Why do we study Cameron-Martin Space and what is the motivation behind it? I'm reading about Gaussian Measures and the chapters always define the Cameron-Martin space shortly after.
Typically they'll define a covariance operator first.  Let $U$ be a separable Banach space, and $\mu$ be a centered Gaussian measure on $U$. $U^*$ is the dual.  The Covariance operator $C:U^*\times U^*\to\mathbb{R}$ is first given by
$$C_{\mu}(f)(g):= \int_Uf(x)g(x)\mu(dx)$$
First define $|x|_{H(\mu)} = \sup_{l\in U^*}\{ l(x) : C_{\mu}(l)(l)\leq 1\}.$  The Cameron-Martin space is then defined as $$H(\mu) := \{x \in U : |x|_{H(\mu)} < \infty \}.$$
(At least that's one of a couple definitions)
Intuitively, I've also heard that the Cameron-Martin space is the set of all elements that make the null sets of $\mu$ translation-invariant (ie you can shift a null-set by that element and it will still be a null set). But I still feel like I'm missing a broader perspective.
 A: The infinite-finite dimension dichotomy , and Fourier coefficients
When we study (multivariate) normal random variables on $\mathbb R^d$, we often have equivalent ways of understanding them. One famous result is that any normal random variable can be specified by providing its mean and covariance matrix. While extending these concepts to infinite dimensions on separable Banach spaces, something analogous occurs : every Gaussian measure $\mu$ on $\mathcal{B}$ is determined by its "mean vector" (an element of $\mathcal{B}$) and a "covariance matrix" (a linear functional on $\mathcal{B}^* \times \mathcal{B}^*$). It is the "covariance matrix" that makes all the difference.
The proof goes via the "Fourier transform" and covers the finite dimensional case as well : let's assume that the mean vector is $0$ (one can easily carry over the argument to anywhere else) and suppose that the covariance operator is given by $C_{\mu}$.
One defines the Fourier transform of $\mu$, called $\hat{\mu} : \mathcal B^* \to \mathbb R$ by $$
\hat{\mu}(l) = \int_{\mathcal{B}}e^{il(x)}d\mu(x)
$$
Now, because $\mu$ is a Gaussian measure, it turns out that one can write $$
\hat{\mu}(l) = \exp\left(-\frac{1}{2} C_{\mu}(l,l)\right)
$$
[This in fact characterizes Gaussian measures on separable Banach spaces] Therefore, one sees that the map $\phi(l) = C_{\mu}(l,l)$ determines $\mu$, since it determines the Fourier transform of $\mu$ via the above formula, and Fourier transforms uniquely determine measures on separable Banach spaces.
The big difference, however, comes as follows. In finite dimensions, if $C_{\mu}$ is the covariance matrix of $\mu$, then the map $\phi$ is determined by the vectors $e_i^TC_{\mu}$ where $e_i$ are the standard basis vectors. But that's true of every other matrix as well : we didn't really use the fact that $C_{\mu}$ is special in any way ,we just used the fact that every finite matrix (and not just positive semidefinite ones) is determined by its rows. So knowing the map $\phi$ still reduces to knowing every single entry of $C_{\mu}$. To put it another way, you're using every possible bit of information you know about $C_{\mu}$, to determine $\mu$.
However, in infinite dimensions, $\phi(l)$ is determined by knowing the maps $l' \to \phi(l, l')$ from $B^*$ to $\mathbb R$ because these will give us the "rows" of $\phi$, hence a description of $C_{\mu}$. These maps belong to $B^{**}$ (linear functionals of $B^*$). It turns out, however, that because the speciality of $C_{\mu}$ counts for something in infinite dimensions, these maps in fact belong to $\mathcal{B}$ (i.e. they belong in the image of the map $\xi : \mathcal B \to \mathcal B^{**}$ given by $[\xi(x)](l) = l(x)$), and not just that : even as elements of $B$, they are very, very special.
That is, thanks to the special properties of $C_{\mu}$, not every element of $B$ comes from a map of the form $l' \to \phi(l, l')$. Only some of them do. A very , very small number of them. And YET, because of what we've said before, these very small number of elements determine $C_{\mu}$ and hence determine $\mu$.
Those elements, they form (i.e. are dense in a specific norm in) the Cameron-Martin space corresponding to $\mu$. A very small space, but a very powerful one, because the elements required to determine $\mu$ come from this space.

Definition and rigorous result

Definition : Given a Gaussian measure $\mu$ with covariance operator $C_{\mu} : \mathcal{B}^* \times \mathcal{B}^* \to \mathbb R$, we already saw that $\hat{C}_{\mu} \in B^{**}$ defined by $[\hat{C}_{\mu}(l)](l') = C_{\mu}(l,l')$ can be interpreted as an element of $B$. The closure of all such elements under the norm $\|h\|^2_{\mu} = C_{\mu}(h^*,h^*)$ (where $h^*\in B$ is any element such that $C_{\mu}(h^*,l) = l(h)$ for all $l \in B^*$) is called the Cameron-Martin space corresponding to $\mu$ and is denoted $\mathcal{H}_{\mu}$.

