I'm reading on kernel mean embeddings, and I got stuck a small detail, but I cannot figure it out, so I'm asking it here :)
Some context, we are given a Reproducing Kernel Hilbert Space on a compact set $\mathcal{X}$ $H$ with kernel $k : \mathcal{X}\times \mathcal{X}\to \mathbb{R}$ and a probability measure $\mathbb{P}$ defined on $\mathcal{X}$. The kernel mean embedding is defined as (we don't worry for now under what conditions this is well defined, we will assume everything is nice at the moment) \begin{align} \mu_{\mathbb{P}} := \int k(x,\cdot)\mathbb{d}\mathbb{P}(x). \end{align} On page 31, section 3.1.1, it is said that if we take $k(x,x') = \langle x,x\rangle$, the linear kernel, that then $\mu_{\mathbb{P}}$ ''becomes just the first moment of $\mathbb{P}$''.
But here is my question: I don't see why this is the case. When I work this out, I get \begin{align} \mu_{\mathbb{P}} &= \int k(x,\cdot)\mathbb{d}\mathbb{P}(x),\\ &= \int \langle x, \cdot\rangle\mathbb{d}\mathbb{P}(x). \end{align} Whereas the first moment, aka the expectation, is defined as \begin{align} \mathbb{E}[X] = \int_{\mathcal{X}} x \mathbb{d}\mathbb{P}(x) \end{align} And I do not see how I can get equality. I have tried to use Lemma 3.1 in the text. Plugging in the identity operator for $f$ seems to do the job, but then I still do not see how $\mu_{\mathbb{P}}$ is the expectation. Another thing that comes to mind, the expectation is an element in a field, but $\mu_{\mathbb{P}}$ is an element in a Hilbert space. So how does it make sense to even state that $\mu_{\mathbb{P}}$ ''becomes'' the first moment, as these elements are in different spaces.
Thanks for any help!
---------EDIT---------
If we assume that $\left\lVert\mathcal{K}(X, \cdot)\right\rVert_H < \infty$, then $\mu_{\mathbb{P}}\in H$. Then, from example 3.1, we can conclude that \begin{align} \mu_{\mathbb{P}}(t) = \int_{\mathcal{X}} \langle x, t\rangle \mathbb{d}\mathbb{P}(x) = \langle \textbf{m}_{\mathbb{P}}(1),\textbf{t}\rangle. \end{align} But still, how to show this?