The question is: $<g, \Phi>=0, \forall \Phi \in C_{c}^{\infty}(X), \text{then } f=0.$ And the map $L_{loc}^{1} \rightarrow D'(X)$ where $D'(X)$ denotes the set of distributions on $X$, and $X \subseteq \mathbb{R}^{n}$ is an open set.
Attempt:
We can assume without loss of generality that $g \in L^{1}$ (since $L^{p} \subseteq L_{loc}^{1}, 1\leq p \leq \infty$). Now we know that $<g, \Psi>=0, \forall,\Psi \in C_{c}^{\infty}(X)$ by assumption.
Here is where I get stuck and don't really know how to proceed...
It is a famous lemma in analysis (which does not depend on the theory of distributions) that if $f\in L^1_{loc}(X)$, where $X$ is an open subset of $\mathbb{R}^n$, and $\int_X f\varphi=0$ for all $\varphi \in C_c^{\infty}(X)$, then $f=0$ almost everywhere. It is not a trivial result, so it is normal that you cannot solve it as an exercise. See e.g. Corollary 4.24 in Brezis's Functional analysis, Sobolev spaces and Partial differential equations.
The idea is that if you can choose $\varphi=\operatorname{sgn}f(x)\chi_K$ where $K\subset X$ is compact, then you get $\int_K |f|=0$ which implies $f=0$ a.e on $K$. Here $\operatorname{sgn}f(x)\chi_K$ is not a test function, but applying convolution with mollifiers you can approximate it with test functions which allows you to obtain the result.