Fair upfront warning: This is not a particularly readable proof section (though much better than Section 2 about belief functions). There's dense notation, logical leaps due to illusion of transparency since I've spent a month getting fluent with these concepts, and a relative lack of editing since it's long. If you really want to read this, I'd suggest PM-ing me to get a link to MIRIxDiscord, where I'd be able to guide you through it and answer questions.

Proposition 1: If $f\in C(X,[0,1\left]\right)$ then $f^{+}:(m,b)\mapsto m\left(f\right)+b$ is a positive functional on $M^{sa}\left(X\right)$.

Proof Sketch: We just check three conditions. Linearity, being nonnegative on $M^{sa}\left(X\right)$, and continuity.

Linearity proof. Using $a,a^{\prime }$ for constants,

$f^{+}\left(a\right(m,b)+a^{\prime }(m^{\prime },b^{\prime }\left)\right)=f^{+}(am+a^{\prime }{m}^{\prime },ab+a{b}^{\prime })=(am+a^{\prime }{m}^{\prime })\left(f\right)+ab+a^{\prime }{b}^{\prime }$

$=a\left(m\right(f)+b)+a^{\prime }\left(m^{\prime }\right(f)+b^{\prime })=a{f}^{+}(m,b)+a^{\prime }{f}^{+}(m^{\prime },b^{\prime })$

So we have verified that $f^{+}(aM+a^{\prime }{M}^{\prime })=a{f}^{+}\left(M\right)+a^{\prime }{f}^{+}\left(M^{\prime }\right)$ and we have linearity.

Positivity proof: An sa-measure $M$, writeable as $(m,b)$ has $m$ uniquely writeable as a pair of finite measures $m^{+}$ (all the positive regions) and a $m^{-}$ (all the negative regions) by the Jordan Decomposition Theorem, and $b+m^{-}\left(1\right)\ge 0$. So,

$f^{+}\left(M\right)=m\left(f\right)+b=m^{+}\left(f\right)+m^{-}\left(f\right)+b\ge 0+m^{-}\left(1\right)+b\ge 0$

The first $\ge$ by $1\ge f\ge 0$, so the expectation of $f$ is positive and $m^{-}$ is negative so taking the expectation of 1 is more negative. The second $\ge$ is by the condition on how $m^{-}$ relates to $b$.

Continuity proof: Fix a sequence $(m_n,b_n)$ converging to $(m,b)$. Obviously the $b$ part converges, so now we just need to show that $m_n\left(f\right)$ converges to $m\left(f\right)$. The metric we have on the space of finite signed measures is the KR-metric, which implies the thing we want. This only works for continuous $f$, not general $f$.

Theorem 1: Every positive functional on $M^{sa}\left(X\right)$ can be written as $(m,b)\mapsto c\left(m\right(f)+b)$, where $c\ge 0$, and $f\in C(X,[0,1\left]\right)$

Proof Sketch: The first part is showing that it's impossible to have a positive functional where the $b$ term doesn't matter, without the positive functional being the one that maps everything to 0. The second part of the proof is recovering our $f$ by applying the positive functional to Dirac-delta measures ${\delta }_x$, to see what the function must be on point $x$.

Part 1:  Let's say $f^{+}$ isn't 0, ie there's some nonzero $(m,b)$ pair where $f^{+}(m,b)>0$, and yet $f^{+}(0,1)=0$ (which, by linearity, means that $f^{+}(0,b)=0$ for all $b$). We'll show that this situation is impossible.

