Let (Ω,F)(\Omega, \mathcal{F}) be a measurable space.

Definition (Additive Set Function). μ:F→R\mu: \mathcal{F} \rightarrow \mathbb{R} is an additive set function, if for all disjoint A1,A2,…∈FA_1, A_2, \ldots \in \mathcal{F},

μ(∪n≥1An)=∑n≥1μ(An). \mu( \cup_{n \ge 1 } A_n ) = \sum_{n \ge 1} \mu (A_n).

Definition (Positive/Negative Set). A set A∈FA \in \mathcal{F} is positive (negative) if μ(E)≥0\mu (E) \ge 0 (resp μ(E)≤0\mu (E) \le 0), for each measurable set EE in AA.

The following theorem states that we can decompose Ω\Omega into a positive set and a negative one.

Theorem (The Hahn Decomposition). There exists a partition Ω=S+∪S−\Omega = S^+ \cup S^- such that

  1. S+S^+ is measurable and positive.
  2. S−S^- is measurable and negative.

Proof.

If Ω\Omega itself is positive, then the proof is trivial.

Otherwise,

s+=sup⁡{μ(E):E∈F}. s^+ = \sup \{ \mu (E) : E \in \mathcal{F} \}.

Suppose that there is an S+∈FS^+ \in \mathcal{F} such that μ(S+)=s+\mu(S^+) = s^+, then

  1. S+S^+ should be positive: if not, there exists a measurable E⊂S+E \subset S^+ such that μ(E)<0\mu(E) < 0. Then μ(S+−E)=μ(S+)−μ(E)>μ(S+)\mu(S^+ - E) = \mu(S^+) - \mu(E) > \mu(S^+), a contradiction.

  2. S−≐Ω−S+S^- \doteq \Omega - S^+ should be negative: if not, there exists a measurable E⊂S−E \subset S^- such that μ(E)>0\mu(E) > 0. Then μ(S+∪E)=μ(S+)+μ(E)>μ(S+)\mu(S^+ \cup E) = \mu(S^+) + \mu(E) > \mu(S^+), a contradiction.

It is left to prove that such S+S^+ exists.

  1. Let An∈FA_n \in \mathcal{F}, n∈N+n \in \mathbb{N}^+ be a sequence of measurable sets such that μ(An)↑s+\mu(A_n) \uparrow s^+.

  2. Let A=∪n≥1AnA = \cup_{n \ge 1} A_n.

  3. For each ω∈A\omega \in A, define the sequence xω=(xω[1],xω[2],…)x_\omega = (x_\omega [1], x_\omega[2], \ldots ), such that

    xω[n]={1,ω∈An,0,ω∈Aˉn.∀n∈N+. x_\omega [n] = \begin{cases} 1, & \omega \in A_n, \\ 0, & \omega \in \bar A_n. \end{cases} \qquad \forall n \in \mathbb{N}^+.

  4. For each n∈N+n \in \mathbb{N}^+, and each y∈{0,1}ny \in \{0, 1 \}^n, define

    Bn,y≐{ω∈A:xω=y}. B_{n, y} \doteq \{ \omega \in A : x_\omega = y \}.

    Then Pn≐{Bn,y:y∈{0,1}n}\mathcal{P}_n \doteq \{ B_{n,y} : y \in \{0, 1 \}^n \} constitutes a partition of AA. For each n∈N+n \in \mathbb{N}^+, the partition Pn+1\mathcal{P}_{n + 1} refines the partition Pn\mathcal{P}_n, in the sense that each Bn,y∈PnB_{n, y} \in \mathcal{P}_n can be decomposed into

    Bn,y=Bn+1,y∘0∪Bn+1,y∘1, B_{n, y} = B_{n + 1, y \circ 0 } \cup B_{n + 1, y \circ 1 },

    where y∘0,y∘1∈{0,1}n+1y \circ 0, y \circ 1 \in \{0, 1 \}^{n + 1} are the concatenation of yy with 00 and 11, respectively. Note that

    μ(Bn,y)=μ(Bn+1,y∘0)+μ(Bn+1,y∘1)≤max⁡{0,μ(Bn+1,y∘0)}+max⁡{0,μ(Bn+1,y∘1)}. \begin{aligned} \mu( B_{n, y} ) &= \mu( B_{n + 1, y \circ 0 } ) + \mu( B_{n + 1, y \circ 1 } ) \\ &\le \max \big\{ 0, \mu( B_{n + 1, y \circ 0 } ) \big\} + \max \big\{ 0, \mu( B_{n + 1, y \circ 1 } ) \big\}. \end{aligned}

  5. For n∈N+n \in \mathbb{N}^+, define

    Cn≐∪Bn,y∈Pn ∧ μ(Bn,y)≥0 Bn,y. C_n \doteq \underset{ B_{n, y} \in \mathcal{P}_n \, \wedge \, \mu( B_{n, y} ) \ge 0 }{\cup} \, B_{n, y}.

  6. For n∈N+n \in \mathbb{N}^+, define

    Dn≐∪k≥nCk. D_n \doteq \cup_{k \ge n} C_k.

    It is easy to see that μ(Dn)≥μ(An)\mu(D_n) \ge \mu(A_n). Further,

    • D1⊃D2⊃D3⊃…D_1 \supset D_2 \supset D_3 \supset \ldots is a decreasing sequence.

    • μ(Dn)≤μ(Dn+1)\mu(D_n) \le \mu(D_{n + 1}). Since if Bn,y⊂DnB_{n, y} \subset D_n, it can be replaced by Bn+1,y∘0∪Bn+1,y∘1B_{n + 1, y \circ 0 } \cup B_{n + 1, y \circ 1 } in Dn+1D_{n + 1}, incurring no lose in the measure. But we can also drop Bn+1,y∘0B_{n + 1, y \circ 0 } or Bn+1,y∘1B_{n + 1, y \circ 1 } if one of them has negative measure.

Finally, define S+S^+ by

S+=∩n≥1Dn=lim⁡nDn. S^+ = \cap_{n \ge 1} D_n = \lim_n D_n.

It holds that Dn↓S+D_n \downarrow S^+ and μ(Dn)↑μ(S+)\mu(D_n) \uparrow \mu(S^+).

□\square

Remark: Let R⊂RR \subset \mathbb{R}, and t∗=sup⁡Rt^* = \sup R. Then it is possible that t∗∉Rt^* \notin R. For example, let R={1−1/n:n∈N+}R = \{ 1 - 1 / n : n \in \mathbb{N}^+ \}. We see t∗=1=sup⁡Rt^* = 1 = \sup R. But 1∉R1 \notin R.

In our analysis, we can an S+∈FS^+ \in \mathcal{F} such that

μ(S+)=s+=sup⁡{μ(E):E∈F}, \mu(S^+) = s^+ = \sup \{ \mu (E) : E \in \mathcal{F} \},

due to 1) F\mathcal{F} is a σ\sigma-field, and 2) μ\mu is a signed measure.

There are other examples where the supremum is achieve.

Example. Let f:[0,1]→Rf: [0, 1] \rightarrow \mathbb{R} be a continuous function. Let R={f(x):x∈[0,1]}R = \{ f(x) : x \in [0, 1] \}, and t∗=sup⁡Rt^* = \sup R. It holds that t∗∈Rt^* \in R, since there exists x∗∈[0,1]x^* \in [0, 1], s.t., t∗=f(x∗)t^* = f(x^*). The claim is based on that 1) [0,1][0, 1] is closed; 2) ff is continuous.

Reference

[1]. H. Royden, P. Fitzpatrick Real Analysis
[2]. Billingsley. Patrick, Probability and measure