Why a Density Is Determined Only Almost Everywhere
A density defines a measure by integrating it over each measurable set. The values of the density on a null set cannot affect any of those integrals, so pointwise uniqueness is not possible in general. The right uniqueness statement is that two integrable densities defining the same measure agree almost everywhere. The essential step is to test the measures on level sets where one density is strictly larger than the other.
Throughout, let \((X,\mathcal{M},\mu)\) be a measure space; no finiteness, sigma-finiteness, or completeness assumption is required. A real-valued or complex-valued measurable function \(f\) is \(\mu\)-integrable when \(\int_X |f|\,d\mu<\infty\). Its density measure assigns \(E\in\mathcal{M}\) the value \(\nu_f(E)=\int_E f\,d\mu\). For real \(f\), this is a finite signed measure; for complex \(f\), it is a finite complex measure. Equality of two such measures means equality of their values on every \(E\in\mathcal{M}\).
The proof reduces the complex case to a real-valued fact. The sets where the real difference is at least \(1/n\), or at most \(-1/n\), have a definite sign and a definite minimum discrepancy. If the integral over every measurable set is zero, each such set must be null.
The Real-Valued Level-Set Argument
Proof. For each positive integer \(n\), define the measurable sets \(A_n=\{x\in X:h(x)\ge 1/n\}\) and \(B_n=\{x\in X:h(x)\le -1/n\}\). On \(A_n\), \(h\ge 1/n\), so monotonicity of the integral for nonnegative functions gives \(\int_{A_n}h\,d\mu\ge \mu(A_n)/n\). The hypothesis says the left side is zero. Therefore \(\mu(A_n)=0\). On \(B_n\), \(-h\ge 1/n\), and the hypothesis applied to \(B_n\) gives \(\int_{B_n}(-h)\,d\mu=0\). Hence \(0\ge\mu(B_n)/n\), so \(\mu(B_n)=0\). Every point where \(h>0\) belongs to some \(A_n\), since for any real \(h(x)>0\) there is an integer \(n\) with \(1/n\le h(x)\). Every point where \(h<0\) belongs to some \(B_n\), by applying the same argument to \(-h(x)>0\). Thus \(\{h\ne0\}=\bigcup_{n=1}^\infty(A_n\cup B_n)\). This is a countable union of null sets, so it is null. Therefore \(h=0\) \(\mu\)-almost everywhere. \(\square\)
There is no hidden finite-measure assumption in this argument. For example, \(A_n\) could initially have infinite measure, but the inequality \(\int_{A_n}h\,d\mu\ge\mu(A_n)/n\) would then force the integral to be infinite. That contradicts both \(\int_{A_n}h\,d\mu=0\) and the integrability of \(h\). The same reasoning applies to \(B_n\).
Proof of the Uniqueness Theorem
Proof. First suppose \(\nu_f(E)=\nu_g(E)\) for every \(E\in\mathcal{M}\). The difference \(h=f-g\) is integrable because \(|f-g|\le |f|+|g|\), and it is measurable. Linearity of the integral gives \(\int_E h\,d\mu=\int_E f\,d\mu-\int_E g\,d\mu=0\) for every \(E\in\mathcal{M}\). If \(f,g\) are real-valued, apply the proposition to \(h\); then \(f=g\) \(\mu\)-almost everywhere.
For complex-valued \(f,g\), write \(h=u+iv\), where \(u=\operatorname{Re}h\) and \(v=\operatorname{Im}h\) are real-valued integrable measurable functions. For every measurable \(E\), \(\int_Eu\,d\mu=\operatorname{Re}\int_Eh\,d\mu=0\) and \(\int_Ev\,d\mu=\operatorname{Im}\int_Eh\,d\mu=0\). The proposition applied separately to \(u\) and \(v\) shows that both vanish almost everywhere. The union of their two exceptional null sets is null, and outside it \(h=u+iv=0\). Thus \(f=g\) \(\mu\)-almost everywhere.
Conversely, suppose \(f=g\) \(\mu\)-almost everywhere. Then \(h=f-g\) vanishes outside a measurable null set, up to a null exceptional set; equivalently, \(|h|=0\) almost everywhere. The integral of a nonnegative function that vanishes almost everywhere is zero, so \(\int_E |h|\,d\mu=0\) for every \(E\in\mathcal{M}\). It follows that \(\int_E h\,d\mu=0\), and hence \(\nu_f(E)=\nu_g(E)\) for every \(E\). This proves both directions. \(\square\)
Worked Example: Identifying Densities on a Counting Space
Let \(X=\{a,b,c\}\), let \(\mathcal{M}\) be the set of all subsets of \(X\), and let \(\mu\) be counting measure, so \(\mu(E)\) is the number of points in \(E\). Define \(f(a)=2,\ f(b)=-1,\ f(c)=0\), and \(g(a)=1,\ g(b)=0,\ g(c)=0\). Compute their density measures and identify the set that detects their discrepancy.
