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 $\epsilon >0$, take the common smaller truncation scale on which each term is within $\epsilon \mathrm{/}2$ of its limit; the triangle inequality then puts their sum within $\epsilon$ of the sum in [L1].

On the real line, split at zero. As $R\to \mathrm{\infty }$, ${\int }_{-R}^{0}f$ and ${\int }_0^{R}f$ tend separately to their two tail values. The same $\epsilon \mathrm{/}2$ 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 $f(x)=1\mathrm{/}x$ 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 ${\int }_0^1{x}^{-p}dx$ 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 $1\mathrm{/}x$ is not one: take $f(x)=x$, which is continuous and therefore locally integrable. For every $R$ the substitution $x\mapsto -x$ gives ${\int }_{-R}^{0}xdx=-{\int }_0^{R}xdx$, so ${\int }_{-R}^{R}xdx=0$ and the whole-line principal value is $0$; but ${\int }_0^{R}xdx=R^2\mathrm{/}2$ is unbounded in $R$, so the tail ${\int }_0^{\mathrm{\infty }}xdx$ does not converge.