= Solution
Use the standard setting of a <compact metrizable convex set> $K$ in a Hausdorff locally convex real topological vector space, with its metrizable topology. Write $A(K)$ for the continuous affine real <functions> on $K$. The <affine upper envelope> of a bounded real <function> is
$$
\boxed{\overline f(x)=\inf\{a(x):a\in A(K),\ a\ge f\text{ on }K\}.}
$$
Constants make this infimum finite, and $f\le\overline f$. An infimum of affine continuous majorants is concave and <upper semicontinuous>, hence Borel. For continuous $f$ it is the upper concave envelope appropriate to barycentric <measures>; it is not the pointwise maximum of $f$ and a selected <affine function>.
For a fixed probability $\mu$, define $p(g)=\int_K\overline g\,d\mu$ on real $C(K)$. Affine majorants show
$$
\overline{g+h}\le\overline g+\overline h,\qquad
\overline{\lambda g}=\lambda\overline g\ (\lambda\ge0),\qquad
\overline{g+a}=\overline g+a\ (a\in A(K)).
$$
Thus $p$ is sublinear and $p(a)=\int a\,d\mu$ for affine $a$. On the span of $f$, the linear functional taking $f$ to $p(f)$ is dominated by $p$: the negative-scalar condition follows from $-p(f)\le p(-f)$. For $f=0$ start from the zero subspace. The <Hahn-Banach theorem> extends it to a linear functional $L$ on $C(K)$ with $L\le p$ and $L(f)=p(f)$.
If $g\ge0$, then $p(-g)\le0$, so $L(g)\ge0$. Also $p(1)=1$ and $p(-1)=-1$, forcing $L(1)=1$. Positivity gives $|L(g)|\le\|g\|_\infty$, and the <Riesz-Markov-Kakutani representation theorem> produces a Borel probability $\nu$ with $L(g)=\int g\,d\nu$. Therefore
$$
\boxed{\int f\,d\nu=\int\overline f\,d\mu,\qquad
\int g\,d\nu\le\int\overline g\,d\mu\quad(g\in C(K)).}
$$
For affine $a$, testing both $a$ and $-a$ gives $\int a\,d\nu=\int a\,d\mu$: the two <measures> have the same <barycenter>. This proves the requested <supporting measure lemma for affine upper envelopes>.
\b[Choquet's theorem:] every $x\in K$ has a <Borel probability measure> $\lambda$ concentrated on the <extreme points> of $K$ whose <barycenter> is $x$; equivalently,
$$
\boxed{\lambda(\operatorname{Ex}K)=1,\qquad
a(x)=\int_K a\,d\lambda\quad(a\in A(K)).}
$$
Concentration is a <measure>-one assertion, not a claim that the topological support must be a closed subset of $\operatorname{Ex}K$.
To prove it, let $\mathcal M_x$ be the <measures> satisfying all the displayed affine equalities. It is nonempty because it contains $\delta_x$, and it is weakly closed in the compact space $P(K)$, hence compact. Let $h\in C(K)$ be strictly convex, as permitted. Choose $\lambda\in\mathcal M_x$ maximizing $\int h\,d\lambda$. Apply the supporting <measure> lemma with $\mu=\lambda$ and $f=h$. It gives $\nu$ with the same affine integrals, so $\nu\in\mathcal M_x$, and
$$
\int h\,d\lambda\ge\int h\,d\nu
=\int\overline h\,d\lambda\ge\int h\,d\lambda.
$$
Thus $\overline h-h\ge0$ has integral zero. If $z$ is not extreme, write $z=ty+(1-t)w$ with $0<t<1$ and distinct $y,w\in K$. Strict convexity and every affine majorant give
$$
h(z)<th(y)+(1-t)h(w)\le\overline h(z).
$$
Therefore the nonnegative gap is strictly positive at every nonextreme point. It is Borel, and $\operatorname{Ex}K$ is Borel by the allowed <G-delta set> assertion. Its zero integral forces $\lambda(K\setminus\operatorname{Ex}K)=0$, proving <Choquet's theorem by strict convexity>. No uniqueness is asserted for the representing <measure>.
For the real $L^\infty$ example, give its closed unit ball $K$ the <weak-star topology> $\sigma(L^\infty,L^1)$, not the norm topology. The <Banach-Alaoglu theorem> makes it compact, and separability of $L^1[0,1]$ makes this ball metrizable. Its extreme points are exactly the classes $h$ satisfying $|h|=1$ almost everywhere. Indeed, if $|h|\le1-\varepsilon$ on a positive-<measure> set, adding and subtracting $\varepsilon$ times its indicator decomposes $h$ nontrivially inside the ball. Conversely, if $|h|=1$ almost everywhere and $h=(u+v)/2$ with $|u|,|v|\le1$, pointwise equality at the endpoints of $[-1,1]$ forces $u=v=h$ almost everywhere. This is the <extreme-point criterion for the L-infinity unit ball>.
For the hinted case $f=1_A-1_B$, put $C=[0,1]\setminus(A\cup B)$ and $h_\pm=1_A-1_B\pm1_C$. Then $h_\pm$ are extreme and
$$
\boxed{\nu=\tfrac12\delta_{h_+}+\tfrac12\delta_{h_-}}
$$
has <barycenter> $f$. If $C$ is null, the two point masses coincide.
For a general real $f$, choose a measurable representative in $[-1,1]$ and set, for $0\le t\le1$,
$$
H_t(s)=2\,1_{\{t\le(1+f(s))/2\}}-1,\qquad
\boxed{\nu=(t\mapsto H_t)_*\operatorname{Leb}_{[0,1]}.}
$$
Every $H_t$ is extreme. The map into the weak-star compact ball is Borel: for each $u\in L^1$, the <function> $t\mapsto\int u(s)H_t(s)\,ds$ is measurable by joint measurability and integration; a countable dense family of such tests generates the ball's topology and Borel sigma-algebra. The <measure> is independent of changes to $f$ on a null set. For each $s$,
$$
\int_0^1 H_t(s)\,dt=2\frac{1+f(s)}2-1=f(s).
$$
The integrand paired with $u$ is dominated by $|u|$, so <Fubini's theorem> yields
$$
\int_K\left(\int_0^1u(s)h(s)\,ds\right)d\nu(h)
=\int_0^1u(s)f(s)\,ds.
$$
These continuous linear tests define the weak-star <barycenter>, hence that <barycenter> is $f$. Together with $\nu(\operatorname{Ex}K)=1$, this is the required <threshold Choquet representation in L-infinity>.
If $L^\infty$ is instead taken over complex scalars, its extreme points satisfy the same unit-modulus condition. Write $f(s)=r(s)\zeta(s)$, with $r=|f|$ and $|\zeta|=1$, choosing $\zeta=1$ where $f=0$. Replace the threshold family by $\zeta(s)(2\,1_{\{t\le(1+r(s))/2\}}-1)$. Its members have unit modulus and its average is $f$, so the same pushforward and Fubini argument gives a representing <measure> in the real locally convex interpretation of the complex ball.
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