The way that Bourbaki defined the measure is highly unusual:
Let $C_{0}(\mathbb{R}^{n})$ be the space of all continuous functions with compact supports.
A linear functional $I$ is said to be a measure if $I:C_{0}(\mathbb{R}^{n})\rightarrow{\mathbb{R}}$ is continuous with respect to the inductive topology for $C_{0}(\mathbb{R}^{n})$: \begin{align*} |I(f)|\leq C_{K}\sup_{x\in K}|f(x)| \end{align*} for each compact set $K$ and $\text{supp}(f)\subseteq K$, $C_{K}$ is a constant depending only on $K$.
Then the integral is simply defined by $I(f)$, but Bourbaki used to write \begin{align*} \left<f,\mu\right>=\int fd\mu \end{align*} instead of $I$.
Folland suggests the name for Bourbaki measure as pseudo-measure, yet, he admits that there is no standard name for Bourbaki one. And he proves that \begin{align*} I(f)=\int fd\mu_{K} \end{align*} where $\mu_{K}$ is really the usual Radon measure that we are familiar in the sense that if the measures that we learned are from, say, Royden, Rudin references. Here $\mu_{K}$ corresponds to each $K$, and $\mu_{K+1}|_{B(K)}=\mu_{K}$, where $B(K)$ is the Borel subsets of $K$.
Nothing really said more about the features of the pseudo-measure by Folland. But I have ever seen some author claims that \begin{align*} I(f)=\int fd\mu \end{align*} for some Radon measure $\mu$. Note that Folland does not show that there is a universal Radon measure to realize the pseudo-measure, instead, he shows that there is an "exhaustion" of Radon measures $\mu_{K}$, each depends on the compact set $K$, to have such a realization.
Therefore I am doubt the author's claim that there is such a universal one. I am looking for counterexample.
The best we can say is the following:
Bourbaki proves that such an $I$ can have positive and negative parts: \begin{align*} I(f)=I^{+}(f)-I^{-}(f). \end{align*} And by usual Riesz Representation Theorem one has for some nonnegative measures $\mu_{1},\mu_{2}$ that \begin{align*} I^{+}(f)&=\int fd\mu_{1},\\ I^{-}(f)&=\int fd\mu_{2}, \end{align*} one may think of the candidate is simply $\mu=\mu_{1}-\mu_{2}$. This may have some flaw. Recall that for a Radon measure $\mu$ one must have either for every $S$, \begin{align*} \mu(S)\in(-\infty,\infty] \end{align*} or for every $S$, \begin{align*} \mu(S)\in[-\infty,\infty) \end{align*} but not both. On the other hand, we cannot even be sure that $\mu_{1}-\mu_{2}$ is well-defined in the sense that $\infty-\infty$ could possibly occur. This is the reason that I think the universal Radon measure may not exist, or else, the genius Folland must have proved that, for what is the point to have less user friendly, that the exhaustion of measures?
I assume that the functions in $C_0(\mathbb R)$ have compact support (this is not clear from your post).
What about the functional (here: $n = 1$) $$ I(f) := \int_0^\infty f \, \mathrm d x - \int_{-\infty}^0 f \, \mathrm d x $$ ?
This should be continuous in Bourbaki's sense, but there does not exists a Radon measure $\mu$ with $$ I(f) = \int_{\mathbb R} f \, \mathrm d\mu, $$ since this measure would have $\mu((-\infty,0)) = -\infty$ and $\mu((0,\infty)) = \infty$, which is not allowed.