Restriction to a Measurable Partition

Splitting a Signed Measure Across Measurable Pieces

A measurable set can be used to isolate the part of a signed measure lying on that set. The resulting restriction is itself a signed measure, and its value on any measurable test set is found by intersecting the test set with the region being retained. For a partition, these intersections divide the test set into disjoint pieces, so countable additivity gives an exact reconstruction of the original value.

Throughout, $(X,\mathcal{M})$ is a measurable space and $\nu$ is a signed measure on it. We use the convention that a signed measure is countably additive and does not take both $+\infty$ and $-\infty$ as values; the usual extended-real restrictions on sums apply. No finiteness, completeness, or sigma-finiteness assumption is needed here. Every set in the formulas below belongs to $\mathcal{M}$, and equalities are equalities of set-function values, not almost-everywhere statements.

Definition: For $A\in\mathcal{M}$, the restriction of $\nu$ to $A$ is the set function $\nu_A$ on $\mathcal{M}$ defined by $\nu_A(E)=\nu(E\cap A)$ for every $E\in\mathcal{M}$. Thus the domain remains $\mathcal{M}$; the restriction records only the part of $E$ inside $A$.

This definition is different from normalizing a measure on $A$: there is no division by $\nu(A)$, and $\nu_A(X)=\nu(A)$, not necessarily $1$. It is also not a claim that $\nu_A(E)=\nu(E)$ for every $E$. Rather, it gives $\nu_A(E)=\nu(E)$ when $E\subseteq A$, and $\nu_A(E)=0$ when $E\cap A=\varnothing$.

Restriction and Partition Formulas

Theorem: For each $A\in\mathcal{M}$, the restriction $\nu_A$ is a signed measure on $(X,\mathcal{M})$. If $A_1,\ldots,A_k$ is a finite measurable partition of $X$, then for every $E\in\mathcal{M}$, $$ \nu(E)=\sum_{j=1}^{k}\nu_{A_j}(E). $$ The same reconstruction holds for a countable measurable partition, with the series interpreted in the countable-additivity sense for signed measures.

Proof. First, $\nu_A(\varnothing)=\nu(\varnothing\cap A)=\nu(\varnothing)=0$. Let $E_1,E_2,\ldots$ be pairwise disjoint sets in $\mathcal{M}$. The sets $E_n\cap A$ are measurable and pairwise disjoint: if a point belonged to both $E_i\cap A$ and $E_j\cap A$ for $i\ne j$, it would belong to both $E_i$ and $E_j$. Also, \[ \left(\bigcup_{n=1}^{\infty}E_n\right)\cap A =\bigcup_{n=1}^{\infty}(E_n\cap A). \] Countable additivity of $\nu$ therefore gives \[ \nu_A\left(\bigcup_{n=1}^{\infty}E_n\right) =\nu\left(\bigcup_{n=1}^{\infty}(E_n\cap A)\right) =\sum_{n=1}^{\infty}\nu(E_n\cap A) =\sum_{n=1}^{\infty}\nu_A(E_n). \] Thus $\nu_A$ is a signed measure. For a finite partition, the sets $E\cap A_j$ are disjoint and their union is $E$, so finite additivity gives $\nu(E)=\sum_{j=1}^{k}\nu(E\cap A_j)=\sum_{j=1}^{k}\nu_{A_j}(E)$. For a countable partition the same sets are countably many disjoint sets with union $E$; countable additivity gives the asserted series identity. This includes empty cells, since their intersections contribute $\nu(\varnothing)=0$. $\square$

In particular, a two-piece partition $X=A\cup A^c$ gives $\nu(E)=\nu_A(E)+\nu_{A^c}(E)$ for every measurable $E$. The complement is essential: restricting to $A$ alone discards the part of $E$ outside $A$.

Worked Example: Restricting an Atomic Signed Measure

Let $X=\{a,b,c,d\}$, let $\mathcal{M}$ be the set of all subsets of $X$, and define $\nu(E)$ as the sum of the weights of the points in $E$, with weights $w(a)=3$, $w(b)=-2$, $w(c)=0$, and $w(d)=4$. Use the partition $A_1=\{a,c\}$, $A_2=\{b,d\}$, and compute the restrictions on $E=\{a,b,d\}$.

By definition, $\nu_{A_1}(E)=\nu(E\cap A_1)$. Here $E\cap A_1=\{a\}$, so $\nu_{A_1}(E)=3$. Likewise $E\cap A_2=\{b,d\}$, and hence $\nu_{A_2}(E)=(-2)+4=2$. Therefore $$ \nu_{A_1}(E)+\nu_{A_2}(E)=3+2=5. $$ Directly, $\nu(E)=w(a)+w(b)+w(d)=3+(-2)+4=5$, as required. Also, $\nu_{A_1}(X)=\nu(\{a,c\})=3+0=3$ and $\nu_{A_2}(X)=\nu(\{b,d\})=-2+4=2$; their sum is $\nu(X)=3+(-2)+0+4=5$. The zero-weight point $c$ lies in the first partition cell but not in $E$, so it contributes to $\nu_{A_1}(X)$ and not to $\nu_{A_1}(E)$.

Worked Example: Restricting a Measure Defined by a Density

On $(\mathbb{R},\mathcal{L})$ with Lebesgue measure $m$, let $g=2\chi_{[0,1]}-\chi_{(1,3]}$, and define $\nu(E)=\int_E g\,dm$ for $E\in\mathcal{L}$. This is a finite signed measure because $g$ is integrable: $\int_{\mathbb{R}}|g|\,dm=2\cdot1+1\cdot2=4$. Restrict to the partition $A=[0,1]$, $A^c=\mathbb{R}\setminus[0,1]$, and evaluate both restrictions on $E=[\tfrac12,\tfrac52]$.

