Derivatives of piecewise smooth functions include jump deltas

By F1, $u_f$ and $u_g$ are distributions. Fix a test $\phi$ and a closed interval $[b,c]$ containing its support in its interior, with endpoints outside $A$. Local finiteness and compactness imply $A\cap [b,c]$ is finite: take a finite subcover of neighborhoods each meeting finitely many points. List these break points in increasing order. On each intervening open interval $(s,t)$ apply F2 to $f,\phi$ on $[s+\epsilon ,t-\epsilon ]$ for sufficiently small positive $\epsilon$. This gives $-{\int }_{s+\epsilon }^{t-\epsilon }f{\phi }^{\mathrm{\prime }}={\int }_{s+\epsilon }^{t-\epsilon }g\phi +f(s+\epsilon )\phi (s+\epsilon )-f(t-\epsilon )\phi (t-\epsilon )\mathrm{.}$.

Let $\epsilon$ decrease to zero, for example through the reciprocal integers once the truncated interval is nonempty. F3 applies to the two integrals, with majorants $\mathrm{\mid }f{\phi }^{\mathrm{\prime }}\mathrm{\mid }$ and $\mathrm{\mid }g\phi \mathrm{\mid }$, integrable on $[b,c]$ by the local integrability assumptions. The boundary values tend to $f(s+)\phi (s)$ and $f(t-)\phi (t)$ by the finite one-sided limits; at $b,c$ the test vanishes. Sum over the finitely many intervals. At each break point $a$, the left interval contributes $-f(a-)\phi (a)$ and the right contributes $f(a+)\phi (a)$. The result is exactly the asserted formula when paired with $\phi$.

On any fixed compact test support $K$, the delta sum is finite and bounded in modulus by $({\sum }_{a\in A\cap K}\mathrm{\mid }f(a+)-f(a-)\mathrm{\mid })p_0(\phi )$. It therefore defines a distribution by F4, so the test equality proves the distribution identity. If there are no break points in $K$ the sum is zero, and a zero jump contributes no delta. If $f$ is $C^1$ up to both sides, $f$ and $g$ are bounded on each of the finitely many compact pieces meeting a compact interval, hence locally integrable, verifying the stated sufficient case. Merely being $C^1$ on the open pieces does not supply local integrability of $g$ at the breaks. $\mathrm{\square }$.