What are the dimensions/units in the kernel density estimate? I've been reading about the kernel density estiamte and the dimensions/units don't make sense to me.
Let $x$ have the dimensions of distance/length (L), i.e.
$$[x] = L,$$
where I am using square brackets to denote the dimensions of the symbol.
According to Wikiepdia, if ($x_1$, $x_2$, .... $x_n$) are indpependendent and identically distributed samples drawn from some univariate distribution with an unknown density $f$ at any given point $x$, then the kernel density estimator is
$$\hat{f}_h(x)=\frac{1}{nh}\sum_{i=1}^n K\left(\frac{x-x_i}{h}\right),$$
where $K$ is a kernel and $h$ is the bandwidth.
I believe the dimensions of $\hat{f}_h(x)$ are one over length, $K(x)$ has no dimensions and the dimension of $h$ are length, i.e.
$$[\hat{f}_h(x)] = L^{-1},$$
$$[K(x)] = 1,$$
$$[h]=L.$$
However, further down on the Wikipedia page, it says
$$\mathrm{AMISE}(h) = \frac{R(K)}{n h} + \mathrm{other\ terms},$$
where
$$R(K)=\int_{-\infty}^\infty K(x)^2 dx.$$
Since
$$[\mathrm{AMISE}(h)] = [\mathrm{MISE}(h)]$$
we know that
$$[\mathrm{AMISE}(h)] = L^{-1}.$$
However,
$$[R(K)]=L,$$
therefore,
$$\left[\frac{R(K)}{nh}\right]=1.$$
But this can't be right as
$$[\mathrm{AMISE}(h)] = \left[\frac{R(K)}{nh}\right].$$
Do you know where I have gone wrong? I must have got the dimensions of one of the symbols wrong?

I think it could be explained if
$$R(K) = \int_{-\infty}^\infty K(\mu)^2d\mu,$$
where $\mu$ is dimensionless, i.e.,
$$[\mu]=1.$$
