Signed and Complex Measures Hahn and Jordan — Examples

By [L1], for every measurable $E$ one has [L1, L2] $\nu (E)={\mathbf{1}}_E(1)-{\mathbf{1}}_E(-1)$. If $E\subseteq P$ is measurable, then $-1\mathrm{\notin }E$, so $\nu (E)={\mathbf{1}}_E(1)\ge 0$. If $E\subseteq N$ is measurable, then $1\mathrm{\notin }E$, so $\nu (E)=-{\mathbf{1}}_E(-1)\le 0$. Thus $P$ is positive and $N$ is negative.

Since $\mathbb{R}=P\bigsqcup N$, step 1.1 gives a Hahn decomposition. [L2, step 1.1] The Jordan measures are therefore ${\delta }_1$ and ${\delta }_{-1}$, so $\mathrm{\mid }\nu \mathrm{\mid }={\delta }_1+{\delta }_{-1}$ by [L2]. ∎

By [A1], the density theorem [L1] makes $(0,\pi )$ positive and $\{0\}\cup [\pi ,2\pi ]$ negative for $\nu$. Thus $P_0,N_0$ is a Hahn decomposition.

The points $0$ and $\pi$ are $\nu$-null because $\sin x=0$ there. Moving those null points from $N_0$ to $P_1$ preserves positivity and negativity, so $P_1,N_1$ is another Hahn decomposition. It is distinct from $P_0,N_0$ and compatible with the uniqueness clause of [L2].

The defining series is absolutely convergent on every subset of [L1, A1] $\mathbb{Z}$, so $\nu$ is countably additive. If $E$ contains only even integers, then every term in the series for $\nu (E)$ is nonnegative; if $E$ contains only odd integers, every term is nonpositive. Thus the even integers form a positive set and the odd integers form a negative set.

The Jordan positive part is therefore the even-atom measure [L1, A1, step 1.1] ${\nu }^{+}(A)={\sum }_{k\in A,k\text{ even}}{2}^{-\mathrm{\mid }k\mathrm{\mid }}$ and the negative part is ${\nu }^{-}(A)={\sum }_{k\in A,k\text{ odd}}{2}^{-\mathrm{\mid }k\mathrm{\mid }}$. Hence [L1] gives $\mathrm{\mid }\nu \mathrm{\mid }(\mathbb{Z})={\nu }^{+}(\mathbb{Z})+{\nu }^{-}(\mathbb{Z})={\sum }_{k\in \mathbb{Z}}{2}^{-\mathrm{\mid }k\mathrm{\mid }}=3$ by [A1].

By [L1], ${\mu }_c$ vanishes on measurable subsets of $\mathbb{R}\setminus C$, [L1, L2] while ${\lambda }_0$ vanishes on measurable subsets of $C$ because $\lambda (C)=0$. Thus ${\mu }_c\perp {\lambda }_0$. Both are positive measures, so $\nu ={\mu }_c-{\lambda }_0$ is a signed measure already written as a difference of mutually singular positive measures.

The uniqueness clause in [L2] now forces [L2, step 1.1] ∎ ${\nu }^{+}={\mu }_c,{\nu }^{-}={\lambda }_0\mathrm{.}$ Hence ${\mu }_c-{\lambda }_0$ is already in Jordan form.

The function $h(x)=e^{ix}$ is bounded on the finite interval $[0,2\pi ]$, [L1, A1] so it lies in $L^1(\lambda )$. Therefore [L1] makes $\nu (E)={\int }_E{e}^{ix}d\lambda$ a complex measure.

Applying [L1] and then [A1] on the whole interval gives [L1, A1, step 1.1] $\mathrm{\mid }\nu \mathrm{\mid }([0,2\pi ])={\int }_0^{2\pi }\mathrm{\mid }{e}^{ix}\mathrm{\mid }dx={\int }_0^{2\pi }1dx=2\pi \mathrm{.}$.

Let $E_1,\dots ,E_m$ be a finite measurable partition of $[0,2\pi ]$. Every [L1, A1] piece of positive measure satisfies the strict inequality from [A1], and the null pieces contribute $0$. Therefore ${\sum }_{j=1}^m\mathrm{\mid }\nu (E_j)\mathrm{\mid }<{\sum }_{j=1}^m\lambda (E_j)=2\pi \mathrm{.}$ So no finite partition attains the total variation value $2\pi$. [L1, A1].

By [A2], countable partitions can produce sums arbitrarily close to $2\pi$. [L1, A2, step 1.1] Combining this with step 1.1 and [L1] shows that the total-variation value $2\pi$ is not attained by any finite partition.