Then, $0<f^{+}(m,b)=f^{+}(m^{+},0)+f^{+}(m^{-},b)$ by our starting assumption, and Jordan decomposition of $m$, along with linearity of positive functionals. Now, $f^{+}(m^{-},b)+f^{+}(-2(m^{-}),0)=f^{+}(-(m^{-}),b)$ because positive functionals are linear, and everything in that above equation is an sa-measure (flipping a negative measure makes a positive measure, which doesn't impose restrictions on the $b$ term except that it be $\ge 0$).  And so, by nonnegativity of positive functionals on sa-measures, $f^{+}(m^{-},b)\le f^{+}(-(m^{-}),b)$. Using this, we get

$f^{+}(m^{+},0)+f^{+}(m^{-},b)\le f^{+}(m^{+},0)+f^{+}(-(m^{-}),b)$

$=f^{+}(m^{+},0)+f^{+}(-(m^{-}),0)+f^{+}(0,b)=f^{+}(m^{+},0)+f^{+}(-(m^{-}),0)$

Another use of linearity was invoked for the first $=$ in the second line, and then the second $=$ made use of our assumption that $f^{+}(0,b)=0$ for all $b$.

At this point, we have derived that $0<f^{+}(m^{+},0)+f^{+}(-(m^{-}),0)$. Both of these are positive measures. So, there exists some positive measure $m^{\prime }$ where $f^{+}(m^{\prime },0)>0$.

Now, observe that, for all $b$, $0=f^{+}(0,b)=f^{+}(m^{\prime },0)+f^{+}(-(m^{\prime }),b)$

Let b be sufficiently huge to make $(-(m^{\prime }),b)$ into an sa-measure. Also, since $f^{+}(m^{\prime },0)>0$, $f^{+}(-(m^{\prime }),b)<0$, which is impossible because positive functionals are nonnegative on all sa-measures. Contradiction. Due to the contradiction, if there's a nonzero positive functional, it must assign $f^{+}(0,1)>0$, so let $f^{+}(0,1)$ be our $c$ term.

Proof part 2: Let's try to extract our f. Let $f\left(x\right):=\frac{f^{+}({\delta }_x,0)}{f^{+}(0,1)}$ This is just recovering the value of the hypothesized $f$ on $x$ by feeding our positive functional the measure ${\delta }_x$ that assigns 1 value to $x$ and nothing else, and scaling. Now, we just have to verify that this $f$ is continuous and in $[0,1]$.

For continuity, let $x_n$ limit to $x$. By the KR-metric we're using, $({\delta }_{x_n},0)$ limits to $({\delta }_x,0)$. By continuity of $f^{+}$, $f^{+}({\delta }_{x_n},0)$ limits to $f^{+}({\delta }_x,0)$. Therefore, $f\left(x_n\right)$ limits to $f\left(x\right)$ and we have continuity.

For a lower bound, $f\ge 0$, because $f\left(x\right)$ is a ratio of two nonnegative numbers, and the denominator isn't 0.

Now we just have to show that $f\le 1$. For contradiction, assume there's an $x$ where $f\left(x\right)>1$. Then $\frac{f^{+}({\delta }_x,0)}{f^{+}(0,1)}>1$, so $f^{+}({\delta }_x,0)>f^{+}(0,1)$, and in particular, $f^{+}(0,1)-f^{+}({\delta }_x,0)<0$.

But then, $f^{+}(-({\delta }_x),1)+f^{+}({\delta }_x,0)=f^{+}(0,1)$, so $f^{+}(-({\delta }_x),1)=f^{+}(0,1)-f^{+}({\delta }_x,0)<0$

However, $(-({\delta }_x),1)$ is an sa-measure, because ${\delta }_x\left(1\right)+1=0$, and must have nonnegative value, so we get a contradiction. Therefore, $f\in C(X,[0,1\left]\right)$.

To wrap up, we can go:

$f^{+}(m,b)=f^{+}(m,0)+f^{+}(0,b)=\frac{f^{+}(0,1)}{f^{+}(0,1)}\left({\int }_X\right(f^{+}({\delta }_x,0))dm+f^{+}(0,b\left)\right)$

$=f^{+}(0,1)({\int }_X\frac{f^{+}({\delta }_x,0)}{f^{+}(0,1)}dm+\frac{f^{+}(0,b)}{f^{+}(0,1)})=c\left({\int }_{X}f\right(x)dm+b)=c\left(m\right(f)+b)$

And $c\ge 0$, and $f\in C(X,[0,1\left]\right)$, so we're done.

