TITLE: Stokes' theorem etc., for non-Hausdorff manifolds
QUESTION [12 upvotes]: This question is prompted by another one. 
I want to motivate the definition of a scheme for people who know about manifolds(smooth, or complex analytic). So I define a manifold in the following way.
Defn: A smooth $n$-manifold is a pair $(X, \mathcal{O}_X)$, where $X$ is a topological space and $\mathcal{O}_X$ is a sheaf of rings on it, such that, every point $x \in X$ has a neighborhood $U_x$ which is homeomorphic to an open set $V_x$ in $\mathbb{R}^n$ and $\mathcal{O}_X$ restricted to  $U_x$ is isomorphic to the sheaf of ring of smooth functions on $V_x$ and its open subsets.
This agrees with the usual definition using charts and atlases, for all except the requirement that a manifold is a (separable) Hausdorff space. But indeed it seems that many things in differential topology can be proven without using the Hausdorff property. In a fleeting conversation in a brief encounter, a personal I shall refer to as O., informed me that even Stokes' theorem can be done this way. But I am unable to ask O. again about this, as he is not physically around.
If the above is true, then this is a really good point to mention when introducing schemes to a person who knows about differentiable manifolds.
So my main question: 

Is it true that the proof of Stokes' theorem for smooth manifolds can be proven without the Hausdorff condition on the manifold? If so, is it done in any well-known reference?

Aside question:
What are some crucial propositions/theorems in differential topology that use Hausdorff condition, except those involving imbedding in some $\mathbb{R}^m$ for high enough $m$, for which of course Hausdorffness is a necessary condition(together with separable property)?
Tom Church answers below that partitions of unity does not work, for instance on the example of the line with the double point. However I believe that one can still make sense of integration of differential forms even with such pathologies, because by introducing a measure for instance, we can ignore sets of measure $0$.

REPLY [2 votes]: If you are a little bit careful as to what you mean by a "non-Hausdorff manifold" then I believe that Stokes' theorem will go through.
The crucial issue is that the smooth functions cannot detect the fact that the manifold is not Hausdorff.  Thus when you try to integrate a smooth function over the manifold, the function that you are integrating does not know that the manifold was not Hausdorff.  Thus the fact that partitions of unity don't work doesn't matter because they do work "up to Hausdorffness".
Thus if "non-Hausdorff manifold" essentially means "becomes a manifold upon Hausdorffification" then you ought to be alright.  I can't think of a counterexample off the top of my head,  but that doesn't mean that one doesn't exist (i.e. an example of a "locally Euclidean space" that does not quotient down to a "Hausdorff locally Euclidean space" - ignore paracompactness for this, that's cheating).
A slightly tangential point is that if you have a "locally Euclidean space" that is not Hausdorff then you probably have the wrong topology on it.  Take, again, the double pointed line.  You probably think that the correct topology on this is the one with basis either $(a,b)$ or $(-a,a)\setminus \{0\}\cup \{\ast\}$.  Wrong!  By doing so, you are artificially imposing the condition that $0$ and $\ast$ can be distinguished without justifying that assertion.  If you try by differential topological means (i.e. not topological, but differential topological) to separate $0$ and $\ast$ you find that you cannot tell the difference between them.  So the correct topology has basis $(a,b)$ with $a$ and $b$ of the same sign, and $(-a,a) \cup \{\ast\}$.  That is, the induced topology on the subset $\{0,\ast\}$ is the indiscrete topology.
At this point I should come clean (especially in the light of Emerton's comment to the original question) and say that when thinking of things that are like-but-not-manifolds then I think of Froelicher spaces.  So a non-Hausdorff manifold really means a non-Hausdorff Froelicher space with nice local properties.  And for Froelicher spaces, the difference between Hausdorff and non-Hausdorff is extremely small.
But if what you are interested in is smooth functions, then that's the right view to take.  And if you are integrating, then you are using smooth functions.  Of course, in other contexts you may want to remember more structure than just what the smooth functions can detect, but that's a different story.