Because every measurable subset of $X$ has $\nu$-value $0$, every [L1] measurable set is both positive and negative. Thus $P_0=\mathrm{\varnothing },N_0=X\text{and}{P}_1=\{0\},N_1=\{1\}$ are both Hahn decompositions.

The two decompositions are different, but the moved set $\{0\}$ is [L1, L2, step 1.1] ∎ $\nu$-null and therefore has total variation $0$ by [L2]. This is exactly the allowed nonuniqueness in [L1].

By [L1], one has $\nu (E)={\delta }_1(E)-{\delta }_{-1}(E)=1-1=0$, so [L1] $\mathrm{\mid }\nu (E)\mathrm{\mid }=0$.

The two singletons $\{1\}$ and $\{-1\}$ form a measurable partition of [L1, L2, step 1.1] $E$, and [L1] gives $\mathrm{\mid }\nu (\{1\})\mathrm{\mid }+\mathrm{\mid }\nu (\{-1\})\mathrm{\mid }=1+1=2.$ Therefore [L2] yields $\mathrm{\mid }\nu \mathrm{\mid }(E)\ge 2>0=\mathrm{\mid }\nu (E)\mathrm{\mid }$, refuting the claim.

By [A2], the value of $\varphi$ on a finite disjoint union of half-open intervals is the sum of the endpoint increments of $g$, so [A1] makes $\varphi$ a finitely additive finite-valued set function on $\mathcal{A}$.

For each $N$, partition the interval $(0,u_1]$ by the ordered points $0<\cdots <v_N<u_N<\cdots <v_1<u_1$. The resulting finite partition sum for $\varphi$ is at least ${\sum }_{n=1}^N\mathrm{\mid }g(u_n)-g(v_n)\mathrm{\mid }={\sum }_{n=1}^N(u_n+v_n)\mathrm{.}$ By [A3], these lower bounds diverge with $N$, so [A4] gives $\mathrm{\mid }\varphi \mathrm{\mid }((0,u_1])=+\mathrm{\infty }$ even though every value of $\varphi$ is finite.

The displayed claim asserts exactly the negation of the second clause in [L1] [L1].

Therefore no signed measure satisfies the claim, and the statement is [L1, step 1.1] ∎ false.

Every measurable subset of $X$ is both positive and negative for the zero [L1] measure, so both $\mathrm{\varnothing }\bigsqcup X\text{and}\{0\}\bigsqcup \{1\}$ are Hahn decompositions.

These decompositions are different, so exact uniqueness fails. This is [L1, step 1.1] ∎ compatible with [L1] because the differing set is null.

By [L1], one has $\nu (E)=0$, so $\mathrm{\mid }\nu (E)\mathrm{\mid }=0$. [L1].

The partition $E=\{1\}\bigsqcup \{-1\}$ gives partition sum [L1, L2, step 1.1] $\mathrm{\mid }\nu (\{1\})\mathrm{\mid }+\mathrm{\mid }\nu (\{-1\})\mathrm{\mid }=2$, so [L2] yields $\mathrm{\mid }\nu \mathrm{\mid }(E)\ge 2$. Hence $\mathrm{\mid }\nu \mathrm{\mid }(E)\mathrm{\ne }\mathrm{\mid }\nu (E)\mathrm{\mid }$.

The family $\mathcal{P}=\{\{0\}\}$ is a pi-system by [L1], and [L1, L2] $\sigma (\mathcal{P})=\mathcal{P}(X)$ because complements and unions recover $\{1\}$ and $X$. Define signed measures $\mu (A):=0,\nu (A):={\mathbf{1}}_A(1)\mathrm{.}$ They agree on the generating pi-system element $\{0\}$.

However $\mu (\{1\})=0$ while $\nu (\{1\})=1$, so the signed measures are not [L2, step 1.1] equal on the generated sigma-algebra. Therefore the claim is false.

The endpoint-increment formula makes $\varphi$ finitely additive on $\mathcal{A}$, and every value of $\varphi$ is a finite real number.

Using the partition points from [A2] gives finite partition sums bounded below by ${\sum }_{n=1}^N(u_n+v_n)$ for arbitrarily large $N$. Because those lower bounds diverge, [A3] gives infinite total variation.

Let $E_1,\dots ,E_m$ be any finite measurable partition of $[0,2\pi ]$. [L1, A1] Applying [A1] on each positive-measure piece and summing gives ${\sum }_{j=1}^m\mathrm{\mid }\nu (E_j)\mathrm{\mid }<{\sum }_{j=1}^m\lambda (E_j)=2\pi \mathrm{.}$

By [L1], the total variation of the whole interval is exactly $2\pi$, so [L1, step 1.1] ∎ step 1.1 shows that no finite partition attains it. Therefore the claim is false.