Lemma 1: Compactness Lemma: Fixing some nonnegative constants ${\lambda }^{◯}$ and $b^{◯}$, the set of sa-measures where $m^{+}\left(1\right)\in [0,{\lambda }^{◯}]$, $b\in [0,b^{◯}]$, is compact. Further, if a set lacks an upper bound on $m^{+}\left(1\right)$ or on $b$, it's not compact.

Proof Sketch: We fix an arbitrary sequence of sa-measures, and then use the fact that closed intervals are compact-complete and the space $\Delta X$ is compact-complete to isolate a suitable convergent subsequence. Since all sequences have a limit point, the set is compact. Then, we go in the other direction, and get a sequence with no limit points assuming either a lack of upper bounds on $m^{+}\left(1\right)$, or a lack of upper bounds on $b$.

Proof: Fix some arbitrary sequence $M_n$ wandering about within this space, which breaks down into $(m_n^{+},0)+(m_n^{-},b_n)$, and then, since all measures are just a probability distribution scaled by the constant $m\left(1\right)$, it further breaks down into $\left(m_n^{+}\right(1)\cdot {\mu }_n,0)+\left(m_n^{-}\right(1)\cdot {\mu }_n^{\prime },b_n)$. Since $b_n+m_n^{-}\left(1\right)\ge 0$, $m_n^{-}\left(1\right)$ must be bounded in $[-b^{◯},0]$.

Now, what we can do is extract a subseqence where $b_n$ ,$m_n^{+}\left(1\right)$, $m_n^{-}\left(1\right)$, ${\mu }_n$, and ${\mu }_n^{\prime }$ all converge, by Tychonoff's Theorem (finite product, no axiom of choice required) Our three number sequences are all confined to a bounded interval, and our two probability sequences are wandering around within $\Delta X$ which is a compact complete metric space if $X$ is. The limit of this subsequence is a limit point of the original sequence, since all its components are arbitrarily close to the components that make up $M_n$ for large enough n in our subsequence.

The limiting value of $m^{+}\left(1\right)$ and $b$ both obey their respective bounds, and the cone of sa-measures is closed, so the limit point is an sa-measure and respects the bounds too. Therefore the set is compact, because all sequences of points in it have a limit point.

In the other direction, assume a set $B$ has unbounded $b$ values. Then we can fix a sequence $(m_n,b_n)\in B$ where $b_n$ increases without bound, so the a-measures can't converge. The same applies to all subsequences, so there's no limit point, so $B$ isn't compact.

Now, assume a set $B$ has bounded $b$ values, call the least upper bound $b^{\odot }$, but the value of $m^{+}\left(1\right)$ is unbounded. Fix a sequence $(m_n,b_n)\in B$ where $m_n^{+}\left(1\right)$ is unbounded above. Assume a convergent subsequence exists. Since $b_n+m_n^{-}\left(1\right)\ge 0$, $m_n^{-}\left(1\right)$ must be bounded in $[-b^{\odot },0]$. Then because $m_n\left(1\right)=m_n^{+}\left(1\right)+m_n^{-}\left(1\right)\ge m_n^{+}\left(1\right)-b^{\odot }$, and the latter quantity is finite, $m_n\left(1\right)$ must be unbounded above. However, in order for the $m_n$ to limit to some $m$, ${lim}_{n\to \infty }{m}_n\left(1\right)=m\left(1\right)$, which results in a contradiction. Therefore, said convergent subsequence doesn't exist, and $B$ is not compact.

Put together, we have a necessary-and-sufficient condition for a closed subset of $M^{sa}\left(X\right)$ to be compact. There must be an upper bound on $b$ and $m^{+}\left(1\right)$, respectively.

Lemma 2: The upper completion of a closed set of sa-measures is closed.

