0
$\begingroup$

Given $\Omega_{n} = \{\alpha:[0,1]\rightarrow[0,1]^{n}\,|\,\alpha\,\,\text{is smooth}\}$, consider the equivalence relation: \begin{align*} & \alpha_{1} \sim \alpha_{2} \Leftrightarrow \int_{0}^{1}\langle G(\alpha_{1}(t)),\alpha^{\prime}_{1}(t)\rangle\,\mathrm{d}t = \int_{0}^{1}\langle G(\alpha_{2}(t)),\alpha^{\prime}_{2}(t)\rangle\,\mathrm{d}t,\\ \end{align*}

Where $\alpha_{1}(0) = \alpha_{2}(0)$ and $\alpha_{1}(1) = \alpha_{2}(1)$. It is assumed that $G:[0,1]^{n}\rightarrow[0,1]^{n}$ is known. Such set can naturally be considered as a metric space (thus a topological space) in accordance to the norm:

\begin{align*} \lVert\alpha\rVert = \max_{0 \leq t \leq 1}\lVert\alpha(t)\rVert_{2} \end{align*}

Let us define $\Omega := \Omega_{n}/G$ as the quotient space according to the above-mentioned equivalence relation. Thus we can introduce a topology on $\Omega$. Precisely speaking, the quotient topology: \begin{align*} \tau_{\Omega} := \{O\subset\Omega\,|\,\pi^{-1}(O)\in\tau_{\Omega_{n}}\} \end{align*}

Where $\pi$ is the map which associates each $\alpha\in\Omega_{n}$ to $[\alpha]\in\Omega$: $\pi(\alpha) = [\alpha]$. Finally, given the topological space $(\Omega,\tau_{\Omega})$, we can construct its associated Borel $\sigma$-algebra $\Sigma$.

Here is my question: how do we introduce a signed finite measure on $(\Omega,\Sigma)$? Precisely, I would like to define a triple $(\Omega,\Sigma,\mathbb{P})$ such that $\mathbb{P}([\alpha]) = -\mathbb{P}([\alpha^{-}]) \geq 0$, where $\alpha^{-}(t) = \alpha(1-t)$.

Such problem makes part of my research project on negative probabilities. I apologize if the question does not fit into Math Overflow context. Any help is appreciated. Thanks in advance.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.