I'm trying to deal with a random set M ⊂ R k of which we...

$\begin{array}{r}E(|M|)={\int }_{\mathrm{\Omega }}|M|dP(M)={\int }_{\mathrm{\Omega }}{\int }_{M}d\omega dP(M)={\int }_{\mathrm{\Omega }}{\int }_{{\mathbb{R}}^k}{\chi }_M(\omega )d\omega dP(M)\\ \stackrel{Tonelli}{=}{\int }_{{\mathbb{R}}^k}{\int }_{\mathrm{\Omega }}{\chi }_M(\omega )dP(M)d\omega (\ast )\end{array}$
But now
${\chi }_M(\omega )=1\iff \omega \in M\iff {\chi }_{A_{\omega }}(M)=1$
where $A_{\omega }:=\{M:\omega \in M\}$
Thus
$(\ast )={\int }_{{\mathbb{R}}^k}{\int }_{\mathrm{\Omega }}{\chi }_{A_{\omega }}(M)dP(M)d\omega ={\int }_{{\mathbb{R}}^k}P([M:\omega \in M])d\omega$

If $P$ is the distribution of $M$ (which you need to be super careful about, you want to define the sigma algebra too), let $\mathcal{P}({\mathbb{R}}^k)$ be the power set of ${\mathbb{R}}^k$ ($P$ is a distribution over $\mathcal{P}({\mathbb{R}}^k)$), you can write :
$\begin{array}{rl}\mathbb{E}[|M|]& ={\int }_{\mathcal{P}({\mathbb{R}}^k)}{\int }_{M}d\omega dP(M)\\ & ={\int }_{\mathcal{P}({\mathbb{R}}^k)}{\int }_{{\mathbb{R}}^k}\mathbf{1}(\omega \in M)d\omega dP(M)\\ \\ & ={\int }_{{\mathbb{R}}^k}{\int }_{\mathcal{P}({\mathbb{R}}^k)}\mathbf{1}(\omega \in M)dP(M)d\omega \\ & ={\int }_{{\mathbb{R}}^k}P(\omega \in M)d\omega \end{array}$
If I were you, I would make sure that you are allowed to exchange the two integrals there.