Proof sketch: We'll take a convergent sequence $(m_n,b_n)$ in the upper completion of $B$ that limits to $(m,b)$, and show that, in order for it to converge, the same sorts of bounds as the Compactness Lemma uses must apply. Then, breaking down $(m_n,b_n)$ into $(m_n^B,b_n^B)+(m_n^{\ast },b_n^{\ast })$, where $(m_n^B,b_n^B)\in B$, and $(m_n^{\ast },b_n^{\ast })$ is an sa-measure, we'll transfer these Compactness-Lemma-enabling bounds to the sequences $(m_n^B,b_n^B)$ and $(m_n^{\ast },b_n^{\ast })$, to get that they're both wandering around in a compact set. Then, we just take a convergent subsequence of both, add the two limit points together, and get our limit point $(m,b)$, witnessing that it's in the upper completion of $B$.

Proof: Let $(m_n,b_n)\in B+M^{sa}\left(X\right)$ limit to some $(m,b)$. A convergent sequence (plus its one limit point) is a compact set of points, so, by the Compactness Lemma, there must be a $b^{◯}$ and ${\lambda }^{◯}$ that are upper bounds on the $b_n$ and $m_n^{+}\left(1\right)$ values, respectively.

Now, for all n, break down $(m_n,b_n)$ as $(m_n^B,b_n^B)+(m_n^{\ast },b_n^{\ast })$, where $(m_n^B,b_n^B)\in B$, and $(m_n^{\ast },b_n^{\ast })$ is an sa-measure.

Because $b_n^B+b_n^{\ast }=b_n\le b^{◯}$, we can bound the $b_n^B$ and $b_n^{\ast }$ quantities by $b^{◯}$. This transfers into a $-b^{◯}$ lower bound on $m_n^{B-}\left(1\right)$ and $m_n^{\ast -}\left(1\right)$, respectively.

Now, we can go:

$m_n^{B+}\left(1\right)+m_n^{B-}\left(1\right)+m_n^{\ast +}\left(1\right)+m_n^{\ast -}\left(1\right)=m_n^B\left(1\right)+m_n^{\ast }\left(1\right)=m_n\left(1\right)$

$=m_n^{+}\left(1\right)+m_n^{-}\left(1\right)\le m_n^{+}\left(1\right)\le {\lambda }^{◯}$

Using worst-case values for $m_n^{B-}\left(1\right)$ and $m_n^{\ast -}\left(1\right)$, we get:

$m_n^{B+}\left(1\right)+m_n^{\ast +}\left(1\right)-2b^{◯}\le {\lambda }^{◯}$

$m_n^{B+}\left(1\right)+m_n^{\ast +}\left(1\right)\le {\lambda }^{◯}+2b^{◯}$

So, we have upper bounds on $m_n^{B+}\left(1\right)$ and $m_n^{\ast +}\left(1\right)$ of ${\lambda }^{◯}+2b^{◯}$, respectively.

Due to the sequences $(m_n^B,b_n^B)$ and $(m_n^{\ast },b_n^{\ast })$ respecting bounds on $b$ and $m^{+}\left(1\right)$ ($b^{◯}$ and ${\lambda }^{◯}+2b^{◯}$ respectively), and wandering around within the closed sets $B$ and $M^{sa}\left(X\right)$ respectively, we can use the Compactness Lemma and Tychonoff's theorem (finite product, no axiom of choice needed) to go "hey, there's a subsequence where both $(m_n^B,b_n^B)$ and $(m_n^{\ast },b_n^{\ast })$ converge, call the limit points $(m^B,b^B)$ and $(m^{\ast },b^{\ast })$. Since $B$ and $M^{sa}\left(X\right)$ are closed, $(m^B,b^B)\in B$, and $(m^{\ast },b^{\ast })\in M^{sa}\left(X\right)$."

