On the one hand, Wikipedia suggests that every distribution defines a Radon measure:
On the other hand, Terry Tao and LK suggest not:
Can someone please clarify this for me?
Do you mean this sentence:
Conversely, essentially by the Riesz representation theorem, every distribution which is non-negative on non-negative functions is of this form for some (positive) Radon measure.
The condition that the distribution be non-negative for non-negative functions is non-trivial. Not every distribution satisfies this, so not every distribution is a Radon measure.
The fundamental examples are the delta function at a point (which is a measure) and its derivatives (which are not measures).
I think the decisive point is continuity with respect to different topologies. Let $C$ be the space of continuous functions of compact support and $D$ the space of smooth functions of compact support. The inclusion $D\hookrightarrow C$ is a continuous map when you give both spaces the corresponding inductive limit topology. That means, that every continuous linear functional of $C$, i.e., each Radon-measure, defines a continuous linear functional on $D$, i.e., a distribution. But not every distribution extends to a continuous linear map on $C$. Examples are the derivatives of the Dirac distribution. The line in Wikipedia relates to an important property of linear functionals on $C$: if such a functional is positive, i.e., if it maps functions $f\ge 0$ to numbers $\ge 0$, then it is AUTOMATICALLY CONTINUOUS. This is a very important fact, though it is not hard to prove.
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
As to:
any way of proving [the derivative of the delta function] is not the difference of two positive distributions:
indeed, $\delta’$ is not even majorized by a positive distribution. Here’s a direct argument.
If we had $\delta’\le u$, for a positive distribution $u\in\mathcal D’(\mathbb R)$, then, for every couple of test functions, $f,g$ in $\mathcal D(\mathbb R)$ with $f\ge g\ge0 $ we would have $$u(f)\ge u(g)\ge g’(0),$$ which is absurd because, for a fixed $f$, $g’(0)$ is certainly not bounded among all non-negative test functions below $f$.
Measures are dual to continuous functions, whereas distributions are derivatives of them.
Let $T$ be a positive distribution. We will show $T$ defines a Radon measure.
For any compact subset $K\subset\Omega$, there exists a constant $C:=T(\eta_K)$, such that for all $\phi\in C_c^\infty(K)$, $$T(\phi)\leq C\lVert\phi\rVert_{C^0} <\infty,$$ where $\eta_K\in C_c^\infty(\Omega)$ is some positive cut-off on $K$. (Since $\lVert\phi\rVert_{C^0}\cdot\eta_K-\phi\geq0$ and $T$ is positive.) This means that $T$ has order $0$ by the definition of order of distributions.
For any $\phi \in C_c(\Omega)$, we can find a sequence $\phi_n \in C^{\infty}_c(\Omega)$ with compact support in a fixed compact neighborhood $K$ of the support of $\phi$ such that $$\lVert\phi-\phi_n\rVert_{C^0}\to 0,\quad n\to \infty.$$ Then we have $$\lvert T(\phi_n) - T(\phi_m)\rvert = \lvert T(\phi_n-\phi_m)\rvert\le C\lVert(\phi_n-\phi_m)\rVert_{C^0}\to 0.$$ So $\lim T(\phi_n)$ exists by the completeness of real numbers. Define $T(\phi):= \lim T(\phi_n)$. This is independent of the choice of $\{\phi_n\}$ since if we have another such sequence $\{\psi_n\}$, then \begin{align*} \lvert T(\phi_n) - T(\psi_n)\rvert &= \lvert T(\phi_n-\psi_n)rvert\le C\lVert(\phi_n-\psi_n)\rVert_{C^0}\\ &\le C\lVert(\phi_n-\phi)\rVert_{C^0}+C\lVert(\psi_n-\phi)\rVert_{C^0}\to 0, \end{align*} thus $T(\phi)$ is well-defined. Thus $T$ can be extended to $C_c(\Omega)$ and so, by Riesz's Theorem (cf. Folland's Real Analysis, Theorem 7.2), it is a Radon measure.
