The following claim and its proof are tailored from a long proof in Evans's Partial Differential Equations (Chapter 8.4).
[Claim.] Let $U\subset{\bf R^3}$ be open, bounded and simply connected. Let $f\in L^2(U;{\bf R}^3)$. Suppose for every $V\Subset U$, $V$ smooth and simply connected, there exists $p\in L^2(V)$ such that $$ \int_V Du:Dv\ dx = \int_V p\text{ div }v+f\cdot v\ dx,\quad\forall v\in H_0^1(V;{\bf R}^3)\tag{1} $$ where $u$ is some function satisfying some assumptions (minimizer of an energy functional, which is not quite relevant here). Then we can conclude that there exists $p\in L_{loc}^2(U)$ such that $$ \int_U Du:Dv\ dx = \int_U p\text{ div }v+f\cdot v\ dx,\quad\forall v\in H^1(U;{\bf R}^3)\text{ with compact suppoert within } U.\tag{2} $$
[Proof.] Choose a sequence of sets $V_k\Subset U$ ($k=1,2,\cdots$) as in (1) with $V_1\subset V_2\subset\cdots$ and $U=\cup_{k=1}^\infty V_k$. We find $p_k\in L^2(V_k)$ so that $$ \int_{V_k} Du:Dv\ dx = \int_{V_k} p_k\text{ div }v+f\cdot v\ dx,\quad\forall v\in H_0^1(V_k;{\bf R}^3)\tag{3} $$
Adding constants as necessary to each $p_k$, we deduce from (3) that if $1\leq l\leq k$, then $p_l=p_k$ on $V_l$. We finally define $p=p_k$ on $V_k$ ($k=1,2,\cdots$). Q.E.D.
Question: Would anyone elaborate the highlighted part in the proof? Using integration by parts on the first term on the RHS of (3), I can see that (3) still holds if one adds some constant to $p_k$ (for a particular $k$). But I don't see how adding a constant would help to get $p_l=p_k$.
Note: If $A= (a_{ij})$ and $B = (b_{ij})$ are $m \times n$ matrices, then $$ A:B = \sum_{i=1}^m\sum_{j=1}^n a_{ij}b_{ij}. $$
I am just expanding Jose27's answer. Assume that $p_{k}$ and $q_{k}$ both satisfy (3). Then by subtracting you get $$ 0=\int_{V_{k}}(p_{k}-q_{k})\operatorname{div}v\,dx $$ for all $v\in H_{0}^{1}(V_{k};\mathbb{R}^{3})$. Take $v=(w,0,0)$. Then $$ \int_{V_{k}}(p_{k}-q_{k})\frac{\partial w}{\partial x_{1}}\,dx=0 $$ for all $w\in H_{0}^{1}(V_{k})$. This says that there exists the weak derivative of $p_{k}-q_{k}$ with respect to $x_{1}$ and $$ \frac{\partial(p_{k}-q_{k})}{\partial x_{1}}=0. $$ Similarly, by taking $v=(0,w,0)$ and $v=(0,0,w)$ you can show that $\frac{\partial(p_{k}-q_{k})}{\partial x_{2}}=0$ and $\frac{\partial (p_{k}-q_{k})}{\partial x_{3}}=0$. This shows that $p_{k}-q_{k}\in H^{1}% (V_{k})$ and has derivatives zero. Since $V_{k}$ is simply connected, (for example by mollifying) this implies that $p_{k}-q_{k}$ is a constant.
Now assume that $p_{k}$ and $p_{l}$ satisfy (3) in $V_{k}$ and in $V_{l}$, respectively. Assume that $k<l$. Since (3) holds in $V_{l}$ for all $v\in H_{0}^{1}(V_{l};\mathbb{R}^{3})$, if you take $v\in H_{0}^{1}(V_{k}% ;\mathbb{R}^{3})$ and extend it to be zero in $V_{l}\setminus V_{k}$ you have that $v\in H_{0}^{1}(V_{l};\mathbb{R}^{3})$. Hence, $p_{l}$ satisfy (3) in $V_{k}$. This shows that $p_{l}$ and $p_{k}$ satisfy (3) in $V_{k}$, but then by what we did above, $p_{k}-p_{l}$ is a constant.