Let $X$ be a measurable space, $\mu$ be a $\sigma$-finite measure on $X$, and $H$ be a separable reproducing kernel Hilbert space over $X$ with a measurable kernel $k$.


At a certain part in a proof I am reading there is the condition


$S\_k : L\_q (\mu) \to H $ has a dense image if and only if $id: H \to L\_p(\mu)$ is injective.


We know (it is shown in the proof) $id$ is the inclusion operator and is continuous. $S\_k$ is the adjoint of the inclusion operator. And $H$ consists if $p$-integrable functions. 


My question is how can the inclusion mapping $id$ *always* be injective? $H$ is a Hilbert space of function and $L\_p(\mu)$ is a space of **equivalence classes** of functions. I am really confused on the intuition here.


Consider the following:


* I take the function $f \in H$. I know $f$ belongs to some element of $L\_p$ and is measurable. Thus the inclusion makes sense.
* The inclusion maps $f$ to its equivalence class $[f]$
* Say I take a measurable $p$-integrable function $g$ which equals $f$ almost everywhere. (I am obviously assuming such a function exists in $H$).
* We see the inclusion maps $g$ to $[f]$.
* Therefore the inclusion map can't be injective


I could see that this can fail if no such function $g$ exists in $H$. (For example if the RKHS $H$ has a *continuous* kernel then every $f \in H$ is continuous and the $g$ in question could not be found.) But I have no reason to believe that such a $g$ can't exist. I am obviously suffering from a knowledge gap. Where am I going wrong in how I am thinking about this.


**PS:** The proof can be found on page 126 here: [here](https://books.google.com/books?id=HUnqnrpYt4IC&lpg=PP1&dq=support%20vector%20machines&pg=PA126#v=onepage&q=support%20vector%20machines&f=false)



Injective inclusion map from RKHS function space to $L_p(\mu)$
It seems, from comments, that the question is based on a misreading. The text does not assert that the inclusion map is always injective; it only gives a necessary and sufficient condition for the map to be injective. It is easy to find examples where it is, and examples where it isn't.


As an incredibly trivial example, let $H = \mathbb{R}$ with its usual inner product, considered as the RKHS of real-valued functions on a one-point set $X = \{x\}$. (The reproducing kernel is $k(x,x) = 1$.) If $\mu$ is a measure on $X$ assigning positive mass to $x$, then the inclusion $H \to L^2(X,\mu)$ is injective. If $\nu$ is the zero measure, then the inclusion $H \to L^2(X,\nu)$ is not injective.


