The Brownian kernels form a semigroup

Fix $m\in \mathbb{R}$ and $\sigma >0$. Write ${\gamma }_s(y):=(2\pi s{)}^{-1\mathrm{/}2}\exp (-y^2\mathrm{/}(2s))$ for $s>0$. Since $N(m,{\sigma }^2)$ is by [F2] the law of $m+\sigma Z$ with $Z\sim \gamma$, its distribution function at $y$ is $P(m+\sigma Z\le y)=P(Z\le (y-m)\mathrm{/}\sigma )={\int }_{-\mathrm{\infty }}^{(y-m)\mathrm{/}\sigma }\phi (u)du$; for $L>\max (0,(m-y)\mathrm{/}\sigma )$, [F3] applied to the increasing affine map $u=(v-m)\mathrm{/}\sigma$ on $[m-\sigma L,y]$ gives ${\int }_{m-\sigma L}^y{\gamma }_{{\sigma }^2}(v-m)dv={\int }_{-L}^{(y-m)\mathrm{/}\sigma }\phi (u)du$, and letting $L=L_0+n\to \mathrm{\infty }$ with $L_0=1+\max (0,(m-y)\mathrm{/}\sigma )$ and $n\in \mathbb{N}$, [F3]'s monotone convergence identifies the distribution function of $m+\sigma Z$ with ${\int }_{-\mathrm{\infty }}^y{\gamma }_{{\sigma }^2}(v-m)dv$, that is with that of the measure with density $v\mapsto {\gamma }_{{\sigma }^2}(v-m)$.

The density measure in step 1.1 exists by [F6]. Its total mass is one: let $y\to \mathrm{\infty }$ through positive integers in the established half-line identity and use [F3] and the mass-one assertion of [F2]. By [F4], with countable choice supplied by [F5], the density of step 1.1 represents $N(m,{\sigma }^2)$ when $\sigma >0$, and by [F6] expectations against that law are integrals of the product with the density.

By [F1] with the one-term list $0<t$, the random variable $B_t-B_0$ has law $N(0,t)$; since $B_0=0$ almost surely, $B_t$ has the same law $N(0,t)$. By steps 1.1-2.1 with $m=0$ and $\sigma =\sqrt{t}$, the law of $B_t$ has density ${\gamma }_t$, and the law of $x+B_t$ has density $v\mapsto {\gamma }_t(v-x)=p_t(x,v)$; hence for bounded Borel $f$, $P_{t}f(x)={\int }_{\mathbb{R}}f(v)p_t(x,v)dv=E[f(x+B_t)]$, while for $t=0$ both sides equal $f(x)$ by the convention $P_{0}f=f$ and $B_0=0$ almost surely. This is assertion 1.

Continuing the kernel analysis of step 3.1, fix $s,t>0$ and $x,z\in \mathbb{R}$ and put $A:=\frac{1}{2s}+\frac{1}{2t}=\frac{s+t}{2st}$, $m:=\frac{tx+sz}{s+t}$ and $C:=\frac{(z-x{)}^2}{2(s+t)}$.

Expanding squares gives $\frac{(y-x{)}^2}{2s}+\frac{(z-y{)}^2}{2t}=A(y-m{)}^2+C$: $A$ is the coefficient of $y^2$, $2Am=x\mathrm{/}s+z\mathrm{/}t$ is the coefficient of $-y$, and the constant coefficient identity is $A{m}^2+C=x^2\mathrm{/}(2s)+z^2\mathrm{/}(2t)$; consequently $p_s(x,y)p_t(y,z)=\frac{1}{2\pi \sqrt{st}}\exp (-A(y-m{)}^2-C)$.

For $L>0$, [F3] applied to the increasing affine map $v=\sqrt{A}(y-m)$ on $[m-L\mathrm{/}\sqrt{A},m+L\mathrm{/}\sqrt{A}]$ converts the Gaussian integral [F7] into ${\int }_{m-L\mathrm{/}\sqrt{A}}^{m+L\mathrm{/}\sqrt{A}}{e}^{-A(y-m{)}^2}dy=A^{-1\mathrm{/}2}{\int }_{-L}^L{e}^{-v^2}dv$; letting $L\to \mathrm{\infty }$ with [F3]'s monotone convergence and [F7] gives ${\int }_{\mathbb{R}}{e}^{-A(y-m{)}^2}dy=\sqrt{\pi \mathrm{/}A}=\sqrt{2\pi st\mathrm{/}(s+t)}$. Multiplying by the constant of step 5.1, ${\int }_{\mathbb{R}}{p}_s(x,y)p_t(y,z)dy=\frac{1}{2\pi \sqrt{st}}\sqrt{\frac{2\pi st}{s+t}}{e}^{-C}=\frac{1}{\sqrt{2\pi (s+t)}}{e}^{-(z-x{)}^2\mathrm{/}(2(s+t))}=p_{s+t}(x,z)$, which is the kernel identity of assertion 3.

Take bounded Borel $f\ge 0$ and $s,t>0$. For fixed $t>0$, Tonelli [F8] on the Borel Lebesgue measure spaces shows that $y\mapsto \int f(z)p_t(y,z)dz$ is Borel, since $(y,z)\mapsto f(z)p_t(y,z)$ is nonnegative and product-measurable. Its absolute value is at most $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ by the density mass in step 2.1. Thus $P_{t}f$ is bounded Borel and $P_s(P_{t}f)(x)={\int }_{\mathbb{R}}({\int }_{\mathbb{R}}f(z)p_t(y,z)dz)p_s(x,y)dy$; the integrand is nonnegative and product-measurable, so [F8] rewrites the iterated integral as ${\int }_{\mathbb{R}}f(z)({\int }_{\mathbb{R}}{p}_s(x,y)p_t(y,z)dy)dz={\int }_{\mathbb{R}}f(z)p_{s+t}(x,z)dz=P_{s+t}f(x)$, using step 6.1 for the inner integral. Applying this to $f^{+}$ and $f^{-}$ and subtracting extends it to general bounded Borel $f$, all four integrals being finite; if $s=0$ or $t=0$ both sides are $P_{t}f$ or $P_{s}f$ by the convention $P_{0}f=f$.

Taking $f=1$ in step 3.1 gives $P_{t}1(x)={\int }_{\mathbb{R}}{p}_t(x,y)dy=1$ for $t>0$, and for $t=0$ it is the convention, so every $P_t$ maps bounded Borel functions to bounded Borel functions with sup norm at most that of its argument; this is assertion 4. Assertion 1 is step 3.1, assertion 2 is step 7.1 and assertion 3 is steps 6.1 and 7.1. AC is used through the Brownian and normal-law interfaces of [F1]-[F2], and [F5] supplies the countable choice required both by the Riemann-to-Lebesgue conversion in [F3] (used in steps 1.1 and 6.1) and by the distribution-function uniqueness in [F4]. The substitution, Gaussian-integral and Tonelli interfaces make no additional choice beyond these declared uses.