Now, does $(m^B,b^B)+(m^{\ast },b^{\ast })=(m,b)$? Well, for any $ϵ$, there's some really large n where $d\left(\right(m_n^B,b_n^B),(m^B,b^B\left)\right)<ϵ$, $d\left(\right(m_n^{\ast },b_n^{\ast }),(m^{\ast },b^{\ast }\left)\right)<ϵ$, and $d\left(\right(m_n,b_n),(m,b\left)\right)<ϵ$. Then, we can go:

$d\left(\right(m,b),(m^B,b^B)+(m^{\ast },b^{\ast }\left)\right)\le d\left(\right(m,b),(m_n,b_n\left)\right)+d\left(\right(m_n,b_n),(m^B,b^B)+(m^{\ast },b^{\ast }\left)\right)$

$=d\left(\right(m,b),(m_n,b_n\left)\right)+d\left(\right(m_n^B,b_n^B)+(m_n^{\ast },b_n^{\ast }),(m^B,b^B)+(m^{\ast },b^{\ast }\left)\right)$

$=d\left(\right(m,b),(m_n,b_n\left)\right)+||\left(\right(m_n^B,b_n^B)+(m_n^{\ast },b_n^{\ast }\left)\right)-\left(\right(m^B,b^B)+(m^{\ast },b^{\ast }\left)\right)||$

$=d\left(\right(m,b),(m_n,b_n\left)\right)+||\left(\right(m_n^B,b_n^B)-(m^B,b^B\left)\right)+\left(\right(m_n^{\ast },b_n^{\ast })-(m^{\ast },b^{\ast }\left)\right)||$

$\le d\left(\right(m,b),(m_n,b_n\left)\right)+||(m_n^B,b_n^B)-(m^B,b^B)||+||(m_n^{\ast },b_n^{\ast })-(m^{\ast },b^{\ast })||$

$=d\left(\right(m,b),(m_n,b_n\left)\right)+d\left(\right(m_n^B,b_n^B),(m^B,b^B\left)\right)+d\left(\right(m_n^{\ast },b_n^{\ast }),(m^{\ast },b^{\ast }\left)\right)<3ϵ$

So, regardless of $ϵ$, $d\left(\right(m,b),(m^B,b^B)+(m^{\ast },b^{\ast }\left)\right)<3ϵ$, so $(m^B,b^B)+(m^{\ast },b^{\ast })=(m,b)$. So, we've written $(m,b)$ as a sum of an sa-measure in $B$ and an sa-measure, certifying that $(m,b)\in B+M^{sa}\left(X\right)$, so $B+M^{sa}\left(X\right)$ is closed.

Proposition 2: For closed convex nonempty $B$,$B+M^{sa}\left(X\right)=\{M|\forall f^{+}\exists M^{\prime }\in B:f^{+}(M)\ge f^{+}(M^{\prime }\left)\right\}$

Proof sketch: Show both subset inclusion directions. One is very easy, then we assume the second direction is false, and invoke the Hahn-Banach theorem to separate a point in the latter set from the former set. Then we show that the separating functional is a positive functional, so we have a positive functional where the additional point underperforms everything in $B+M^{sa}\left(X\right)$, which is impossible by the definition of the latter set.

Easy direction: We will show that $B+M^{sa}\left(X\right)\subseteq \{M|\forall f^{+}\exists M^{\prime }\in B:f^{+}(M)\ge f^{+}(M^{\prime }\left)\right\}$

This is because a $M\in (B+M^{sa}(X\left)\right)$, can be written as $M=M^B+M^{\ast }$. Let $M^B$ be our $M^{\prime }$ of interest. Then, it is indeed true that for all $f^{+}$, $f^{+}\left(M\right)=f^{+}\left(M^B\right)+f^{+}\left(M^{\ast }\right)\ge f^{+}\left(M^B\right)$

Hard direction: Assume by contradiction that

$B+M^{sa}\left(X\right)\subset \{M|\forall f^{+}\exists M^{\prime }\in B:f^{+}(M)\ge f^{+}(M^{\prime }\left)\right\}$

Then there's some $M$ where $\forall f^{+}\exists M^{\prime }\in B:f^{+}\left(M\right)\ge f^{+}\left(M^{\prime }\right)$ and $M\notin B+M^{sa}\left(X\right)$. $B+M^{sa}\left(X\right)$ is the upper completion of a closed set, so by the Compactness Lemma, it's closed, and since it's the Minkowski sum of convex sets, it's convex.