For counting measure, the integral over a set is the sum of the function values at its points. The values on the singleton sets are therefore \(\nu_f(\{a\})=2,\ \nu_f(\{b\})=-1,\ \nu_f(\{c\})=0\), while \(\nu_g(\{a\})=1,\ \nu_g(\{b\})=0,\ \nu_g(\{c\})=0\). In particular, \(\nu_f(\{a\})\ne\nu_g(\{a\})\), and \(\{a\}\) detects the strict discrepancy \(f(a)-g(a)=1\).
Both total integrals happen to equal \(1\): \(\int_X f\,d\mu=2+(-1)+0=1\), and \(\int_Xg\,d\mu=1+0+0=1\). But the measures are not equal, because equality must hold on every subset, not just on \(X\). Check: the level set \(\{x:f(x)-g(x)\ge1\}\) is exactly \(\{a\}\); its integral of \(f-g\) is \(1\), in agreement with the singleton calculation.
Worked Example: Changing a Density at One Point
On \(X=[0,1]\) with the Lebesgue sigma-algebra and Lebesgue measure \(m\), let \(f(x)=x\) for every \(x\), and let \(g(x)=x\) for \(x\ne 1/2\), with \(g(1/2)=7\). Show that the densities define the same measure even though they are not equal everywhere.
The singleton \(\{1/2\}\) has Lebesgue measure zero. The functions agree on its complement, so \(f=g\) \(m\)-almost everywhere. More explicitly, for every measurable \(E\subseteq[0,1]\), the difference \(g-f\) is zero on \(E\setminus\{1/2\}\), and its only possible nonzero contribution is at \(1/2\). If \(1/2\notin E\), then \(\int_Eg\,dm=\int_Ef\,dm\) because the functions agree throughout \(E\). If \(1/2\in E\), changing the integrand at that one point still contributes zero to the Lebesgue integral, since that point has measure zero. Thus \(\int_Eg\,dm=\int_Ef\,dm\) for every \(E\).
Check: the exceptional set is exactly \(\{1/2\}\), which is null, while \(f(1/2)=1/2\) and \(g(1/2)=7\) are unequal. The theorem gives equality almost everywhere, not equality at every point.
Worked Example: A Complex Discrepancy Detected by a Level Set
On \([0,1]\) with Lebesgue measure, define \(f(x)=i\) on \([0,1/2]\) and \(f(x)=-i\) on \((1/2,1]\); let \(g(x)=0\) everywhere. Both functions are complex-valued and integrable. Show why equal total integrals do not make their density measures equal.
The interval \([0,1/2]\) has measure \(1/2\), and \((1/2,1]\) also has measure \(1/2\). Therefore \(\int_{[0,1]}f\,dm=i(1/2)-i(1/2)=0\), while \(\int_{[0,1]}g\,dm=0\). Now choose the measurable set \(E=[0,1/2]\). On this set \(f=i\), so \(\int_Ef\,dm=i(1/2)=i/2\), whereas \(\int_Eg\,dm=0\). The measures differ on \(E\).
The level-set proof detects the same failure through the imaginary part: \(\operatorname{Im}(f-g)=1\) on \([0,1/2]\) and \(-1\) on \((1/2,1]\). For \(n=1\), the set where \(\operatorname{Im}(f-g)\ge1/n\) is \([0,1/2]\), whose integral of that imaginary part is \(1/2\), not zero. Check: the total integral cancels, but the integral over the positive level set does not; this is precisely why all measurable sets, rather than just \(X\), must be tested.
A Common Mistake: Testing Only the Whole Space
A frequent error is to infer equality of density measures from equality of total integrals. The counting-space example shows the incorrect calculation concretely: \(\int_X f\,d\mu=2-1+0=1\) and \(\int_Xg\,d\mu=1+0+0=1\), so the whole-space integrals agree. The wrong conclusion would be \(\nu_f=\nu_g\). But for \(E=\{a\}\), \(\int_Ef\,d\mu=2\), whereas \(\int_Eg\,d\mu=1\). The correction is to use the hypothesis on every measurable set. In the uniqueness proof, this supplies zero integral for the difference on each level set, not merely cancellation over the whole space.
There is a related sign error to avoid. A zero integral of a real function over \(X\) does not imply that the function vanishes: positive and negative values may cancel. The level-set argument avoids cancellation by restricting to a set on which the function has one sign and is bounded away from zero. For \(A_n=\{h\ge1/n\}\), every value of \(h\) on \(A_n\) is at least \(1/n\); hence a positive measure for \(A_n\) would force a strictly positive integral there.
Check Your Understanding
Use the level-set argument and distinguish carefully between equality everywhere and equality almost everywhere.
- Suppose \(h\) is real-valued, integrable, and \(\int_Eh\,d\mu=0\) for every measurable \(E\). Define the sets used to prove that \(\{h>0\}\) is null.
- Why does \(\int_X f\,d\mu=\int_Xg\,d\mu\) alone not imply \(\nu_f=\nu_g\)? Give a finite counting-measure example.
- If complex-valued integrable \(f,g\) define the same measure, which two real-valued functions should be tested with the proposition?
- On \([0,1]\) with Lebesgue measure, can two functions that differ only at \(x=1/3\) define the same density measure? Explain the role of the singleton \(\{1/3\}\).
- In the proof, why is it necessary to test both \(\{h\ge1/n\}\) and \(\{h\le-1/n\}\) for real \(h\)?