Borel Representatives of Lebesgue-Measurable Functions

Replacing Lebesgue-Measurable Functions by Borel Functions

On Euclidean space, a Lebesgue-measurable set need not be Borel. Consequently, a finite Lebesgue-measurable function need not itself be Borel-measurable. Nevertheless, its values can be represented almost everywhere by a Borel-measurable function. The construction encodes the function through its strict sublevel sets at rational thresholds, replaces each such set by a Borel representative, and then handles the exceptional points carefully.

Throughout, the measure space is \((\mathbb{R}^n,\mathcal{L},m)\), where \(\mathcal{L}\) is the complete Lebesgue sigma-algebra and \(m\) is Lebesgue measure. This space is sigma-finite, but not finite. A function described as finite here is real-valued everywhere, not extended-real-valued. A Borel function means a function measurable from \((\mathbb{R}^n,\mathcal{B}_{\mathbb{R}^n})\) to \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\). Agreement almost everywhere means equality outside a set of Lebesgue measure zero.

We use the completion property of Lebesgue measure: for every \(E\in\mathcal{L}\), there is a Borel set \(B\) such that \(m(E\triangle B)=0\). Also, every Lebesgue null set is contained in a Borel null set. These are standard consequences of the fact that \(\mathcal{L}\) is the completion of the Borel sigma-algebra with respect to Lebesgue measure. In particular, a Borel representative of a measurable set means a Borel set equal to it up to a null set; it need not equal it pointwise.

The Borel Representative Theorem

Theorem: If \(f:\mathbb{R}^n\to\mathbb{R}\) is finite everywhere and Lebesgue-measurable on \((\mathbb{R}^n,\mathcal{L},m)\), then there is a finite Borel-measurable function \(h:\mathbb{R}^n\to\mathbb{R}\) such that \(h=f\) \(m\)-almost everywhere.

Proof. For each rational \(q\), let \(E_q=\{x:f(x)<q\}\). Since \(f\) is Lebesgue-measurable, \(E_q\in\mathcal{L}\). Choose a Borel set \(B_q\) such that \(m(E_q\triangle B_q)=0\). There are only countably many rational numbers, so the set

N_0=\bigcup_{q\in\mathbb{Q}}(E_q\triangle B_q)

is Lebesgue-measurable and null. For every \(x\notin N_0\) and every rational \(q\), membership agrees: \(x\in B_q\) if and only if \(f(x)<q\).

Define an extended-real-valued function \(g\) by

g(x)=\inf\{q\in\mathbb{Q}:x\in B_q\},

where the infimum of the empty set is \(+\infty\), and an infimum unbounded below is \(-\infty\). For every real \(a\),

\{x:g(x)<a\} = \bigcup_{\substack{q\in\mathbb{Q}\\q<a}}B_q.

Indeed, if \(g(x)<a\), the definition of infimum gives some \(q\in\mathbb{Q}\) with \(x\in B_q\) and \(q<a\). Conversely, if such a \(q\) exists, then \(g(x)\le q<a\). The displayed union is Borel. The strict sublevel sets \(\{g<a\}\) being Borel for all real \(a\) imply that \(g\) is Borel-measurable as an extended-real-valued function.

We next check its values outside \(N_0\). Fix \(x\notin N_0\). Then \(\{q\in\mathbb{Q}:x\in B_q\}=\{q\in\mathbb{Q}:f(x)<q\}\). Since \(f(x)\) is finite, every rational in this set is greater than \(f(x)\), so its infimum is at least \(f(x)\). For each \(\varepsilon>0\), density of the rationals supplies a rational \(q\) with \(f(x)<q<f(x)+\varepsilon\). Thus the infimum is at most \(f(x)+\varepsilon\) for every \(\varepsilon>0\), and hence equals \(f(x)\). Therefore \(g(x)=f(x)\) and is finite for every \(x\notin N_0\).

The set \(N_0\) is null but need not be Borel. By the completion property, choose a Borel null set \(Z\) containing \(N_0\). Define \(h(x)=g(x)\) for \(x\notin Z\), and \(h(x)=0\) for \(x\in Z\). Outside \(Z\), \(g=f\) is finite; on \(Z\), \(h=0\), so \(h\) is finite everywhere. For each real \(a\), its strict sublevel set is

\{x:h(x)<a\} = \bigl(Z^c\cap\{x:g(x)<a\}\bigr) \cup \begin{cases} Z,&a>0,\\ \varnothing,&a\le 0. \end{cases}

Both parts are Borel, so \(h\) is Borel-measurable. Since \(N_0\subseteq Z\) and \(g=f\) outside \(N_0\), we have \(h=f\) outside \(Z\). As \(m(Z)=0\), this is equality \(m\)-almost everywhere, as required. \(\square\)

Worked Constructions

Worked Example: A Non-Borel Null Modification

Let \(A\subseteq\mathbb{R}^n\) be a Lebesgue null set that is not Borel, and let \(f=\chi_A\). This is finite and Lebesgue-measurable: for every Borel set of values, its preimage is one of \(\varnothing,A,A^c,\mathbb{R}^n\), all of which are Lebesgue-measurable. Choose a Borel null set \(C\) containing \(A\).

For rational \(q\le0\), \(E_q=\{f<q\}=\varnothing\). For \(0<q\le1\), \(E_q=A^c\), since \(f=0\) off \(A\) and \(f=1\) on \(A\). For \(q>1\), \(E_q=\mathbb{R}^n\). Take \(B_q=\varnothing\) for \(q\le0\), \(B_q=C^c\) for \(0<q\le1\), and \(B_q=\mathbb{R}^n\) for \(q>1\). These are Borel representatives: in the middle range, \(E_q\triangle B_q=A^c\triangle C^c=C\setminus A\), which is null.

