Variation Under Restriction and Addition
Total variation records the full size of a signed measure, including positive and negative mass. Restricting a signed measure to a measurable set removes its values outside that set; in this case the variation restricts exactly as well. By contrast, adding signed measures can produce cancellation, so the variation of a sum is bounded by, but need not equal, the sum of the variations.
Throughout, \((X,\mathcal{M})\) is a measurable space. All signed measures in this tutorial are real-valued finite signed measures: they are countably additive and have finite total variation on \(X\). In particular, their total variation is a finite positive measure. For a measurable set \(E\), a finite measurable partition of \(E\) means finitely many pairwise disjoint sets in \(\mathcal{M}\) whose union is \(E\). The empty set has the empty partition, whose sum is zero.
The partition definition measures the largest amount of signed mass that can be exposed by splitting a set into measurable pieces. It immediately makes restriction an exact operation on variation, while the triangle inequality for sums follows by applying the ordinary triangle inequality on every piece of a partition.
The Restriction Identity
Proof. Write \(\rho=\nu|_A\). If \(E=\varnothing\), then \(\rho(E)=0\) and both sides of the claimed identity are zero. Now let \(E\in\mathcal{M}\), and take any finite measurable partition \(E_1,\ldots,E_k\) of \(E\). The sets \(E_j\cap A\) partition \(E\cap A\), with empty pieces allowed. Therefore, $$ \sum_{j=1}^{k}|\rho(E_j)| =\sum_{j=1}^{k}|\nu(E_j\cap A)| \le |\nu|(E\cap A). $$ Taking the supremum over all partitions of \(E\) gives \(|\rho|(E)\le|\nu|(E\cap A)\). For the reverse inequality, take any finite measurable partition \(F_1,\ldots,F_l\) of \(E\cap A\). The sets \(F_1,\ldots,F_l,E\setminus A\) partition \(E\); if \(E\setminus A=\varnothing\), that last piece is simply omitted. Since \(\rho(F_i)=\nu(F_i)\) and \(\rho(E\setminus A)=\nu((E\setminus A)\cap A)=0\), this partition gives $$ |\rho|(E)\ge\sum_{i=1}^{l}|\nu(F_i)|. $$ Taking the supremum over partitions of \(E\cap A\) gives \(|\rho|(E)\ge|\nu|(E\cap A)\). Thus equality holds for every \(E\). The formula at \(E=X\) gives \(\|\rho\|_{\mathrm{TV}}=|\nu|(A)\le|\nu|(X)<\infty\), so the restriction is finite. \(\square\)
The identity includes the edge cases: if \(A=\varnothing\), the restricted measure is zero and its variation is zero; if \(A=X\), the restriction and its variation are unchanged. It also gives the setwise estimate \(|\nu|_{\text{variation of }\nu|_A}(E)\le|\nu|(E)\), since \(E\cap A\subseteq E\).
Worked Example: Restricting a Three-Point Signed Measure
Let \(X=\{a,b,c\}\), let \(\mathcal{M}\) be all subsets of \(X\), and define \(\nu=4\delta_a-3\delta_b+2\delta_c\), where \(\delta_x(E)=1\) if \(x\in E\) and \(0\) otherwise. Take \(A=\{a,c\}\). Compute the variation before and after restriction.
For a measure supported on finitely many points, the partition into singleton sets gives the sum of the absolute point masses. No partition can give a larger sum: for each partition piece, the absolute value of the sum of its point masses is at most the sum of their absolute values. Thus $$ |\nu|(X)=|4|+|-3|+|2|=4+3+2=9. $$ The restriction removes the mass at \(b\), so \(\nu|_A=4\delta_a+2\delta_c\). Its variation is $$ |\nu|_{\text{variation of }\nu|_A}(X)=|4|+|2|=4+2=6. $$ The restriction identity gives the same result because \(|\nu|(A)=|4|+|2|=6\). Check: for \(E=\{a,c\}\), the singleton partition gives variation \(4+2=6\), while the one-piece partition gives \(|\nu(E)|=|4+2|=6\); hence the supremum is exactly \(6\).
Bounds for Sums and Differences
Proof. The sum and difference of two real-valued countably additive measures are real-valued and countably additive, so \(\nu+\lambda\) and \(\nu-\lambda\) are signed measures. For any finite measurable partition \(E_1,\ldots,E_k\) of \(E\), the real-number triangle inequality gives $$ \sum_{j=1}^{k}|(\nu+\lambda)(E_j)| =\sum_{j=1}^{k}|\nu(E_j)+\lambda(E_j)| \le\sum_{j=1}^{k}|\nu(E_j)|+\sum_{j=1}^{k}|\lambda(E_j)| \le|\nu|(E)+|\lambda|(E). $$ Taking the supremum over partitions proves the first setwise inequality. For the difference, use \(|\nu(E_j)-\lambda(E_j)|\le|\nu(E_j)|+|\lambda(E_j)|\) on every piece and take the same supremum. Taking \(E=X\) gives both norm inequalities. Since the right sides at \(X\) are finite, the variations of the sum and difference are finite; hence both are finite signed measures, as asserted. \(\square\)
The setwise form is stronger than the norm form: it controls variation separately on every measurable set, not just on all of \(X\). The difference estimate is also a sum estimate, because \(\nu-\lambda=\nu+(-\lambda)\), and the partition definition gives \(|-\lambda|(E)=|\lambda|(E)\).
