Separate improper convergence implies convergence of the principal value

At an interior point $c$, the two truncated terms in [L3] tend separately to the two finite one-sided values. Given $\varepsilon>0$, take the common smaller truncation scale on which each term is within $\varepsilon/2$ of its limit; the triangle inequality then puts their sum within $\varepsilon$ of the sum in [L1].

On the real line, split at zero. As $R\to\infty$, $\int_{-R}^0f$ and $\int_0^Rf$ tend separately to their two tail values. The same $\varepsilon/2$ estimate shows that their sum tends to the mixed value. Split-point invariance [L2] removes any dependence on zero.

The converses fail, and a witness is available on this page rather than assumed. Take $f(x)=1/x$ on $[-1,1]$ with the interior singularity at $0$. For every $\delta\in(0,1)$ the substitution $x\mapsto-x$ gives $\int_{-1}^{-\delta}x^{-1}\,dx=-\int_{\delta}^{1}x^{-1}\,dx$, so the symmetric truncations cancel exactly and the principal value exists and is $0$. But $\int_0^1x^{-p}\,dx$ converges exactly when $p<1$ by [L4], so at $p=1$ the right-hand one-sided integral diverges, and by the same reflection so does the left-hand one. Hence the principal value can exist while neither one-sided improper integral converges, and the converse of the first claim fails. For the whole line a separate witness is needed, because [L5] admits only a function locally Riemann integrable on all of $\mathbb R$ and $1/x$ is not one: take $f(x)=x$, which is continuous and therefore locally integrable. For every $R$ the substitution $x\mapsto-x$ gives $\int_{-R}^0x\,dx=-\int_0^Rx\,dx$, so $\int_{-R}^Rx\,dx=0$ and the whole-line principal value is $0$; but $\int_0^Rx\,dx=R^2/2$ is unbounded in $R$, so the tail $\int_0^\infty x\,dx$ does not converge.