I'm betting `yes, sure!', but don't see it. Could someone please point me toward, 
or construct for me, a Lagrangian submanifold immersed in 
standard symplectic ${\mathbb R}^{2n}$ for $n > 1$, whose closure is all of ${\mathbb R}^{2n}$?


(For an $n =1$ example, one can use the leaves arising from 
[this modification by Panov](http://www2.imperial.ac.uk/~dpanov/TORUS.PDF) of irrational flow on the two-torus.) 


Strong preference given to analytic immersions of ${\mathbb R}^n$.


Holomorphically immersed complex lines which are dense in complex 2-space -- i.e. dense ${\mathbb C}$'s in ${\mathbb C}^2$ -- are well-known. Ilyashenko in 1968 showed that the typical solution of the typical polynomial ODE (in complex time) yields such a curve. Following his line
of thought, it might be easier to construct an entire singular Lagrangian foliation of ${\mathbb R}^{2n}$ whose typical leaf is dense, rather than the one submanifold. 


Motivation: I have a certain unstable manifold related to a Hamiltonian system.
It is Lagrangian. I would like to be ``as dense as can be'', so I'd like to know 
how dense can that be. 



Can a  Lagrangian submanifold of  ${\mathbb R}^{2n}$ be dense ($n>1$)?
Your question already has the answer in it for $n=2$. Take a connected complex curve $L\subset\mathbb{C}^2$ that is dense in $\mathbb{C}^2$. Then $L$ is Lagrangian for the real part of the holomorphic $2$-form $\Upsilon = dz^1\wedge dz^2$. This real part of $\Upsilon$ is equivalent to the standard symplectic structure on $\mathbb{R}^4$ by a linear change of variables.


**Added comment about injectivity:** Note, by the way, that one can easily arrange for such an $L$ to be a submanifold, not just the image of an immersion (i.e., the immersion is *injective*). One simple explicit way to do this is to select constants $\lambda\_1,\ldots,\lambda\_k$ such that the subgroup in $\mathbb{C}^\times$ generated by the numbers $\mathrm{e}^{2\pi i\lambda\_1},\ldots,\mathrm{e}^{2\pi i\lambda\_k}$ is dense in $\mathbb{C}^\times$ and consider the linear differential equation
$$
\frac{dy}{dx} = \left(\frac{\lambda\_1}{x-x\_1}+\cdots + \frac{\lambda\_k}{x-x\_k}\right)\ y
$$
where $x\_1,\ldots,x\_k\in \mathbb{C}$ are distinct. The graph of any nonzero multi-valued solution $y(x)$ over $\mathbb{C}\setminus\{x\_1,\ldots,x\_k\}$ will then be dense in $\mathbb{C}^2$. (Consider the holonomy around the punctures $x\_j$.) Of course, these graphs are the Riemann surfaces associated to the multivalued functions
$$
y = y\_0 (x{-}x\_1)^{\lambda\_1}\cdots(x{-}x\_k)^{\lambda\_k}
$$
(when $y\_0\not=0$). These are obviously integral curves (leaves) of the polynomial $1$-form
$$
\omega = (x{-}x\_1)\cdots(x{-}x\_k)\ dy - q(x) y\ dx
$$
for some polynomial $q$ of degree at most $k{-}1$ in $x$. Aside from the obvious closed leaves $x-x\_j=0$ and $y=0$, the rest of the leaves are dense submanifolds. (This just gives a simple, explicit example of the general theorem that Richard quoted.)


**Dense analytic curves in $\mathbb{R}^2$:** It is not hard to construct dense, connected analytic curves in $\mathbb{R}^2$: There exist analytic metrics on the $2$-sphere that have geodesics that wander densely over the surface. Now take such a geodesic and remove a point from $S^2$ through which the geodesic doesn't pass. What's left is a dense analytic curve in $\mathbb{R}^2$. If you are willing to use Finsler metrics, you can even do this with a rotationally invariant real analytic Finsler metric on the $2$-sphere (eg. Katok's examples), so that you can write down the dense analytic curve very explicitly.


I'll think about the case $n>2$. I don't see it yet either, but maybe it's not too hard.


