- First we show that the union of the line segments is in $\sigma$. Let $x=\left \{ \sum\limits_{i=1}^n s_ia_i | s_i \geq 0, \sum\limits_{i=1}^n s_i=1 \right \}$ be a point on the simplex spanned by $\left \{a_1,...,a_n \right \}$. Then the line segment $\left \{t_0a_0+(1-t_0)\sum\limits_{i=1}^n s_ia_i | 0 \leq t_0 \leq 1 \right \}$ is in $\sigma$ because $t_0s_i \geq 0$ and $t_0+(1-t_0)(s_1+...+s_n)=1$.
Now we show that $\sigma$ is in the union of the line segments.
Given a point $\sum\limits_{i=0}^n t_ia_i$ in $\sigma$ with $t_0 \neq 1$, set $s_i=\frac{t_i}{1-t_0}$ for $i=1,...,n$. This shows that every point in the simplex spanned by $a_0,...,a_n$, except $a_0$, is in the union of the line segments. Clearly $a_0$ is also in the union, since it is in each line segment. Therefore $\sigma$ is the union of all line segments from $a_0$ to the simplex spanned by $a_1,...,a_n$.
Furthermore, we show that two such line segments intersect only at $a_0$. Clearly two line segments from $a_0$ to the simplex spanned by $a_1,...,a_n$ do intersect at $a_0$. Now suppose they intersect at some other point $y$. Then the two segments must lie on the same line. The other endpoints of the line segments lie on the simplex spanned by $a_1,...,a_n$. But then the line must be contained in the simplex spanned by $a_1,...,a_n$. This contradicts the fact that $a_0$ does not lie on the simplex spanned by $a_1,...,a_n$, since $\left \{a_0,a_1,...,a_n \right \}$ are geometrically independent.
- We demonstrate compactness by showing that $\sigma$ is closed and bounded. The simplex $\sigma$ is bounded since
\begin{equation}
\left \Vert \sum\limits_{i=0}^n t_ia_i \right \Vert \leq \max \left \{ t_0,...,t_n \right \} \max \left \{\|a_o\|,...,\|a_n\| \right \} \leq \max \left \{\|a_0\|,...,\|a_n\| \right \}
\end{equation}
The standard $n$-simplex is closed since it is the inverse image of $\left \{1 \right \}$ under the continuous map $\left(x_i \right) \rightarrow \sum\limits_{i}{x_i}$.
For convexity, suppose that $\left(t_0,..,t_n\right)$ and $\left(s_0,...,s_n\right)$ are two distinct $n$-tuples satisfying $t_i \geq 0$, $s_i \geq 0$, $\sum\limits_{i}{t_i}=1$ and $\sum\limits_{i}{s_i}=1$. Then $\lambda \left(t_0,...,t_n\right) + (1-\lambda)\left(s_0,...,s_n\right)$ parametrizes a line segment between these two points for $\lambda \in \left[ 0,1 \right]$. The corresponding curve
\begin{equation}
\lambda\sum\limits_{i=0}^n{t_ia_i}+(1-\lambda)\sum\limits_{i=0}^n{s_ia_i}=\sum\limits_{i=0}^n{(\lambda t_i + (1-\lambda) s_i) a_i}
\end{equation}
is a line segment connecting $\sum\limits_{i=0}^n{t_ia_i}$ to $\sum\limits_{i=0}^n{s_ia_i}$. Since $t_i$ and $s_i$ are non-negative and sum to $1$, and since $\lambda \in \left[0,1\right]$, we have $\lambda t_i + (1-\lambda) s_i$ is also non-negative and sums to $1$. So $\sigma$ is convex.
To show that $\sigma$ is the convex hull of $\left\{a_0,...,a_n\right\}$, we apply Theorem \ref{thm:convexsets} and show that $\sigma$ is the intersection of all convex sets containing $\left\{a_0,...,a_n \right\}$. Suppose $S$ is a convex set containing $a_0,...,a_n$. Then for every point $x=\sum\limits_{i=0}^n{t_ia_i} \in \sigma$, we can show that $x \in S$ as follows. By convexity, since $a_0$ and $a_1$ are in $S$, we have $b_1=\frac{t_0}{t_0+t_1}a_0+\frac{t_1}{t_0+t_1}a_1$ must also be in $S$. Therefore,
\begin{equation}
b_2=\frac{t_0+t_1}{t_0+t_1+t_2}b_1+\frac{t_2}{t_0+t_1+t_2}a_2
\end{equation}
\begin{equation}
b_2=\frac{t_0}{t_0+t_1+t_2}a_0 + \frac{t_1}{t_0+t_1+t_2}a_1 + \frac{t_2}{t_0+t_1+t_2}a_2
\end{equation}
is also in $S$. Continuing in this manner, we find
\begin{equation}
b_n=\frac{t_0}{t_0+...+t_n}a_0 + ... + \frac{t_n}{t_0+...+t_n}a_n \in S,
\end{equation}
but $\sum\limits_{i=0}^n{t_i}=1$, so $x \in S$. Then $\sigma$ is contained in every convex set containing $\left\{a_0,...,a_n \right\}$
- We begin by showing that if $x \in \sigma$ and $x \neq a_0,...,a_n$ then $x$ lies in some open line segment contained in $\sigma$. Suppose $x=\sum\limits_{i=0}^n{t_ia_i}$ is not a vertex. Then $t_i \neq 1$ for any $i$. It follows that at least two of the $t_i$ are nonzero, say $t_0$ and $t_1$, without loss of generality. Choose $\epsilon > 0$ such that $(t_0-\epsilon, t_0+\epsilon) \subset (0,1)$ and $(t_1-\epsilon, t_1+\epsilon) \subset (0,1)$. Then for $t \in (-\epsilon, \epsilon)$, $(t_0+t)a_0+(t_1-t)a_1+\sum\limits_{i=2}^n{t_ia_i}$ is in $\sigma$. Then it follows that every point in $\sigma$ which is not a vertex is on an open line segment contained in $\sigma$.
