= Solution
For $0<\delta<1$, define the compensated small-jump <Poisson integral>
$$
Y_t^\delta=\int_{(0,t]\times\{\delta<|y|\leq1\}}y\,(\mu-\nu)(dy,ds).
$$
This is an ordinary compensated finite-intensity sum. <Independent> <Poisson random variables> give mean zero and the <L2 construction of compensated Poisson integrals> gives, for $0<\eta<\delta$,
$$
\mathbb E|Y_t^\eta-Y_t^\delta|^2=t\int_{\{\eta<|y|\leq\delta\}}y^2K(dy)=2ct(\delta-\eta).
$$
Thus $Y_t^\delta$ is <Cauchy> in <L2 space> as $\delta\downarrow0$, defining the first integral. It is not the difference of two separately convergent integrals. On a compact time interval $[0,L]$, the <Doob L2 maximal inequality> bounds the squared supremum of the same difference by $8cL(\delta-\eta)$. Hence the truncated <martingales> are Cauchy in the expected squared supremum norm. A sufficiently rapidly decreasing sequence of cutoffs converges uniformly <almost surely>; its limit has <càdlàg> paths, so the construction also supplies a consistent process.
For large jumps, $K(\{|y|>1\})=2c<\infty$. There are only finitely many such atoms before any fixed time, <almost surely>, and each atom has a finite mark. The uncompensated large-jump integral is therefore a finite random sum, even though its absolute first moment is infinite. The two terms are well-defined for these different reasons.
There is also a literal signed-integral defect in the PDF: $\int yK(dy)$ on either symmetric region is not $+\infty$. Its positive and negative parts both have infinite mass, so the ordinary signed <Lebesgue integral> is undefined. The correct divergence statement is
$$
\int_{0<|y|\leq1}|y|K(dy)=\int_{|y|>1}|y|K(dy)=\infty.
$$
A symmetric principal value may be zero, but is not an ordinary signed integral. \b[Square-integrable compensation defines the small jumps; finite activity defines the large jumps.]
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