The intersection $E\cap A=[\tfrac12,1]$ has Lebesgue measure $1-\tfrac12=\tfrac12$, and $g=2$ there. Thus $\nu_A(E)=2\cdot\tfrac12=1$. The other intersection is $E\cap A^c=(1,\tfrac52]$, which has measure $\tfrac52-1=\tfrac32$; on it $g=-1$. Therefore $\nu_{A^c}(E)=(-1)\cdot\tfrac32=-\tfrac32$. Their sum is $1-\tfrac32=-\tfrac12$.

Check: On $E$, the density is $2$ on $[\tfrac12,1]$, whose contribution is $2(\tfrac12)=1$, and $-1$ on $(1,\tfrac52]$, whose contribution is $-(\tfrac32)=-\tfrac32$. The direct integral is $1-\tfrac32=-\tfrac12=\nu_A(E)+\nu_{A^c}(E)$. The shared endpoint $1$ is included in $A$, while its singleton has Lebesgue measure zero, so the endpoint convention does not alter either computed value.

Sign-Definite Cells

A set $P\in\mathcal{M}$ is called a positive set for $\nu$ if $\nu(E)\ge0$ for every measurable $E\subseteq P$. A set $N\in\mathcal{M}$ is a negative set if $\nu(E)\le0$ for every measurable $E\subseteq N$. These definitions concern all measurable subsets of the set, not just the value on the set itself.

Proposition: If $P$ is a positive set for $\nu$, then $\nu_P$ is a nonnegative measure. If $N$ is a negative set, then $-\nu_N$, defined by $(-\nu_N)(E)=-\nu_N(E)$, is a nonnegative measure.

Proof. For any $E\in\mathcal{M}$, $E\cap P$ is a measurable subset of $P$. The positive-set condition gives $\nu_P(E)=\nu(E\cap P)\ge0$, and the restriction theorem gives countable additivity. Thus $\nu_P$ is a nonnegative measure. For a negative set $N$, the set $E\cap N$ is a measurable subset of $N$, so $\nu_N(E)=\nu(E\cap N)\le0$. It follows that $(-\nu_N)(E)\ge0$; multiplying the countable-additivity identity for $\nu_N$ by $-1$ proves countable additivity of $-\nu_N$. This argument also covers empty sets and sets whose restricted values are zero. $\square$

Worked Example: A Partition into Sign-Definite Pieces

On $X=\{a,b,c,d\}$, use the atomic signed measure with weights $3,-2,0,4$ from the earlier calculation. Let $P=\{a,c,d\}$ and $N=\{b\}$. Verify that these are positive and negative sets, respectively, and calculate the signed decomposition on $E=\{a,b,d\}$.

Every subset of $P$ has value equal to a sum chosen from the nonnegative weights $3,0,4$, so its value is nonnegative; hence $P$ is positive. The subsets of $N$ are $\varnothing$ and $\{b\}$, with values $0$ and $-2$, so $N$ is negative. Since $P,N$ partition $X$, the restrictions give $\nu_P(E)=\nu(\{a,d\})=3+4=7$ and $\nu_N(E)=\nu(\{b\})=-2$. Consequently $\nu(E)=\nu_P(E)+\nu_N(E)=7+(-2)=5$. Equivalently, the nonnegative measure $-\nu_N$ has value $2$ on $E$, and $\nu(E)=\nu_P(E)-(-\nu_N)(E)=7-2=5$.

Check: Direct summation on $E$ gives $3+(-2)+4=5$, matching both restriction formulas. This example uses a partition already known to have sign-definite cells; it does not assert that every signed measure comes with such a partition.

Common Mistakes and Scope

A frequent error is to write $\nu_A(E)=\nu(E)$ for arbitrary $E$. For example, in the atomic example, take $A=\{a,c\}$ and $E=\{a,b,d\}$. The incorrect calculation would give $\nu_A(E)=\nu(E)=3-2+4=5$. The definition instead gives $E\cap A=\{a\}$, so the correct value is $\nu_A(E)=\nu(\{a\})=3$. The missing contribution $2$ is precisely $\nu_{A^c}(E)$, and $3+2=5$.

Another error is to infer that a restriction is nonnegative merely because it is a measure in the signed-measure sense. A general restriction can still take negative values: in the atomic example, restricting to $A=\{a,b\}$ gives $\nu_A(\{b\})=-2$. Nonnegativity follows from a sign condition on every measurable subset of the restricting set, as in the proposition, not from restriction alone.

Restriction is useful whenever a signed measure must be evaluated piece by piece. On a finite or countable measurable partition, first intersect the test set with each cell, then evaluate the corresponding restricted values and add them using the partition theorem. For countable partitions, use the signed measure's countable-additivity convention; do not reorder a series of signed values unless a separate result justifies doing so.

Check Your Understanding

For each question, write the restriction values using intersections with the stated measurable sets.

For a signed measure $\nu$ and measurable $A$, show directly from the definition that $\nu_A(X)=\nu(A)$ and that $\nu_A(E)=0$ whenever $E\cap A=\varnothing$.
If $X=A\cup A^c$, express $\nu(E)$ as the sum of the two restrictions for an arbitrary $E\in\mathcal{M}$.
In the atomic example with weights $3,-2,0,4$, calculate $\nu_{\{b,d\}}(\{a,b,c\})$.
Why does $\nu_P(E)\ge0$ for every measurable $E$ follow when $P$ is a positive set?
Give a concrete example showing that a restriction to a measurable set need not be nonnegative.

Restriction to a Measurable Partition

What You'll Learn