Two Different Meanings of “Zero”
For a positive measure, a set of measure zero has no measurable subset of positive measure. A signed measure allows both positive and negative values, so a set can have signed measure zero because its positive and negative contributions cancel. These are different notions: knowing only the total value of a set does not tell us the values of its subsets.
Throughout, \((X,\mathcal{M})\) is a measurable space and \(\nu\) is a finite signed measure on it: \(\nu\) is countably additive, takes finite real values, and has finite total variation. All sets considered are in \(\mathcal{M}\). For \(A\subseteq E\), the phrase “measurable subset” means \(A\in\mathcal{M}\). Set equalities and signed-measure values below are exact, not almost-everywhere statements.
The distinction is local versus total. To test whether \(E\) is null, one must test every measurable part of \(E\), not just \(E\) itself. The examples below make that difference explicit.
Worked Example: Cancellation on a Two-Point Set
Let \(X=\{a,b,c\}\), let \(\mathcal{M}\) be the power set of \(X\), and define the finite signed measure \(\nu\) by assigning weights \(3,-3,0\) to \(a,b,c\), respectively. Thus, for every \(A\subseteq X\), \(\nu(A)\) is the sum of the weights of the points in \(A\).
Take \(E=\{a,b\}\). Its value is \(\nu(E)=3+(-3)=0\), so \(E\) is a zero-value set. But the measurable subset \(\{a\}\subseteq E\) has value \(\nu(\{a\})=3\). Therefore \(E\) is not a null set. The cancellation in \(\nu(E)\) does not persist on all subsets.
For completeness, we can identify every null set in this example. A set containing \(a\) is not null because its subset \(\{a\}\) has value \(3\). A set containing \(b\) is not null because its subset \(\{b\}\) has value \(-3\). Any subset of \(\{c\}\) is null: its only possibilities are \(\varnothing\) and \(\{c\}\), both of which have value zero. Hence the null sets are exactly \(\varnothing\) and \(\{c\}\).
Check: The total value of \(E\) is \(3-3=0\), whereas the subset values \(3\) and \(-3\) are nonzero. This verifies directly that \(E\) is zero-value but not null.
Positive and Negative Sets
Recall the sign-definite terminology for the finite signed measure \(\nu\) on \((X,\mathcal{M})\): a measurable set \(E\) is a positive set if \(\nu(A)\geq 0\) for every measurable \(A\subseteq E\), and it is a negative set if \(\nu(A)\leq 0\) for every measurable \(A\subseteq E\). A null set is both positive and negative. Conversely, if a set is both, every subset has value both at least and at most zero, so every subset has value zero.
Proof. Suppose first that \(E\) is null. Every measurable \(A\subseteq E\) satisfies \(\nu(A)=0\), and therefore satisfies both \(\nu(A)\geq0\) and \(\nu(A)\leq0\). Thus \(E\) is both positive and negative.
Conversely, suppose \(E\) is both positive and negative. For any measurable \(A\subseteq E\), positivity gives \(\nu(A)\geq0\), while negativity gives \(\nu(A)\leq0\). Hence \(\nu(A)=0\). Since this holds for every measurable subset \(A\) of \(E\), \(E\) is null.
Now suppose \(E\) is positive and \(\nu(E)=0\). For any measurable \(A\subseteq E\), both \(A\) and \(E\setminus A\) are measurable subsets of \(E\), so positivity gives \(\nu(A)\geq0\) and \(\nu(E\setminus A)\geq0\). Finite additivity gives \(\nu(E)=\nu(A)+\nu(E\setminus A)\). Since the left side is zero and both terms on the right are nonnegative, each term is zero. In particular, \(\nu(A)=0\). Thus \(E\) is null. If \(E\) is negative instead, apply the same argument to the nonpositive values: two nonpositive numbers summing to zero must both be zero. This again gives \(\nu(A)=0\) for every measurable \(A\subseteq E\). \(\square\)
Worked Example: A Zero-Value Set with a Density
On \(([0,2],\mathcal{L})\), where \(\mathcal{L}\) is the Lebesgue sigma-algebra, define the real-valued function \(f\) by \(f(x)=1\) on \([0,1)\) and \(f(x)=-1\) on \([1,2]\). Define the finite signed measure \(\nu\) by \(\nu(A)=\int_A f\,dm\) for \(A\in\mathcal{L}\). The function is integrable because \(\int_{[0,2]}|f|\,dm=1+1=2\).
Let \(E=[0,2]\). Its signed measure is \(\nu(E)=\int_{[0,1)}1\,dm+\int_{[1,2]}(-1)\,dm=1-1=0\). Thus \(E\) is a zero-value set. It is not null: the measurable subset \(A=[0,1)\) has \(\nu(A)=1\).
In this example we can characterize all null sets exactly. If \(E\) is null, then its measurable subsets \(E\cap[0,1)\) and \(E\cap[1,2]\) must each have value zero. Their values are, respectively, \(m(E\cap[0,1))\) and \(-m(E\cap[1,2])\). Thus both intersections have Lebesgue measure zero. Conversely, if both intersections have Lebesgue measure zero, then for any measurable \(A\subseteq E\), \(\nu(A)=m(A\cap[0,1))-m(A\cap[1,2])=0-0=0\). So \(E\) is null exactly when both of its intersections with the two sign regions are Lebesgue null.