Now, we can use the variant of the Hahn-Banach theorem from the Wikipedia article on "Hahn-Banach theorem", in the "separation of a closed and compact set" section. Our single point $M$ is compact, convex, nonempty, and disjoint from the closed convex set $B+M^{sa}\left(X\right)$. Banach spaces are locally convex, so we can invoke Hahn-Banach separation.

Therefore, there's some continuous linear functional $\varphi$ s.t. $\varphi \left(M\right)<{inf}_{M^{\prime }\in (B+M^{sa}(X\left)\right)}\varphi \left(M^{\prime }\right)$

We will show that this linear functional is actually a positive functional!

Assume there's some sa-measure $M^{\ast }$ where $\varphi \left(M^{\ast }\right)<0$. Then we can pick a random $M^B\in B$, and consider $\varphi (M^B+c{M}^{\ast })$, where $c$ is extremely large. $M^B+c{M}^{\ast }$ lies in $B+M^{sa}\left(X\right)$, but it would also produce an extremely negative value for \phi which undershoots $\varphi \left(M\right)$ which is impossible. So $\varphi$ is a positive functional.

However, $\varphi \left(M\right)<{inf}_{M^{\prime }\in (B+M^{sa}(X\left)\right)}\varphi \left(M^{\prime }\right)$, so $\varphi \left(M\right)<{inf}_{M^{\prime }\in B}\varphi \left(M^{\prime }\right)$. But also, $M$ fulfills the condition $\forall f^{+}\exists M^{\prime }\in B:f^{+}\left(M\right)\ge f^{+}\left(M^{\prime }\right)$, because of the set it came from. So, there must exist some $M^{\prime }\in B$ where $\varphi \left(M\right)\ge \varphi \left(M^{\prime }\right)$. But, we have a contradiction, because $\varphi \left(M\right)<{inf}_{M^{\prime }\in B}\varphi \left(M^{\prime }\right)$.

So, there cannot be any point in $\{M|\forall f^{+}\exists M^{\prime }\in B:f^{+}(M)\ge f^{+}(M^{\prime }\left)\right\}$ that isn't in $B+M^{sa}\left(X\right)$. This establishes equality.

Lemma 3: For any closed set $B\subseteq M^{sa}\left(X\right)$ and point $M\in B$, the set $\left(\right\{M\}-M^{sa}(X\left)\right)\cap B$ is nonempty and compact.

Proof: It's easy to verify nonemptiness, because $M$ is in the set. Also, it's closed because it's the intersection of two closed sets. $B$ was assumed closed, and the other part is the Minkowski sum of $\left\{M\right\}$ and $-M^{sa}\left(X\right)$, which is closed if $-M^{sa}\left(X\right)$ is, because it's just a shift of $-M^{sa}\left(X\right)$ (via a single point). $-M^{sa}\left(X\right)$ is closed because it's -1 times a closed set.

We will establish a bound on the $m^{+}\left(1\right)$ and $b$ values of anything in the set, which lets us invoke the Compactness Lemma to show compactness, because it's a closed subset of a compact set.