The exceptional union is \(N_0=C\setminus A\). The associated infimum \(g\) equals \(0\) on \(C^c\), because there \(x\in B_q\) for every rational \(q>0\) and for no \(q\le0\), and the infimum of the positive rationals is \(0\). On \(C\), \(x\notin B_q\) for \(q\le1\) and \(x\in B_q\) for \(q>1\), so \(g(x)=1\). Taking \(Z=C\) and resetting \(g\) to zero on \(Z\) gives \(h=0\) everywhere. This is Borel and agrees with \(f\) outside \(A\). Check: \(m(A)=0\), so \(f=h\) \(m\)-almost everywhere, even though \(f\) is not Borel-measurable: \(\{f>1/2\}=A\) is not Borel.

Worked Example: An Unbounded Finite Function

Let \(A\subseteq\mathbb{R}^n\) be a non-Borel null set, let \(C\) be a Borel null set containing \(A\), and define \(f(x)=x_1+\chi_A(x)\), where \(x_1\) is the first coordinate. This function is finite everywhere and Lebesgue-measurable. For rational \(q\), choose \(B_q=\{x:x_1<q\}\), a Borel half-space.

If \(x\notin A\), then \(f(x)=x_1\), and \(f(x)<q\) is equivalent to \(x_1<q\). If \(x\in A\), then \(f(x)=x_1+1\), so any difference between the conditions \(f(x)<q\) and \(x_1<q\) occurs at a point of \(A\). Thus \(E_q\triangle B_q\subseteq A\) for every rational \(q\), and \(N_0\subseteq A\subseteq C\). For any \(x\), the infimum of the rational numbers \(q\) satisfying \(x_1<q\) is \(x_1\): all such \(q\) exceed \(x_1\), and there is one between \(x_1\) and \(x_1+\varepsilon\) for every \(\varepsilon>0\). Hence \(g(x)=x_1\). Set \(h=x_1\) off \(C\) and \(h=0\) on \(C\). Its sublevel sets are Borel by the same piecewise formula in the theorem, and \(h\) is finite everywhere. Outside \(C\), \(x\notin A\), so \(h=f=x_1\). Therefore \(h=f\) \(m\)-almost everywhere. Check: the exceptional set is contained in \(C\), and \(m(C)=0\), which verifies the asserted almost-everywhere equality.

Worked Example: Two Measurable Sets in a Finite-Valued Function

Let \(A\) be a non-Borel null set contained in a Borel null set \(C\), and put \(E=[0,1]\cup A\) and \(F=[1,2]\setminus A\) in \(\mathbb{R}\). Both sets are Lebesgue-measurable. The function \(f=3\chi_E-2\chi_F\) is finite and Lebesgue-measurable. Replacing \(E,F\) by the Borel sets \(E'=[0,1]\) and \(F'=[1,2]\) changes each set only on a subset of \(A\): \(E\triangle E'=A\setminus[0,1]\), and \(F\triangle F'=A\cap[1,2]\). Thus \(h=3\chi_{[0,1]}-2\chi_{[1,2]}\) is Borel and equals \(f\) off \(A\), hence \(m\)-almost everywhere.

Explicitly, \(h=3\) on \([0,1)\), \(h=1\) at \(1\), \(h=-2\) on \((1,2]\), and \(h=0\) outside \([0,2]\). The value at \(1\) is \(3-2=1\), since \(1\) belongs to both intervals. These endpoint choices do not affect the almost-everywhere claim, but they do specify the representative everywhere. Check: \(f=h\) at every point outside \(A\), and \(m(A)=0\).

The Exceptional Set Must Be Repaired

A common mistake is to call \(N_0=\bigcup_{q\in\mathbb{Q}}(E_q\triangle B_q)\) Borel merely because it is a countable union. Countable unions preserve Borel measurability only when the sets being united are Borel. Here \(E_q\triangle B_q\) is known to be Lebesgue-measurable and null, but need not be Borel. For example, take a non-Borel null set \(A\subseteq C\), where \(C\) is Borel and null, and use the choices from the first worked example. For \(0<q\le1\), \(E_q\triangle B_q=C\setminus A\). This set is not Borel: if it were, then \(A=C\setminus(C\setminus A)\) would be Borel. Thus \(N_0\) need not be Borel.

The correction is essential to the proof: first note that \(N_0\) is Lebesgue-measurable and null, then choose a Borel null set \(Z\supseteq N_0\), and reset the function on \(Z\), not merely on \(N_0\). The resulting strict sublevel sets are Borel by an explicit formula, so the final function really is Borel-measurable. The theorem concerns existence of an almost-everywhere equal representative; it does not assert that the original function is Borel, or that the two functions agree at every point.

Check Your Understanding

In each question, distinguish carefully between equality everywhere and equality almost everywhere.

Why are rational thresholds sufficient to recover a finite real value as the infimum of the rational numbers strictly above it?
In the theorem's construction, show that \(\{g<a\}=\bigcup_{q\in\mathbb{Q},\,q<a}B_q\), including the case where the defining set for the infimum is empty.
Why does the completion property give a Borel null set \(Z\) containing the measurable null set \(N_0\)?
For \(f=x_1+\chi_A\), explain why the selected half-space \(B_q=\{x:x_1<q\}\) differs from \(\{f<q\}\) only on \(A\).
Give the precise flaw in asserting that a countable union of Lebesgue-measurable null sets must be Borel.

Borel Representatives of Lebesgue-Measurable Functions

What You'll Learn