It's well known that $PV(\frac{1}{x})$ defined as $PV(\frac{1}{x})(u)=\int_{0}^{+\infty} \frac{u(x)-u(-x)}{x} \, dx$ is a distribution over $\mathbb{R}$. I think that this distribution is not of the form $u \mapsto \int_{\mathbb{R}} u \, d\mu$ for some Borel measure $\mu$ on real numbers (and so this is not even associated to some $f \in L^1_{loc}({\mathbb{R}}$).
I found this topic Principal value of $1/x$ does not arise from either a locally integrable function or a Radon measure but I couldn't understand the solution (I got why the inequality should hold but how can we find that test function and how we get to the contradiction?).
The Riesz-Markov-Kakutani theorem says that Borel measures on $[a,b]$ are continuous linear functionals on continuous functions with sup norm. (On the whole real line, we'd need a "strict colimit topology", but the question here is really about what happens near $0$, so we can ignore that complication.)
For continuity, that'd mean that there is a constant $C$ such that $\Big|PV\int_{-1}^1 {f(x)\over x}\,dx\Big|\le C\cdot \sup|f|$ for all continuous $f$ on $[-1,1]$. It is an exercise to see that there is no such constant.