 Solution

Recall that $g_i\in L_{\mathrm{loc}}^1(\Omega )$ is the weak derivative $D_{i}u$ if $u\in L_{\mathrm{loc}}^1(\Omega )$ and $\int u{D}_i\varphi =-\int g_i\varphi$ for all $\varphi \in C_c^{\infty }(\Omega )$. A global weak derivative restricts to every open neighborhood simply by taking test functions supported there.

Conversely, suppose local weak derivatives exist near each point. local integrability of $u$ follows by a finite cover of each compact subset. On overlapping neighborhoods the two candidates for $D_{i}u$ have identical integrals against every smooth compactly supported test function. The fundamental lemma for locally integrable functions therefore makes them equal almost everywhere. Choose a countable local cover, possible in Euclidean space; the countably many exceptional overlap sets can all be discarded. The derivatives consequently patch to measurable $g_i\in L_{\mathrm{loc}}^1(\Omega )$.

For a test function $\varphi$, take a smooth partition of unity $({\chi }_j)$ subordinate to the local cover and locally finite on its support. Only finitely many terms occur there. Because ${\sum }_j{\chi }_j=1$ near the support,The derivatives of the partition sum to zero, explaining the first equality. Thus the patched functions are the global weak derivatives. This proves both directions of locality of weak differentiability.