# Showing a "signed Markov transition density" will lead to a trivial measure on path space.

Let for all $t>0$, $x\mapsto p(t,x)$ be a Schwartz function on $\mathbb R$, satisfying $\int_{\mathbb R}p(t,x)\mathrm dx=1$ and $\int_\mathbb{R}|p(t,x)|\mathrm dx\equiv C>1$ for all $t>0$ (so $p(t,x)$ is not positive) and the Chapman-Kolmogorov condition $\int_\mathbb Rp(t,\mathrm dz -x)p(s,y-z)=p(t+s,y-x)$. Let $\Omega =(C^0_0([0,1],\mathbb R),\|\cdot\|_{\infty})$ denote the space of continuous paths started at $0$, equipped with the Borel sigma algebra.

Let $C\subset\Omega$ be a cylinder set $$C=\{\omega: \omega(t_i)\in A_i, \,i=1,\dots,n\}\quad\text{some}\,A_i\in\mathcal B(\mathbb R).$$

Then we may consistently define a signed, finitely additive measure on paths $$\mu(C):=\int_{A_1}\dots\int_{A_n}\Pi_{i=1}^np(t_i-t_{i-1},x_i-x_{i-1})\mathrm dx_i,$$ with $t_0=x_0=0$.

Note that \begin{align} \mathrm{TV}(\mu)\geq&\int_{\mathbb R}\dots\int_{\mathbb R}\Pi_{i=1}^n|p(t_i-t_{i-1},x_i-x_{i-1})|\mathrm dx_i \\=\left(\int_\mathbb{R}|p(1,x)|\mathrm dx\right)^n=C^n. \end{align}

For any $n\in \mathbb N$ and $C>1$ so the total variation of $\mu$ is $\infty$.

I suspect and would like to prove that $\mu$ is essentially a useless measure by which I mean that its total variation is $\infty$ not only on the whole space (as I have shown above) but also on a large class of compact sets $K\subset\subset\Omega$. Is this even true and if so, could you please give me a hint how to prove it?

Hocberg constructs a Markovian signed measure (of unbounded variation) on the space of continuous paths, associated with the operator $Lu:={\partial^4u\over\partial x^4}$. The transition density $p$ is the fundamental solution of ${\partial p\over \partial t}+Lp = 0$.