Note that if $M^{\prime }\in \left(\right\{M\}-M^{sa}(X\left)\right)\cap B$, then $M^{\prime }=M-M^{\ast }$, so $M^{\prime }+M^{\ast }=M$. Rewrite this as $(m^{\prime },b^{\prime })+(m^{\ast },b^{\ast })=(m,b)$

Because $b^{\prime }+b^{\ast }=b$, we can bound $b^{\prime }$ and $b^{\ast }$ by $b$. This transfers into a $-b$ lower bound on $m^{{}^{\prime }-}\left(1\right)$ and $m^{\ast -}\left(1\right)$. Now, we can go:

$m^{{}^{\prime }+}\left(1\right)+m^{{}^{\prime }-}\left(1\right)+m^{\ast +}\left(1\right)+m^{\ast -}\left(1\right)=m^{\prime }\left(1\right)+m^{\ast }\left(1\right)=m\left(1\right)$

$=m^{+}\left(1\right)+m^{-}\left(1\right)\le m^{+}\left(1\right)$

Using worst-case values for $m^{{}^{\prime }-}\left(1\right)$ and $m^{\ast -}\left(1\right)$, we get:

$m^{{}^{\prime }+}\left(1\right)+m^{{}^{\prime }+}\left(1\right)-2b\le m^{+}\left(1\right)$

$m^{{}^{\prime }+}\left(1\right)\le m^{{}^{\prime }+}\left(1\right)+m^{\ast +}\left(1\right)\le m^{+}\left(1\right)+2b$

So, we have an upper bound of $m^{+}\left(1\right)+2b$ on $m^{{}^{\prime }+}\left(1\right)$, and an upper bound of $b$ on $b^{\prime }$. Further, $(m^{\prime },b^{\prime })$ was arbitrary in $\left(\right\{M\}-M^{sa}(X\left)\right)\cap B$, so we have our bounds. This lets us invoke the Compactness Lemma, and conclude that said closed set is compact.

Lemma 4: If $\ge$ is a partial order on $B$ where $M^{\prime }\ge M$ iff there's some sa-measure $M^{\ast }$ where $M=M^{\prime }+M^{\ast }$, then

$\exists M^{\prime }>M\leftrightarrow (M\in B\wedge \exists M^{\prime }\ne M:M^{\prime }\in \{M\}-M^{sa}(X\left)\right)\cap B)\leftrightarrow M\text{is not minimal in}B$

Proof: $\exists M^{\prime }>M\leftrightarrow \exists M^{\prime }\ne M:M^{\prime }\ge M$

Also, $M^{\prime }\ge M\leftrightarrow (M^{\prime },M\in B\wedge \exists M^{\ast }:M=M^{\prime }+M^{\ast })$

Also, $\exists M^{\ast }:M=M^{\prime }+M^{\ast }\leftrightarrow \exists M^{\ast }:M-M^{\ast }=M^{\prime }\leftrightarrow M^{\prime }\in \left(\right\{M\}-M^{sa}(X\left)\right)$

Putting all this together, we get

$(\exists M^{\prime }>M)\leftrightarrow (M\in B\wedge \exists M^{\prime }\ne M:M^{\prime }\in (\left\{M\right\}-M^{sa}\left(X\right))\cap B)$

And we're halfway there. Now for the second half.

$M\text{is not minimal in}B\leftrightarrow M\in B\wedge (\exists M^{\prime }\in B:M^{\prime }\ne M\wedge (\exists M^{\ast }:M=M^{\prime }+M^{\ast }\left)\right)$

Also, $\exists M^{\ast }:M=M^{\prime }+M^{\ast }\leftrightarrow \exists M^{\ast }:M-M^{\ast }=M^{\prime }\leftrightarrow M^{\prime }\in \left(\right\{M\}-M^{sa}(X\left)\right)$

Putting this together, we get

$M\text{is not minimal in}B\leftrightarrow (M\in B\wedge \exists M^{\prime }\ne M:M^{\prime }\in (\left\{M\right\}-M^{sa}\left(X\right))\cap B)$

And the result has been proved.

Theorem 2: Given a nonempty closed set $B$, the set of minimal points $B^{min}$ is nonempty and all points in $B$ are above a minimal point.

Proof sketch: First, we establish an partial order that's closely tied to the ordering on $B$, but flipped around, so minimal points in $B$ are maximal elements. We show that it is indeed a partial order, letting us leverage Lemma 4 to translate between the partial order and the set $B$. Then, we show that every chain in the partial order has an upper bound via Lemma 3 and compactness arguments, letting us invoke Zorn's lemma to show that that everything in the partial order is below a maximal element. Then, we just do one last translation to show that minimal points in $B$ perfectly correspond to maximal elements in our partial order.

