A flat prior gives a proper known-variance normal-mean posterior

By [A1] and [F6], countable choice is available for [F3]. The boxes $E_k=(-k,k]$ have finite measure $2k$ and cover $\mathbb{R}$ by [F3], so $\lambda$ is sigma-finite. Since $E_k\subseteq \mathbb{R}$, [F4] gives $\lambda (\mathbb{R})\ge 2k$ for every $k$, hence $\lambda (\mathbb{R})=+\mathrm{\infty }$ and the flat prior is improper.

Fix $\mu$ and set $T(x)=(x-\mu )\mathrm{/}\sigma$. This is a $C^1$ diffeomorphism of $\mathbb{R}$ with inverse $t\mapsto \mu +\sigma t$ and $\mathrm{\mid }{T}^{\mathrm{\prime }}(x)\mathrm{\mid }=1\mathrm{/}\sigma$. The formula for $q_{\mu }$ is continuous in $x$, hence Borel by [F12] and Lebesgue measurable by [F15]. Since $\varphi$ is also Lebesgue measurable by [F5, F15], [F7] gives ${\int }_{\mathbb{R}}{q}_{\mu }d\lambda ={\int }_{\mathbb{R}}\varphi d\lambda =1$. The set function $P_{\mu }(A)={\int }_A{q}_{\mu }d\lambda$ is a measure by [F16], and its total mass one makes it a probability by [F17]. For a Borel $A$, apply [F7] to $f={\mathbf{1}}_A{q}_{\mu }$ and $S(z)=\mu +\sigma z$: $N(\mu ,{\sigma }^2)(A)=\int {\mathbf{1}}_A(S(z))\varphi (z)d\lambda (z)={\int }_A{q}_{\mu }(x)d\lambda (x),$ since $\sigma q_{\mu }(S(z))=\varphi (z)$ and [F14] defines $N(\mu ,{\sigma }^2)$ as this affine pushforward. Thus each coordinate law has the stated density.

Put $S={\sum }_{i=1}^n(y_i-\bar{y}{)}^2$. Expanding and using ${\sum }_i(y_i-\bar{y})=0$ gives ${\sum }_i(y_i-\mu {)}^2=S+n(\mu -\bar{y}{)}^2$. Hence $L_{\mathbf{y}}(\mu )=(2\pi {\sigma }^2{)}^{-n\mathrm{/}2}{e}^{-S\mathrm{/}(2{\sigma }^2)}{e}^{-n(\mu -\bar{y}{)}^2\mathrm{/}(2{\sigma }^2)}$. This is continuous in $\mu$, hence Borel by [F12] and $\mathcal{L}(\mathbb{R})$-measurable by [F15]; therefore its evidence integral is defined by [F1].

For $k\ge 1$, recursively form the finite product data measure ${\nu }_k={\lambda }^{\otimes k}$ on $\mathcal{L}(\mathbb{R}{)}^{\otimes k}$ using [F8] and [F9]. Starting from the sigma-finite $\lambda$, [F10] ensures that each ${\nu }_k$ exists and is sigma-finite. The coordinate projections are measurable by [F8], so each pullback of $q_{\mu }$ is measurable by step 1.2; finite multiplication is continuous, making $f_k(y_1,\dots ,y_k)={\prod }_{i=1}^k{q}_{\mu }(y_i)$ product-measurable. Step 1.2 gives $\int f_{1}d{\nu }_1=1$. If $\int f_{k}d{\nu }_k=1$, the recursive product definition and Tonelli [F11] give $\int f_{k+1}d{\nu }_{k+1}=\int f_{k}d{\nu }_k\int q_{\mu }d\lambda =1$. Thus $P_{\mu ,n}(E):={\int }_E{f}_{n}d{\nu }_n$ is a measure by [F16] and a probability by [F17]. For measurable rectangles $E=A_1\times \cdots \times A_n$, repeated Tonelli gives $P_{\mu ,n}(E)={\prod }_{i=1}^n{P}_{\mu }(A_i)$. The product measure theorem [F10], iterated over the finite factors, identifies $P_{\mu ,n}$ as the product of the one-coordinate laws. Therefore these coordinates are independent and $L_{\mathbf{y}}(\mu )$ is their joint density.

Set $T(\mu )=\sqrt{n}(\mu -\bar{y})\mathrm{/}\sigma$, a $C^1$ diffeomorphism of $\mathbb{R}$ with inverse $u\mapsto \bar{y}+\sigma u\mathrm{/}\sqrt{n}$ and derivative $\sqrt{n}\mathrm{/}\sigma$. Since $e^{-u^2\mathrm{/}2}=\sqrt{2\pi }\varphi (u)$ is nonnegative and Lebesgue measurable by [F5, F15], [F7] gives ${\int }_{\mathbb{R}}{e}^{-n(\mu -\bar{y}{)}^2\mathrm{/}(2{\sigma }^2)}d\lambda (\mu )=\frac{\sigma }{\sqrt{n}}\sqrt{2\pi }$. Integrating the factorization in step 1.3 against the flat prior gives $Z(\mathbf{y})=(2\pi {\sigma }^2{)}^{-n\mathrm{/}2}{e}^{-S\mathrm{/}(2{\sigma }^2)}\frac{\sigma \sqrt{2\pi }}{\sqrt{n}}$, which is strictly positive and finite for every finite data vector.

Dividing step 1.3 by the finite positive evidence in step 2.2 gives $h(\mu )=\frac{\sqrt{n}}{\sigma }\varphi \text{\hspace{0.17em}⁣}\left(\frac{\sqrt{n}(\mu -\bar{y})}{\sigma }\right)$, the displayed Lebesgue density. Put $s=\sigma \mathrm{/}\sqrt{n}>0$; $S(z)=\bar{y}+sz$ is a $C^1$ diffeomorphism with inverse $\mu \mapsto (\mu -\bar{y})\mathrm{/}s$. For every Borel $A$, [F14] defines $N(\bar{y},s^2)$ as this pushforward; [F7] applied to $f={\mathbf{1}}_{A}h$, which is Lebesgue measurable by [F12, F13, F15], gives ${\int }_{A}h(\mu )d\lambda (\mu )=\int {\mathbf{1}}_A(S(z))h(S(z))sd\lambda (z)=\int {\mathbf{1}}_A(S(z))\varphi (z)d\lambda (z)=N(\bar{y},s^2)(A)\mathrm{.}$ The update on the full Lebesgue sigma-algebra is a probability by [F2]. If $n=0$ were allowed, its empty-product likelihood would be $1$ and the evidence would be $\lambda (\mathbb{R})=+\mathrm{\infty }$ by step 1.1, so no normalized update would exist.