• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $T>0$
  • $I:=(0,T]$
  • $d\in\mathbb N$
  • $M:\Omega\times\overline I\times\mathbb R^d\to\mathbb R$ such that $M(\;\cdot\;,\;\cdot\;,x)$ is $\mathcal A\otimes\mathcal B(\overline I)$-measurable for all $x\in\mathbb R^d$

Now, let $i\in\left\{1,\ldots,d\right\}$ and $$N(\omega,t,x,\theta):=\frac{M(\omega,t,x+\theta e_i)-M(\omega,t,x)}\theta$$ for $(\omega,t,x,\theta)\in\Omega\times\overline I\times\mathbb R^d\times\left(\mathbb R\setminus\left\{0\right\}\right)$.

Let $\delta\in(0,1]$. Assuming that for all $p\ge2$, there is a $C>0$ such that $$\int\sup_{t\in\overline I}\left|N(\omega,t,x,\theta)-N(\omega,t,y,\vartheta)\right|^p\operatorname P\left[{\rm d}\omega\right]\le C\left(|x-y|^{\delta p}+|\theta-\vartheta|^{\delta p}\right)\tag1$$ for all $(x,\theta),(y,\vartheta)\in\mathbb R^d\times\left(\mathbb R\setminus\left\{0\right\}\right)$, how can we conclude that $M(\omega,t,\;\cdot\;)$ is partially differentiable with respect to the $i$th variable for $\operatorname P$-almost all $\omega\in\Omega$ for all $t\in\overline I$?

This should be an application of a Kolmogorov-type theorem, but which version of that theorem do we need and how do we need to apply it exactly?


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