A proper Gaussian likelihood gives a proper posterior

The set $D=U=\{0,1\}$ is Borel and nonempty. The function ${\mu }_0(A)=\frac{1}{2}{\mathbf{1}}_{\{0\in A\}}+\frac{1}{2}{\mathbf{1}}_{\{1\in A\}}$ is countably additive: in a disjoint sequence of Borel sets each of the two points occurs at most once, so the sum of assigned masses equals the mass of the union. Also ${\mu }_0(\mathbb{R})=1$ and ${\mu }_0(D)=1$, making it a prior on the stated domain by [F8]. The identity $F:D\to \mathbb{R}$ is measurable because every subset of finite $D$ is Borel, and it satisfies the forward-map interface [F9].

For any $u\in D$ and $y\in \mathbb{R}$, put $a:=2\pi$; since $\log a$ occurs in the real-valued potential, [F15] gives $a>0$. Put $b:=\frac{1}{2}(y-u{)}^2$ and $c:=\frac{1}{2}\log a$. By [F15], $\exp (2c)=a$. By [F16], $\exp (c{)}^2=\exp (2c)=a$; by [F18], $\exp (c)>0$. Thus uniqueness of the nonnegative square root [F17] gives $\exp (c)=\sqrt{2\pi }$, and [F18] gives $\exp (-c)=1\mathrm{/}\sqrt{2\pi }$. Applying [F16] again, $e^{-\mathrm{\Phi }(u;y)}=\exp (-(b+c))=\exp (-b)\exp (-c)=\frac{1}{\sqrt{2\pi }}\exp \text{\hspace{0.17em}⁣}(-\frac{(y-u{)}^2}{2})=L(u;y)$. The last expression is the displayed Gaussian density, so the likelihood and potential agree pointwise.

By [A1] and [F1], each $N(u,1)$ is a probability law with the displayed density; since $D$ is finite, every evaluation $u\mapsto N(u,1)(B)$ is measurable, so [F10] gives a probability kernel. By [F2] and [F3], Lebesgue measure is sigma-finite. The potential and likelihood are jointly Borel: for each of the two $u$ values their data sections are continuous, each singleton $\{u\}$ is Borel in $D$, and each preimage of a Borel set is a finite union of rectangles $\{u\}\times B_u$ in $\mathcal{B}(D)\otimes \mathcal{B}(\mathbb{R})$ by [F13]. Thus [F4] gives a proper Bayesian model with the stated prior and sampling kernel.

The finite subspace $D=\{0,1\}$ is discrete: the open balls of radius $1\mathrm{/}2$ about $0$ and $1$ isolate the two points, so every real-valued function on $D$ is continuous. For every $u\in D$ and $y\in \mathbb{R}$, $\mathrm{\Phi }(u;y)$ is finite. By [F11], $(y-u{)}^2\ge 0$ and $a\ge -\mathrm{\mid }a\mathrm{\mid }$ for $a:=\frac{1}{2}\log (2\pi )$, so $\mathrm{\Phi }(u;y)\ge -C$ where $C:=\mathrm{\mid }a\mathrm{\mid }\ge 0$. The constant envelope $M_1(r)=C$ is Borel, and [F12] gives ${\int }_D{e}^{M_1(\mathrm{\parallel }u\mathrm{\parallel })}{\mu }_0(du)=e^C{\mu }_0(D)=e^C<\mathrm{\infty }$. Applying [F5] with ${\mathsf{Y}}_0=\mathbb{R}$ proves $0<Z(y)<\mathrm{\infty }$ and propriety of the normalized update for every datum.

By [F14], integrate the likelihood section against the prior. Step 1.2 gives $L(u;y)=e^{-\mathrm{\Phi }(u;y)}$ for both atoms, so this likelihood evidence equals $Z_{\mathrm{\Phi }}(y)$; using the two prior atoms gives $Z(y)=\frac{1}{2}L(0;y)+\frac{1}{2}L(1;y)$. Step 3.1 gives $0<Z(y)<\mathrm{\infty }$. Since $D$ is finite, the likelihood section is simple, so [F12] gives the numerator for any $A\subseteq D$ as $\frac{1}{2}L(0;y){\mathbf{1}}_{\{0\in A\}}+\frac{1}{2}L(1;y){\mathbf{1}}_{\{1\in A\}}$. Dividing by $Z(y)$ yields the stated posterior weights.

Apply [F6] with parameter $u\in D$, prior ${\mu }_0$, dominating measure Lebesgue, and the jointly Borel density $L(u;y)$. Its evidence is exactly the $Z(y)$ computed in step 4.1, so the displayed ${\mathrm{\Pi }}^y$ is a data-indexed posterior-kernel version. By [F7], changing its values on a data-marginal-null set leaves the conditional identities unchanged.