That is, given $\mu$, we collect all possible values in $B$ that the map $l' \to C_{\mu}(l,l')$ can correspond to, and close it under a norm corresponding to $\mu$. The closure is essentially done to ensure that the Cameron-Martin space obtains a Hilbert-space structure. Note that $\mathcal H_{\mu} \subset \mathcal{B}$ because of the relation $\|h\|_{\mu} \geq \|C_{\mu}\|^{-1} \|h\|^2$ for $h \in \mathcal B$, so the completion continues to have elements of finite $\mathcal B$ norm.
By making everything before a little more rigorous, we obtain the following result.

If $\mathcal B$ is a Banach space and $\mu,\nu$ are two Gaussian measures such that $\mathcal H_{\mu} = \mathcal H_{\nu}$ and $\|h\|_{\mu} = \|h\|_{\nu}$ for every $h \in \mathcal H_{\mu}$, then $\mu=\nu$ as measures on $\mathcal {B}$. In other words, a Gaussian measure is determined by its Cameron-Martin space and the norm on that space.

This is perhaps the most "broad" motivation for the Cameron-Martin space. How it ties in with the rest of the theory is also fascinating.

An equivalent description of $\mathcal {H}_{\mu}$
Recalling the Bochner integral, let $L^2(\mathcal B, \mu) = \{l : \int_{\mathcal B} (l(x))^2\mu(dx) < \infty\}$, where $l$ isn't an element of $B^*$, but instead is a $\mu$-a.e. equivalence class of functionals (i.e. $l_1 = l_2$ $\mu$-a.e. precisely when $l_1(x) \neq l_2(x)$ has $\mu$ measure zero).
While defining the CM space, we showed that every $h\in \mathcal H_{\mu}$ corresponded to some $h^* \in B^*$ in the way that $C_{\mu}(h^*,l) = l(h)$ for every $l\in B^*$. It's true that the map $h \to h^*$ is not unique. However, if we identify functionals up to $\mu$ a.e. equivalence, we have the following result.

$\mathcal H_{\mu}$ is isomorphic to the closure of $\mathcal{B}^*$ in $L^2(\mathcal{B},\mu)$. The latter space is called the reproducing kernel Hilbert space, or RKHS.

That is, the Cameron-Martin space can also be identified with the set of all ($\mu$ a.e. equivalence classes of) linear functionals $l$ such that knowing $C_{\mu}(l,l)$ is enough to determine $C_{\mu}$ completely.
The upshot of both the sections I've just written, is that one can associate to every Gaussian measure on a separable Banach space, another Hilbert space whose elements and their norms "capture" all possible entries of the covariance matrix $C_{\mu}$, thereby determining $C_{\mu}$. That space can either interpreted as a vector space in $\mathcal B$ under a different norm, or as a subspace of $L^2(\mathcal{B},\mu)$.
Another remarkable result which shows how important the Cameron Martin space is, lies in the pushforward of Gaussian measures. As Gaussian measures are determined by the Cameron-Martin space, one can specify how to push forward a Gaussian measure onto another Banach space, simply by specifying how the Cameron-Martin space is pushed forward under some linear map. The formal result is

Let $\mu$ be a Gaussian measure on $\mathcal B_1$ with CM space $\mathcal H_{\mu}$. Let $A : H_{\mu} \to \mathcal B_2$ be a bounded linear operator such that for some Gaussian measure $\nu$ on $\mathcal B_2$, it is true that $C_{\nu}(h,k) = \langle A^*h,A^*k\rangle_{\mu}$. Then, there exists a measurable map $\hat{A} : \mathcal B_1 \to \mathcal B_2$ such that $\nu$ is the pushforward of $\mu$ under $A^*$, $\hat{A} = A$ on $\mathcal H_{\mu}$, and $A$ is linear on a subspace of $\mu$-measure $1$. Furthermore, $\hat{A}$ is unique $\mu$ a.e.

That is, a linear map from the Cameron-Martin space that produces a covariance operator on the other Banach space, leads to a (unique) map which pushes forward the Gaussian measure. This is perhaps the broadest interpretation of the CM space : a space that is used not only to determine a Gaussian measure, but is used to push it forward onto other spaces.
Remember, the Hahn-Banach theorem tells you that there can be many, many possible extensions of a linear map $A$ on $\mathcal H_{\mu}$ to the domain $\mathcal B_1$. The Cameron-Martin space is so specially linked with $\mathcal B_1$, that defining how it transforms under a particular linear map is tantamount to saying how the entire Gaussian measure is pushed forward according to a particular extension of that map.
I hope I have provided a more "high-level" demonstration of how the CM space is linked very, very deeply with the underlying Gaussian measure. It is possible to deduce some other interesting properties of the CM space (including the Cameron-Martin theorem), but these are a little more specific compared to the above "by definition" exposition of the CM space.
A: As far as I know the reasoning goes as follows: a multivariate normal vector with non singular covariance matrix is absolutely continuous wrt to Lebesgue. If we translate this law we get a new law absolutely continuous to the original law.
This concept is then extented to the Brownian Motion by Girsanov Theorem. Through this we can prove the Cameron-Martin formula which tells us by translating  the Wiener measure $\mathcal{W}$ with a function $f \in D[0,T]$, we get a measure $\tilde{\mathcal{W}}$ absolutely continuous wrt $\mathcal{W}$. Here we have defined the following space: $D[0,T] := \{ f:[0,T] \to \mathbb{R}: \exists g \in L^2[0,T]: f(t) = \int_{0}^{t}g(s) ds, \quad \forall t \in [0,T] \}$.
Hence CM gives the infinite-dimensional counterpart of the Gaussian Measures.
Hope this helps.