Next we'll show that the vertices are not on any open line segments contained in $\sigma$.
If any vertex, say $a_0$, were on an open segment contained in
$\sigma$, then there would exist points $x,y\in \sigma$ with $a_0 = tx+(1-t)y$. If $x=\sum\limits_{i=0} ^n t_i a_i$ and $y=\sum\limits_{i=0}^n s_i a_i$, then this means $a_0 = \sum\limits_{i=0}^n \left( t t_i + (1-t) s_i\right) a_i$. This violates the geometric independence of $a_0,
..., a_n$, unless $t_0=s_0=1$ and the others are zero, in which case
$x=y$. Thus $a_0$ is not on an open line segment.
- The barycentric coordinates $t_i$ are continuous functions from the plane $P$ spanned by vertices of $\sigma$ to $\mathbb{R}$. The preimage of the open interval $(0, \infty)$ under each $t_i$ is therefore open in $P$. The interior of $\sigma$ is the intersection of these finitely many open sets and therefore it is open.
The interior of $\sigma$ is convex since the convex combination $(1-t)x+ty$ of any $x,y \in Int(\sigma)$ is in $Int(\sigma)$.
\begin{equation}
(1-t)x+ty\, =\, (1-t)\sum\limits_{i=0}^nt_ia_i + t \sum\limits_{i=0}^n s_ia_i
\, =\, \sum\limits_{i=0}^n ((1-t)t_i+ t s_i)a_i
\end{equation}
and if $t_i$ and $s_i$ are positive, so is $(1-t)t_i+ t s_i$.
Since $\sigma$ is closed, the closure of $Int(\sigma)$ is a subset of
$\sigma$. On the other hand if a point $x \in Bd(\sigma)$ is not in the
closure of $Int(\sigma)$, one can construct a sequence of points
in $Int(\sigma)$ which converges to $x$.
Without loss of generality, assume that
$x= \sum\limits_{i=0}^n t_i a_i,$ and $ t_i(x) > 0$ for $0 \le i \le k$
and $ t_{k+1}(x)= ... = t_n(x)=0.$
Let $r=\min\{ t_i(x)\, \vert \, 0\le i \le k\, \}/2$ and let $x_m\, =\, \sum\limits_{i=0}^k (t_i(x) - \frac{r}{m(k+1)})a_i
+ \sum\limits_{i=k+1}^n \frac{r}{m(n-k)}$. \newline
Then $x_m \in Int(\sigma)$ and the sequence converges to $x$.
Finally, we show that $Int(\sigma)$ equals the union of open line segments
which join a vertex $a_0$ to points in the interior of the opposite
face. Let $x$ be an interior point of $\sigma$ so that $x = \sum\limits_{i=0}^{n} t_ia_i$ with $t_i > 0$ and $\sum{t_i} =1.$
Rewrite this as $x= t_0a_o + (1-t_0)
\sum\limits_{i=1}^{n} \frac{t_i}{1- t_0} a_i.$ Therefore $x$ is on an open line segment joining $a_0$ to a point on the face opposite $a_0$.
Best Answer
I think that your statement about the empty set is a bit shaky. When $X$ is empty, the function $\mu$ would have to be a triple $(\emptyset, \mathbb R, Q)$, where $Q$ is a subset of $\emptyset \times \mathbb R$ having the usual "function" properties...and then only such subset is the empty set. But for that function, the required sum is not 1. (Yeah, this is pedantic nonsense about functions defined on the empty set, but I happen, in my own way, to think it matters). Even ignoring the summation question, the notation $\mu(\emptyset)$ is not defined.
For compactness, you have to decide what topology you're putting on the space of functions.
Since $X$ is finite -- let's say it has $n$ elements -- the set of maps from $X$ to $\mathbb R$ looks a lot like $\mathbb R^n$. Compactness these is the same as "closed and bounded," by the Heine-Borel theorem. "Bounded" for $\Delta$ is easy: the set lies entirely in the unit cube in $\mathbb R^n$. What about "closed"? Suppose that $h$ is in the complement of $\Delta$. Then $h$ is a function on $X$ with one of two properties:
The sum of $h$'s values is $K \ne 1$, or
One of $h$'s value is outside the interval $[0, 1]$. I'll consider the case where the value is $K > 1$.
In each case, taking a ball of radius $\frac{K-1}/2$ around $h$, we find that every element of that ball also has property 1 or 2 (depending on which one held for $h$), so the entire ball is in the complement. Hence the complement of $\Delta$ is open, so $\Delta$ is closed.
(You still have to do the $K < 0$ case for condition 2, but I suspect that you can handle that.)