# Why is this measure absolutely continuous?

Let $$U \subset \mathbb R^3$$ be an open, bounded and connected set with a $$C^2-$$regular boundary $$\Gamma$$. If

• $$f \in \mathcal M_{+}(\Gamma)$$ i.e belongs to the space of non negative, Radon measures on $$\Gamma$$ with finite mass
• $$\phi \in C^2(\Gamma)$$ is an arbitary test function
• $$g \in L^{\infty}(\Gamma)$$ then for constants $$a,b \gt 0$$ the following:

$$\int_{\Gamma} af\phi = b\int_{\Gamma} \phi g+ \int_{\Gamma} \phi \int_{\Gamma} f$$

implies that $$af$$ is an absolutely continuous measure.

I'm having a really hard time understanding the above implication. I know the definition of absolutely continuity for measures:

Let $$\mu$$ and $$\nu$$ be two measure on a $$\sigma-$$algebra $$\mathcal B$$ of subsets of $$X$$.Recall that $$\nu$$ is absolutely continuous with respect to $$\mu$$ if $$\nu(A)=0$$ for any $$A \in \mathcal B$$ such that $$\mu(A)=0$$.

Since there is no reference with respect to which measure $$af$$ is absolute continuous, I suppose it's with respect to the Lebesgue measure. However I can not proceed with this proof.

Any help or hint is appreciated. Thanks a lot in advance!