Integral of $L1$ function against compactly supported smooth function: I have the following intuition, which I am not sure about if it is true or not, since I cannot come up with a proof or with a counterexample.

Given an $f \in L^1(\mathbb{R})$, if $$\int_\mathbb{R}f \cdot \phi = 0, \forall\phi \in C_c^{\infty}(\mathbb{R})$$
  then we have $f = 0$ (in the sense of equivalence classes).

Does anyone know a proof of this? And if the integral evaluates to for example $\phi(1)$ does it still hold that $f=0$?
Thanks a lot in advance!
 A: It is easy to prove with this condition that for all $[a,b]$ (it is easy to construct a sequence $\phi_n\rightarrow 1_{[a,b]} $, with $\phi_n$ smooth with compact support, by convolution and then use dominating theorem (since $f$ is in $L^1$))
$$ \int f1_{[a,b]} =0.$$
And then since the borelian $\sigma$-algebra is generated by $\{[a,b],\ a,b \in \mathbb{R} \}$ you will have
$$ \int f 1_A =\int_A f =0,\quad \forall A \in \mathcal{B}(\mathbb{R})$$
which implies easily (take $A=\{f>0\}$ and then $A'=\{f<0\}$, you will have $\lambda(A)=\lambda(A')=0$, where $\lambda$ is the Lebesgue measure) the result.
A: Let $r>0.$ Then there exists a sequence $\phi_n$ of smooth functions, supported in $[-r,r],$ with $0 \le \phi_n\le 1$ for all $n,$ such that $\phi_n(x) \to \text { sgn }f(x)$ for a.e. $x\in [-r,r].$ It follows from the DCT that
$$0=\int_{-r}^r \phi_n f \to \int_{-r}^r(\text {sgn }f)\cdot f =\int_{-r}^r |f|.$$
Thus $ \int_{-r}^r |f|=0.$ Since $r$ is arbitrary, $\int_{\mathbb R} |f|= 0.$ Hence $f=0$ a.e. 
A: Choose a sequence of uniformly bounded smooth functions $\phi_n$ with support in $[-1,2]$ that converge pointwise to $1_{[0,1]}$. Then
$0=\int f \phi_n \to \int f 1_{[0,1]} = 0$.
Hence $\int_a^t f= 0$ for any $a,t$, hence Lebesgue's differentiation theorem shows that we have $f=0$ ae.