Worked Example: Addition Can Cancel Mass
On \(X=\{a,b\}\) with all subsets measurable, let \(\nu=2\delta_a-\delta_b\) and \(\lambda=-\delta_a+3\delta_b\). Calculate the variations of the measures and their sum and difference, then check both norm bounds.
The point masses of \(\nu\) are \(2\) at \(a\) and \(-1\) at \(b\), so $$ \|\nu\|_{\mathrm{TV}}=|2|+|-1|=2+1=3. $$ Those of \(\lambda\) are \(-1\) at \(a\) and \(3\) at \(b\), so $$ \|\lambda\|_{\mathrm{TV}}=|-1|+|3|=1+3=4. $$ Adding the coefficients point by point gives \(\nu+\lambda=\delta_a+2\delta_b\), and therefore $$ \|\nu+\lambda\|_{\mathrm{TV}}=|1|+|2|=1+2=3\le3+4=7. $$ Subtracting gives \(\nu-\lambda=3\delta_a-4\delta_b\), so $$ \|\nu-\lambda\|_{\mathrm{TV}}=|3|+|-4|=3+4=7=3+4. $$ Check: the singleton partition \(\{\{a\},\{b\}\}\) gives \(3\) for the variation of the sum and \(7\) for the variation of the difference. The one-piece partition of \(X\) gives respectively \(|1+2|=3\) and \(|3-4|=1\), so neither one-piece sum exceeds the displayed singleton-partition value.
Worked Example: Setwise Bounds on a Restricted Set
Let \(X=\{p,q,r\}\), with all subsets measurable, and set \(\nu=5\delta_p-2\delta_q+\delta_r\) and \(\lambda=-3\delta_p+4\delta_q-\delta_r\). Take \(E=\{p,r\}\). Compute the variation of the sum and difference on \(E\), and compare them with the setwise bound.
On \(E\), the masses of \(\nu\) are \(5\) at \(p\) and \(1\) at \(r\), so \(|\nu|(E)=|5|+|1|=6\). The masses of \(\lambda\) there are \(-3\) and \(-1\), so \(|\lambda|(E)=|-3|+|-1|=3+1=4\). Thus the right side of either setwise inequality is \(6+4=10\).
On \(E\), the sum has masses \(5+(-3)=2\) at \(p\) and \(1+(-1)=0\) at \(r\). Hence \(|\nu+\lambda|(E)=|2|+|0|=2\le10\). The difference has masses \(5-(-3)=8\) at \(p\) and \(1-(-1)=2\) at \(r\). Hence \(|\nu-\lambda|(E)=|8|+|2|=10\le10\). Check: the partition into \(\{p\}\) and \(\{r\}\) gives \(2\) for the sum and \(10\) for the difference; the one-piece partition gives \(|2+0|=2\) and \(|8+2|=10\), respectively. The values agree with the partition definition in both cases.
What the Inequality Does—and Does Not—Say
The triangle inequality is a bound, not a claim of equality. When positive and negative contributions from the two measures oppose one another on the same region, addition can reduce variation. The restriction identity has a different character: it is always an equality because restricting simply removes the part outside \(A\), without combining the remaining values of the measure.
Common mistake: assuming variation is additive under addition. In the preceding two-point example, an incorrect calculation would be $$ \|\nu+\lambda\|_{\mathrm{TV}}=\|\nu\|_{\mathrm{TV}}+\|\lambda\|_{\mathrm{TV}}=3+4=7. $$ But the actual sum is \(\delta_a+2\delta_b\), whose variation is \(|1|+|2|=3\). The correction is the triangle inequality \(\|\nu+\lambda\|_{\mathrm{TV}}\le7\), not an equality in general. The difference in that example happens to attain the bound, but this does not make equality automatic for differences either.
These estimates are useful whenever signed measures are combined or restricted: they ensure the resulting measure remains finite and quantify how large its variation can be. They also clarify why cancellation must be checked rather than ignored. If either measure is zero, the inequalities remain valid, with equality for the sum or difference as appropriate; if \(X=\varnothing\), all the variations are zero. No exceptional case needs a separate convention beyond the empty partition having sum zero.
Check Your Understanding
Use the partition definition of total variation to justify each answer.
- For a finite signed measure \(\nu\), why does every partition of \(E\) induce a partition of \(E\cap A\) when proving the upper bound in the restriction identity?
- Suppose \(X=\{u,v\}\), \(\nu=3\delta_u-\delta_v\), and \(\lambda=-2\delta_u+4\delta_v\). Compute the variations of \(\nu+\lambda\) and \(\nu-\lambda\), and compare each with \(\|\nu\|_{\mathrm{TV}}+\|\lambda\|_{\mathrm{TV}}\).
- State the setwise triangle inequality for \(|\nu+\lambda|(E)\). Why does it imply the corresponding norm inequality?
- Give a reason that \(\|\nu+\lambda\|_{\mathrm{TV}}\) need not equal \(\|\nu\|_{\mathrm{TV}}+\|\lambda\|_{\mathrm{TV}}\).
- If \(A=\varnothing\), what are \(\nu|_A\) and \(\|\nu|_A\|_{\mathrm{TV}}\), and how does the restriction identity verify the result?