Choquet's theorem (Exercise 7 of Chapter 13 in Lax's functional analysis) I have a problem with Exercise 7 of Chapter 13 in Lax's functional analysis. The definitions in Exercise 7 comes from Theorem 5. I post both of them below.  My attempt is to get an sequence of points $v_n\in K_e$ and $v_n\rightarrow v$ in $\bar{K}_e$, so that for each $v_n$, we have a measure $m_n$, according to Theorem 5. But I don't know how to prove $m_n$ converge to some measure $m$ and $m(v)=0$. 


 A: 
What do we know?  $v \in\overline{K_e}\subseteq K$ but $v \notin K_e$ ... that is what we have to work with.  

$v$ is not an extreme point of $K$.  That is:
There exist $v_1, v_2 \in K$, $v_1 \ne v_2$, $v = \frac{1}{2}(v_1+v_2)$.  Note also $v_1 \ne v \ne v_2$.  By Theorem 5, there exist probability measures $m_1, m_2$ on $\overline{K_e}$ that represent $v_1, v_2$, respectively:
$$
v_1 = \int_{\overline{K_e}} e\;dm_1(e),\qquad
v_2 = \int_{\overline{K_e}} e\;dm_2(e)
$$
Since $v \ne v_1$ we have $m_1(\{v\}) < 1$; similarly $m_2(\{v\}) < 1$.  We can define probability measure $m_3$ on $\overline{K_e}$ by
$$
m_3 = \frac{1}{2}(m_1+m_2)
$$
so that $m_3$ represents $v$:  

$$
\int_{\overline{K_e}} e\;d m_3(e) =
\frac{1}{2}\int_{\overline{K_e}} e\;d m_1(e)+
\frac{1}{2}\int_{\overline{K_e}} e\;d m_2(e)
=\frac{1}{2}\;v_1 + \frac{1}{2}\;v_2 = v
$$

There is also probability measure $m_4$, the Dirac measure at $v$, which represents $v$:

$$
v = \int_{\overline{K_e}} e\;d m_4(e)
$$

The measure $m$ we want will be a linear combination of $m_3$ and $m_4$.  It has to be a probability measure, it has to vanish on $\{v\}$, and
it has to represent $v$.  
Write $\alpha = m_3(\{v\})$.  We have $0 \le \alpha < 1$.  Define (signed) measure $m$ by
$$
m = \frac{1}{1-\alpha}m_3 - \frac{\alpha}{1-\alpha}m_4
$$
Then $m$ is a nonnegative measure.  

If $A \subseteq K$ then
  take two cases: $v \in A$ and $v \notin A$.  If $v \notin A$, then
  $$
m(A) = \frac{1}{1-\alpha}m_3(A)+0 \ge 0 .
$$
  If $v \in A$, then $m_3(A) \ge m_3(\{v\}) = \alpha$ and thus
  $$
m(A) = \frac{1}{1-\alpha}m_3(A) - \frac{\alpha}{1-\alpha}m_4(A)
\ge \frac{1}{1-\alpha}\;\alpha - \frac{\alpha}{1-\alpha}\;1 = 0
$$

And $m$ is a probability measure

$$
m(\overline{K_e}) = \frac{1}{1-\alpha}m_3(\overline{K_e}) 
- \frac{\alpha}{1-\alpha}m_4(\overline{K_e})
= \frac{1}{1-\alpha}\;1 - \frac{\alpha}{1-\alpha}\;1 = 1
$$

And $m$ vanishes on $\{v\}$:

$$
m(\{v\}) = 
\frac{1}{1-\alpha}\;m_3(\{v\}) - \frac{\alpha}{1-\alpha}\;m_4(\{v\}) =
\frac{1}{1-\alpha}\;\alpha - \frac{\alpha}{1-\alpha}\;1 = 0
$$

And $m$ represents $v$

$$
\int_{\overline{K_e}} e\;dm(e) =
\frac{1}{1-\alpha}\int_{\overline{K_e}} e\;dm_3(e) -
\frac{\alpha}{1-\alpha}\int_{\overline{K_e}} e\;dm_4(e) =
\frac{1}{1-\alpha}v - \frac{\alpha}{1-\alpha}v = v
$$

