This is a summary of what I've learned about this question based on the answers of the other commenters.

[*] Any positive distribution defines a positive Radon measure.

I had naively assumed a result for distributions like The Hahn Decomposition Theorem[1] for measures, i.e. I assumed that a distribution could be expressed as the difference of two positive distributions. If it could be, then applying Theorem [*] would yield the result that any distribution is a signed measure.

However, this is not the case. The derivative of the delta function, i.e. δ', satisfies δ'(f) = -f'(0). This is not a measure. I can't find any way of proving it's not the difference of two positive distributions, other than by contradiction using the above result.

[1] http://en.wikipedia.org/wiki/Hahn_decomposition_theorem

This is a summary of what I've learned about this question based on the answers of the other commenters.

[*] Any positive distribution defines a positive Radon measure.

I had naively assumed a result for distributions like The Hahn Decomposition Theorem[1] for measures, i.e. I assumed that a distribution could be expressed as the difference of two positive distributions. If it could be, then applying Theorem [*] would yield the result that any distribution is a signed measure.

However, this is not the case. The derivative of the delta function, i.e. δ', satisfies δ'(f) = -f'(0). This is not a measure. I can't find any way of proving it's not the difference of two positive distributions, other than by contradiction using the above result.

[1] https://en.wikipedia.org/wiki/Hahn_decomposition_theorem