Proof: first, impose a partial order on $B$, where $M^{\prime }\ge M$ iff there's some sa-measure $M^{\ast }$ where $M=M^{\prime }+M^{\ast }$. Notice that this flips the order. If an sa-measure is "below" another sa-measure in the sa-measure addition sense, it's above that sa-measure in this ordering. So a minimal point in $B$ would be maximal in the partial order. We will show that it's indeed a partial order.

Reflexivity is immediate. $M=M+(0,0)$, so $M\ge M$.

For transitivity, assume $M^{\prime \prime }\ge M^{\prime }\ge M$. Then there's some $M^{\ast }$ and $M^{{}^{\prime }\ast }$ s.t. $M=M^{\prime }+M^{\ast }$, and $M^{\prime }=M^{\prime \prime }+M^{{}^{\prime }\ast }$. Putting these together, we get $M=M^{\prime \prime }+(M^{\ast }+M^{{}^{\prime }\ast })$, and adding sa-measures gets you an sa-measure, so $M^{\prime \prime }\ge M$.

For antisymmetry, assume $M^{\prime }\ge M$ and $M\ge M^{\prime }$. Then $M=M^{\prime }+M^{\ast }$, and $M^{\prime }=M+M^{{}^{\prime }\ast }$. By substitution, $M=M+(M^{\ast }+M^{{}^{\prime }\ast })$, so $M^{{}^{\prime }\ast }=-M^{\ast }$. For all positive functionals, $f^{+}\left(M^{{}^{\prime }\ast }\right)=f^{+}(-M^{\ast })=-f^{+}\left(M^{\ast }\right)$, and since positive functionals are always nonnegative on sa-measures, the only way this can happen is if $M^{\ast }$ and $M^{{}^{\prime }\ast }$ are 0, showing that $M=M^{\prime }$.

Anyways, since we've shown that it's a partial order, all we now have to do is show that every chain has an upper bound in order to invoke Zorn's lemma to show that every point in $B$ lies below some maximal element.

Fix some ordinal-indexed chain $M_{\gamma }$, and associate each of them with the set $S_{\gamma }=\left(\right\{M_{\gamma }\}+(-M^{sa}\left(X\right)\left)\right)\cap B$, which is compact by Lemma 3 and always contains $M_{\gamma }$.

The collection of $S_{\gamma }$ also has the finite intersection property, because, fixing finitely many of them, we can consider a maximal ${\gamma }^{\ast }$, and $M_{{\gamma }^{\ast }}$ is in every associated set by:

Case 1: Some other $M_{\gamma }$ equals $M_{{\gamma }^{\ast }}$, so $S_{\gamma }=S_{{\gamma }^{\ast }}$ and $M_{{\gamma }^{\ast }}\in S_{{\gamma }^{\ast }}=S_{\gamma }$.

Case 2: $M_{{\gamma }^{\ast }}>M_{\gamma }$, and by Lemma 4, $M_{{\gamma }^{\ast }}\in \left(\right\{M_{\gamma }\}-M^{sa}(X\left)\right)\cap B$.

Anyways, since all the $S_{\gamma }$ are compact, and have the finite intersection property, we can intersect them all and get a nonempty set containing some point $M_{\infty }$. $M_{\infty }$ lies in $B$, because all the sets we intersected were subsets of $B$. Also, because $M_{\infty }\in (M_{\gamma }-M^{sa}(X\left)\right)\cap B$ for all $\gamma$ in our chain, then if $M_{\infty }\ne M_{\gamma }$, Lemma 4 lets us get $M_{\infty }>M_{\gamma }$, and if $M_{\infty }=M_{\gamma }$, then $M_{\infty }\ge M_{\gamma }$. Thus, $M_{\infty }$ is an upper bound for our chain.