Is it possible to integrate a function over a null hypersurface? For $f$ a compactly supported function on a spacetime $(\mathcal{M},g)$, one may define its integral over $\mathcal{M}$ by 
$$f\longmapsto \int_\mathcal{M}\star f,$$
where $\star$ is the Hodge star of $\mathcal{M}$. Since $\star$ is unique, this map is canonical. Suppose that $\Sigma$ is a null hypersurface of $\mathcal{M}$. (Assume $\Sigma\cap\operatorname{supp}f\ne\emptyset$.) We cannot define the integral of $f$ over $\Sigma$ by $$f\longmapsto \int_\Sigma\star_\Sigma f$$
because the induced metric on $\Sigma$ is degenerate, and hence $\star_\Sigma$ is not defined. Is there a way to "canonically" integrate $f$ over a null hypersurface?
 A: You cannot without other pieces of information. With a vector field J -- typically a timelike vector field-- you can: Defining the volume form $\omega = *J$ and next integrating $f\omega$.
I wrote timelike because in GR, one usually has some notion of time and that notion can be used. However one could also take the future-directed null normal $J= \ell$ to $\Sigma$.
ADDENDUM. If $\Sigma$ is an embedded null 3-submanifold in the 4D spacetime, locally it is parametrized by coordinates $x^1,x^2,U$ where $\partial_{x^1},\partial_{x^2}$ are spacelike and $\partial_U$ is light-like. It is possible to complete this set of coordinates with another lightlike coordinate $V$, constant over $\Sigma$, such that $g(\partial_U,\partial_V)=-1$ and $g(\partial_{x^1}\partial_V)= g(\partial_{x^2},\partial_V)= g(\partial_{x^1}\partial_U)= g(\partial_{x^2},\partial_U)=0$. 
The normal co-vector to $\Sigma$ is $dV$, however its contravariant form is $-\partial_U$ which is tangent to $\Sigma$. 
What I meant above is (a bit improperly perhaps)
$$\ell := \partial_V$$ 
so that, with may conventions, $*$ transforms $1$-vector ($\ell$) into a $3$-form $\omega$. The latter turns out to be proportional to $\sqrt{|\det g|} \:dx^1\wedge dx^2 \wedge dU$, where
$$g = -dU\otimes dV -dV\otimes dU + \sum_{i,j=1}^2h_{ij}dx^i \otimes dx^j$$ 
so that $\sqrt{|\det g|}= \sqrt{\det h}$. 
The form $\omega$ is not canonical as there are many ways to fix coordinates $x^1,x^2, U,V$.
