Xrfgtjtk

I've seen proofs that Dirac measures are the extreme points of probability measures, but how do we prove it for general complex Borel measures with total variation norm 1?

I only want to know why they're in the set of all extreme points.

$newcommand{norm}[1]{leftlVert#1rightrVert}$
Let $(X, mathcal{A})$ by a measurable space and denote by $mathcal{M}$ the space of (finite) complex measures on $(X, mathcal{A})$, endowed with the variation norm. Let $B$ be the closed unit ball in $mathcal{M}$. I will show the following:

For any $x in X$ and $a in mathbb{C}$ with $|a| =1$, the measure $a delta_x$ is an extreme point of $B$ (where $delta_x$ is the Dirac measure centered at $x$).

Proof:

Let $mu := a delta_x$ for some $|a| =1$. The idea of the proof is that the variation measure $| mu |$ is equal to $delta_x$, so that $| mu |$ is an extreme point of the convex set $mathcal{P}$ of probability measures on $(X, mathcal{A})$.

Suppose that $mu = s nu_1 + (1-s) nu_2$, where $nu_1, nu_2 in B$ and
$0 <s <1$. We want to prove that $nu_1 = nu_2 = mu$.

First, we can assume that $norm{ nu_1 } = norm{ nu_2 } = 1$ (so that $| nu_1 |$ and $| nu_2 |$ are probabilities). Otherwise, we would have $norm{ mu } = norm{ s nu_1 + (1-s) nu_2 } leq s norm{nu_1} +(1-s) norm{ nu_2} < s + (1-s) = 1$, a contradiction.

We have by definition of $mu$ that:
$$
delta_x = | mu | = | s nu_1 +(1-s) nu_2 | leq s |nu_1| +(1-s) |nu_2|.
$$

The measure $nu := s | nu_1 | + (1-s) |nu_2|$ is a probability measure, so the inequality $delta_x leq nu$ is in fact an equality. Indeed, if $A$ is measurable and contains $x$ we must have
$$
1 = delta_x (A) leq nu(A) leq 1,
$$
so $nu(A)=1$. On the other hand, if $x$ is not in $A$, then $nu(A) = nu(X) - nu(X setminus A) = 1 - 1 = 0$. This means that $nu = delta_x$.

Therefore we have $delta_x = s | nu_1 | + (1-s) | nu_2 |$. Since $| nu_1 |$ and $| nu_2 |$ are probabilites and $delta_x$ is an extreme point of $mathcal{P}$, we must have $| nu_1 | = | nu_2 | = delta_x$.

The equality $| nu_1 | = | nu_2 | = delta_x$ implies that $nu_1$ and $nu_2$ are multiples of $delta_x$. Indeed, if $A$ is a measurable set that does not contain $x$, then
$$
| nu_i (A) | leq |nu_i|(A) = delta_x (A) = 0,
$$
so that $nu_i(A) =0$. On the other hand, if $A$ contains $x$ then $nu_i(A) = nu_i(X) - nu_i(X setminus A ) = nu_i(X)$. Therefore we have $nu_i = b_i delta_x$, where $b_i = nu_i(X)$. Furthermore, we have $| nu_i | = |b_i delta_x | = |b_i| delta_x$, which implies that $|b_i|=1$.

Our assumption that $mu = s nu_1 + (1-s) nu_2$ thus gives
$$
a delta_x = (s b_1 + (1-s) b_2) delta_x,
$$
and therefore $a = s b_1 +(1-s) b_2$, where $0< s <1$ and $a$, $b_1$ and $b_2$ are complex numbers of modulus $1$. This is only possible if $a = b_1 = b_2$, and we conclude that $nu_1 = nu_2 = a delta _x$. $square$

Here is part of the proof. Can you finish it?

Notation. $K$ is a compact Hausdorff space. $C(K)$ is the Banach space of continuous, complex-valued, functions on $K$ with norm
$$
|f| = sup {|f(x);:; x in K}
$$
An element $f in C(K)$ is called nonnegative if $f(x) ge 0$ for all $x in K$. We write $f ge 0$ if that happens.

Write $mathbb{1}$ for the constant function with value $1$. Then of course $mathbb{1} in C(K)$ and $|mathbb{1}| = 1$ and $mathbb{1} ge 0$.

$C(K)^*$ is the Banach space of bounded linear functionals on $C(K)$ with norm
$$
|alpha| = sup {|alpha(f)|;:; f in C(K), |f| le 1}
$$
A functional $alpha in C(K)^*$ is called nonnegative iff $alpha(f) ge 0$ for all nonnegative $f in C(K)$. We write $alpha ge 0$ if that happens.

For $a in K$ let $delta_a$ be the "Dirac functional" defined by
$$
delta_a (f) = f(a)qquadtext{for all } f in C(K).
$$
Then $delta_a in C(K)^*$ and $|delta_a| = 1$ and $delta_a ge 0$.

Lemma. Let $alpha in C(K)^*$. Suppose $|alpha| le 1$ and $alpha(mathbf{1}) = 1$. Then $alpha ge 0$.

Proof goes here

Proposition. Let $B = {alpha in C(K)^*;:;|alpha|le 1}$ be the unit ball of $C(K)^*$. Let $a in K$. Then $delta_a$ is an extreme point of $B$.

Proof.
Suppose $alpha, beta in B$ and $frac{1}{2}(alpha+beta) = delta_a$. We must show that $alpha = beta= delta_a$. First we apply the Lemma to $alpha, beta$. Compute
$$
1 = |1| = left|delta_a(mathbb{1})right|
=left|frac{1}{2}big(alpha(mathbb{1})+beta(mathbb{1})big)right|
le frac{1}{2}left(big|alpha(mathbb{1})big|+big|beta(mathbb{1})big|right)
le frac{1}{2}(1+1) = 1 .
$$
Equality holds in the triangle inequality, so in fact $alpha(mathbb{1}) = 1 = beta(mathbb{1})$. Also, $alpha, beta in B$ by hypothesis. So by the Lemma, we conclude that $alpha ge 0, beta ge 0$.

To be continued...