumerate the rationals in \([0,1]\) as \(q_1,q_2,\ldots\). For each \(n\ge1\), let $$ I_n=(q_n-2^{-(n+4)},q_n+2^{-(n+4)}), \qquad K=[0,1]\setminus\bigcup_{n=1}^{\infty}I_n. $$ The set \(K\) is closed in \([0,1]\), hence closed and bounded in \(\mathbb{R}\) and compact. Every \(q_n\) lies in \(I_n\), so \(K\) contains no rational. Consequently \(f(x)=0\) at every \(x\in K\), and \(f|_K\) is the constant, therefore continuous, function zero.
Countable subadditivity of Lebesgue measure gives $$ m(A\setminus K)\le \sum_{n=1}^{\infty}m(I_n) =\sum_{n=1}^{\infty}2^{-(n+3)} =\frac{1/16}{1-1/2} =\frac{1}{8}<\frac{1}{4}. $$ The intervals may extend outside \([0,1]\), which can only make this upper bound larger than the measure of the part actually removed from \(A\). Check: \(m(K)=1-m(A\setminus K)\ge 1-1/8=7/8\). This explicit compact set has the required measure, and the restriction is zero everywhere on it.
Worked Example: A Jump Discontinuity Removed by a Small Gap
Let \(A=[-1,1]\), and define \(f(x)=\chi_{[0,1]}(x)\). For \(\varepsilon=1/4\), take $$ K=[-1,-1/10]\cup[1/10,1]. $$ This is a finite union of compact intervals and is therefore compact. On the first component \(f=0\), and on the second component \(f=1\). Each point of either component has a relative neighborhood in \(K\) that does not meet the other component, so these constant values show that \(f|_K\) is continuous.
The omitted set is \(A\setminus K=(-1/10,1/10)\), whose measure is \(1/10-(-1/10)=1/5\). Thus \(m(A\setminus K)=1/5<1/4\). Check: the lengths of the two components of \(K\) are \(9/10\) and \(9/10\), so \(m(K)=9/5\); since \(m(A)=2\), the difference is \(2-9/5=1/5\). The jump at zero is excluded from \(K\), as required for continuity of the restriction.
What Compactness Adds
The theorem’s continuity conclusion has useful consequences because the set on which continuity holds is compact. The next elementary result makes two of them explicit. These conclusions concern the restriction to \(K\); they do not imply that \(f\) is bounded or uniformly continuous on all of \(A\).
Proof. If \(K=\varnothing\), boundedness and uniform continuity hold vacuously. Otherwise, for each \(x\in K\), continuity at \(x\) gives \(r_x>0\) such that \(|h(y)-h(x)|<1\) whenever \(y\in K\) and \(|y-x|<r_x\). The open intervals \((x-r_x,x+r_x)\), for \(x\in K\), cover \(K\). Compactness gives a finite subcover centered at \(x_1,\ldots,x_N\). For any \(y\in K\), some \(j\) satisfies \(|y-x_j|<r_{x_j}\), and therefore \(|h(y)|\le |h(x_j)|+1\). The finite number \(1+\max_{1\le j\le N}|h(x_j)|\) bounds \(|h|\) on \(K\).
To prove uniform continuity, suppose it fails. Then there is an \(\eta_0>0\) such that for every \(n\ge1\) there are \(x_n,y_n\in K\) with \(|x_n-y_n|<1/n\) but \(|h(x_n)-h(y_n)|\ge\eta_0\). By compactness, some subsequence \(x_{n_j}\) converges to \(x\in K\). Since \(|x_{n_j}-y_{n_j}|<1/n_j\to0\), the same subsequence \(y_{n_j}\) converges to \(x\). Continuity at \(x\) gives \(h(x_{n_j})\to h(x)\) and \(h(y_{n_j})\to h(x)\). Thus \(|h(x_{n_j})-h(y_{n_j})|\to0\), contradicting its lower bound \(\eta_0>0\). Hence \(h\) is uniformly continuous. \(\square\)
Applying this proposition to Lusin’s compact set shows that, for every \(\varepsilon>0\), \(f\) is bounded and uniformly continuous on some compact \(K\subseteq A\) with \(m(A\setminus K)<\varepsilon\). The bound and the uniform-continuity scale may depend on \(K\) and on \(\varepsilon\); the theorem does not provide a single bound or a single scale working across all choices.
Common Misreadings and Scope
Confusing continuity on \(K\) with continuity on \(A\). For \(f=\chi_{[0,1]}\) on \([-1,1]\), the function jumps at zero. The compact set in the preceding example omits a neighborhood of zero, so the restriction is continuous. It would be incorrect to conclude that \(f\) has become continuous at zero: zero is not in \(K\).
Confusing small exceptional measure with small pointwise error. Lusin’s theorem does not state that \(|f-g|<\varepsilon\) on all of \(A\) for some continuous \(g\). Its conclusion concerns the measure of \(A\setminus K\), while the guaranteed continuity is for \(f|_K\). A separate approximation statement would need to specify a function \(g\), an error, and the set on which that error is controlled.
Dropping the finite-measure assumption. The theorem is stated for \(m(A)<\infty\), so the desired conclusion is that the discarded set has small finite measure. For \(A=\mathbb{R}\), no compact \(K\subseteq\mathbb{R}\) can satisfy \(m(\mathbb{R}\setminus K)<\varepsilon\): a compact subset of \(\mathbb{R}\) is bounded, and its complement contains an unbounded interval of infinite Lebesgue measure. This does not contradict Lusin’s theorem, because \(\mathbb{R}\) does not meet its finite-measure hypothesis.
Allowing infinite values without changing the statement. The theorem here assumes \(f(x)\in\mathbb{R}\) at every point of \(A\). A function taking values in the extended real line requires a version of the result with additional care about the set where it is infinite. That is not part of this finite-valued statement.
Check Your Understanding
In each question, distinguish the measure estimate from the continuity conclusion, and state which hypotheses are being used.
- State Lusin’s theorem, specifying the measure space, the domain and range of the function, and the sense of continuity on the compact set.
- For \(f(x)=1/x\) on \((0,1)\), compute the measure omitted by \(K=[1/10,9/10]\), and check whether it is less than \(1/4\).
- Why is the restriction of \(\chi_{[0,1]}\) continuous on \([-1,-1/10]\cup[1/10,1]\), although the original function is discontinuous at zero?
- What boundedness and uniform-continuity conclusions follow after applying Lusin’s theorem and the compactness proposition? What do they not imply about all of \(A\)?
- Explain why the finite-measure hypothesis is not satisfied when \(A=\mathbb{R}\), and why no compact \(K\subseteq\mathbb{R}\) can have \(m(\mathbb{R}\setminus K)<1\).