Check: For the whole interval, the two contributions are \(m([0,1))=1\) and \(-m([1,2])=-1\), giving total \(1-1=0\). For the subset \([0,1)\), the value is \(m([0,1))=1\), confirming that the whole interval is not null.
Null Sets Behave Differently from Zero-Value Sets
Proof. Let \(E\) be null and let \(F\in\mathcal{M}\) satisfy \(F\subseteq E\). If \(A\in\mathcal{M}\) and \(A\subseteq F\), then \(A\subseteq E\). Nullity of \(E\) gives \(\nu(A)=0\). This holds for every measurable \(A\subseteq F\), so \(F\) is null.
Now let \(E_1,E_2,\ldots\) be null sets and put \(U=\bigcup_{n=1}^{\infty}E_n\). To show \(U\) is null, fix any measurable \(A\subseteq U\). Define disjoint measurable sets \(D_1=A\cap E_1\) and \(D_n=(A\cap E_n)\setminus\bigcup_{j=1}^{n-1}E_j\) for \(n\geq2\). Each \(D_n\subseteq E_n\), so nullity of \(E_n\) gives \(\nu(D_n)=0\). The \(D_n\) are disjoint and their union is \(A\). Countable additivity therefore gives \(\nu(A)=\sum_{n=1}^{\infty}\nu(D_n)=\sum_{n=1}^{\infty}0=0\). This is true for every measurable \(A\subseteq U\), so \(U\) is null. \(\square\)
A related consequence is that removing a null set does not change the signed value of any measurable set. If \(N\) is null and \(F\in\mathcal{M}\), then \(F\cap N\) is measurable and \(\nu(F\cap N)=0\). Finite additivity on the disjoint sets \(F\setminus N\) and \(F\cap N\) gives \(\nu(F)=\nu(F\setminus N)+0\). A zero-value set, by contrast, cannot generally be removed from another set without changing its value: its subsets need not have value zero.
Worked Example: Zero Value on a Sign-Definite Set
Let \(X=\{a,b,c,d\}\) with its power-set sigma-algebra, and let \(\nu\) assign weights \(0,0,4,-4\) to \(a,b,c,d\), respectively. Again, the value of any set is the sum of the weights of its points. Consider \(E=\{a,b\}\). Its value is \(\nu(E)=0+0=0\).
The measurable subsets of \(E\) are \(\varnothing,\{a\},\{b\},\{a,b\}\). Their values are, respectively, \(0,0,0,0\), so \(E\) is null. Also, \(E\) is a positive set: every subset has value zero, which is nonnegative. The proposition explains why a zero-value positive set must be null. In contrast, the set \(\{c,d\}\) also has total value \(4-4=0\), but its subset \(\{c\}\) has value \(4\), so it is not null.
Check: For \(E=\{a,b\}\), the four subset values are all zero, as required for nullity. For \(\{c,d\}\), the total is \(4+(-4)=0\), but the subset value \(4\) rules out nullity. The distinction is the presence or absence of cancellation within the set.
A Common Mistake and Why the Distinction Matters
A tempting but invalid argument is: “\(\nu(E)=0\), so every subset of \(E\) has value zero.” In the first example, this would claim that \(\nu(\{a\})=0\), even though direct calculation gives \(\nu(\{a\})=3\). The correction is to use the definition: a null set requires zero value on every measurable subset. A zero total value supplies only the single equality \(\nu(E)=0\).
There is one useful condition that makes the tempting inference valid: sign-definiteness. If all measurable subsets of \(E\) have nonnegative values and \(\nu(E)=0\), then a subset \(A\) and its complement within \(E\) have nonnegative values whose sum is zero. Both must therefore be zero. The same reasoning applies when all subset values are nonpositive. Without that sign restriction, positive and negative contributions can cancel, as in the atomic and density examples.
This distinction matters whenever signed measures are used to describe mass with opposing signs. A null set has no signed mass on any measurable part, so it can be ignored when computing values on other sets. A zero-value set may still contain substantial positive and negative contributions. Treating these two kinds of sets as interchangeable loses information about how the signed measure is distributed.
Check Your Understanding
For each question, distinguish the value of a whole set from the values of its measurable subsets.
- For the atomic measure with weights \(3,-3,0\) on \(\{a,b,c\}\), list the zero-value sets and identify which are null.
- Prove directly from the definitions that every null set is both a positive set and a negative set.
- Suppose \(E\) is a negative set for a finite signed measure and \(\nu(E)=0\). Why must every measurable subset of \(E\) have value zero?
- For the density \(f=1\) on \([0,1)\) and \(f=-1\) on \([1,2]\), compute the signed value of \([0,\tfrac12)\cup[1,\tfrac32]\). Is this set null?
- Give an example of a zero-value set that is not null, and name a measurable subset that verifies it is not null.