show that $f = c$ a.e. Let $f \in L_{loc}^1 (\mathbb{R}) $ such that $ \int_{\mathbb{R}} f(x) \phi ' (x) dx = 0$, for every $\phi \in C_c ^{\infty}( \mathbb{R}) $. Show that there exist a $c \in \mathbb{R}$ such that  $f(x) = c$ a.e. 
What I did is: 
Assume that $f \in C^ {\infty} $. Therefore, for every $a,b \in \mathbb{R}$, $ 0=\int _a ^b ff' = 1/2 (f(b)^2-f(a)^2)$. Considering that $f$ is continuous, we can conclude that $f(a)  = f(b)$.
but i couldn't prove the general case. Any clue?
 A: Suppose that
$$
\int_{\mathbb{R}}f(x)\phi'(x)\,\mathrm{d}x=0\tag{1}
$$
for all $\phi\in C_c^\infty$. Then consider the functions
$$
\psi_{a,\lambda}(x)=\left\{\begin{array}{}
0&\text{for $x\lt a$}\\
\frac{x-a}\lambda&\text{for $x\in[a,a+\lambda]$}\\
1&\text{for $x\gt a+\lambda$}\tag{2}
\end{array}\right.
$$
By approximating $\psi_{a,\lambda}-\psi_{b,\mu}$ as close as we want (in $L^\infty$) by $\phi\in C_c^\infty$, we get that
$$
\int_{\mathbb{R}}f(x)\phi'(x)\,\mathrm{d}x\tag{3}
$$
approximates
$$
\frac1\lambda\int_a^{a+\lambda}f(x)\,\mathrm{d}x-\frac1\mu\int_b^{b+\mu}f(x)\,\mathrm{d}x\tag{4}
$$
Since $(3)$ is $0$, we get that $(4)$ must be as well. This means that
$$
\frac1\lambda\int_a^{a+\lambda}f(x)\,\mathrm{d}x=\frac1\mu\int_b^{b+\mu}f(x)\,\mathrm{d}x\tag{5}
$$
Since almost every point is a Lebesgue point of a locally integrable function, $(5)$ means that
$$
f(a)=\int_0^1f(x)\,\mathrm{d}x\tag{6}
$$
for almost all $a\in\mathbb{R}$.
A: clearly $f \in L_{loc}^{1}(\mathbb{R})$ is weak differentiable with weak derivation $Df$since $$-\int_{\mathbb{R}}f\phi' dx =0=\int_{\mathbb{R}}(Df)\phi dx, $$ $ ϕ∈C_{0}^\infty(\mathbb{R}). $ Applying the fundamental theorem of calculus of variation to the right hand side  this yields to $Df=0 $ a.e. and therefore $$f \equiv c \in \mathbb{R}$$ a.e. 
