$\textbf{Exercise 4.14}$ (Fundamental lemma of the Calculus of Variations) Let $A$ an open set and $\nu$ be a Radon measure on $\mathbb{R}^n$ with values in $\mathbb{R}^m$. If
$$\int_{\mathbb{R}^n} \varphi d\nu = 0, \forall \varphi \in C_c^{\infty}(A;\mathbb{R}^m),$$
then $|\nu|(A) = 0$. In particular, if $u \in L_{\text{loc}}^1(\mathbb{R}^n,\mathbb{R}^m)$ and
$$\int_{\mathbb{R}^n} (\varphi(x) \cdot u(x)) dx = 0, \forall \varphi \in C_c^{\infty}(A;\mathbb{R}^m),$$
then $u = 0$ a.e. on $A$.
I would like to know if what I did is right:
WLOG, we supposed that $|\varphi| = 1$ for all $\varphi \in C_c^{\infty}(A;\mathbb{R}^m)$.
Observe that if the functional $\nu$ defined by $\langle \nu,\varphi \rangle = \int_{\mathbb{R}^n} \varphi d\nu$ (see remark $4.11$ here to understand that I'm denoting the fucntional by $\nu$) is bounded, then it follows from Riesz's theorem, which can be read in the previous link, that
$$|\nu|(A) = \sup \{ \langle \nu, \varphi \rangle ; \varphi \in C_c^0(A,\mathbb{R}^m), |\varphi| \leq 1 \} = 0$$
by hypothesis. Thus, it sufficient prove that $\nu$ is bounded, but this is equivalent to show that $\nu$ is continuous, then we will show that $\nu$ is continuous with respect to the convergence $C_c^0(\mathbb{R}^n,\mathbb{R}^m)$, which is described below:
By the exposition above, it sufficient show that given a sequence in $\varphi_h \in C_c^{\infty}(A;\mathbb{R}^m)$, exists $\varphi \in C_c^{\infty}(A;\mathbb{R}^m)$ such that $\varphi_h \rightarrow \varphi$. WLOG, we can assume $A$ is convex, restricting $\varphi_h$ and $\varphi$ to an open ball $B$ contained in $A$ if necessary, then $\varphi_h$ and all higher derivatives are equicontinuous in $A$ by Mean Value Inequality and also are bounded because $\varphi_h$ and its all higher derivatives are continuous with support compact. The existence of $\varphi \in C_c^{\infty}(A,\mathbb{R}^m)$ such that $\varphi_h \rightarrow \varphi$ follows now from the Arzela-Ascoli's theorem. $\square$
I was thinking better and I don't sure if $\varphi_h$ and all higher derivatives are bounded, because I know that exist $M_h^k > 0$ such that $|\varphi_h^{(k)}| \leq M_h^k$ for each $h$ by compactness and I want $M^k > 0$ such that $|\varphi_h^{(k)}| \leq M^k$ for each $h$. I would like to use the Uniform Boundedness Principle, but I think it doesn't work because $\varphi_h^{(k)}$ is a $k$-linear map and not a linear map unless that exists a generalization of the Uniform Boundedness Principle which I don't know if it exists.