A question regarding $L^1$and $C^\infty_C(\Omega)$ spaces

$f,g \in L^1 (\Omega)$, $\Omega\subset\mathbb{R}^n$ is a Lipschitz-domain.

Prove that $$(\forall\phi\in C^{\infty}_{C}(\Omega))\Big(\int_{\Omega}^{}f*\phi=\int_{\Omega}^{}g*\phi\Big)\Rightarrow f=g$$

where $\phi\in C^\infty_C(\Omega) \Rightarrow f\in C^\infty(\Omega)$ and $supp f \subset\Omega$ is compact.

I'm working on a project, and I'm stuck on this proof here,so if anyone could help me I would be most grateful

-
You should write out in words what exactly is the assumption and what you are trying to prove. Writing all in symbols only makes the formulas unreadable, especially when you're mixing up formal and informal notation. That, and you should say why you think it's true and what you have tried to do to show it. –  tomasz Oct 19 '12 at 19:31
@TTT, i wass thinking on this question and i think you have to change $\Omega$ by $\mathbb{R}^{n}$, because $f(x-y)$ is not well defined for all $x\in\Omega$. –  Tomás Oct 20 '12 at 11:04

If $\Omega$ is not all $\mathbb{R}^{n}$, the function $$\int_{\Omega}f(x-y)\phi(y)dy$$ is not well defined for all $x$, so im gonna assume that $\Omega=\mathbb{R}^{n}$.

Let $F(x,y)=f(x-y)\phi(y)$. Note that \begin{eqnarray} \int_{\mathbb{R}^{n}}|F(x,y)|dx &=& \int_{\mathbb{R}^{n}}|f(x-y)|\phi(y)|dx \nonumber \\ &=& |\phi(y)|\|f\|_{L^{1}(\mathbb{R}^{n})} \end{eqnarray}

Hence, $$\int_{\mathbb{R}^{n}} \int_{\mathbb{R}^{n}}|F(x,y)|dxdy=\|\phi\|_{L^{1}(\mathbb{R}^{n})}\|f\|_{L^{1}(\mathbb{R}^{n})}$$

By Tonelli's theorem (see Brezis - Functional Analyis, Sobolev Spaces and PDE, page 91) $F\in L^{1}(\mathbb{R}^{n}\times\mathbb{R}^{n})$.

Then by Fubini's theorem (see Brezis - Functional Analyis, Sobolev Spaces and PDE, page 91), we have for all $\phi\in C^{\infty}_{C}(\mathbb{R}^{n})$

\begin{eqnarray} \int_{\mathbb{R}^{n}} \int_{\mathbb{R}^{n}}f(x-y)\phi(y)dydx &=& \int_{\mathbb{R}^{n}} \int_{\mathbb{R}^{n}}f(x-y)\phi(y)dxdy \nonumber \\ &=& \|\phi\|_{L^{1}(\mathbb{R}^{n})}\|f\|_{L^{1}(\mathbb{R}^{n})}\\ &=& 0 \end{eqnarray}

Therefore, $f=0$.

-