If $\mu_F\ll\beta_1$ and $\mu_F$ is finite then $F$ is absolutely continuous.
Note: here $\beta_1$ is the Lebesgue measure restricted to the Borel $\sigma$-algebra in the real line.
Im not sure if my solution is right, because I didnt used the fact that $\mu_F\ll\beta_1$, that is, that $\beta_1(A)=0\implies\mu_F(A)=0$ also.
Here $F(x):=\int_{-\infty}^x f(t)\, dt$ for some $f\in\mathcal L_0(\Bbb R,\lambda_1,\overline{\Bbb R}^+)$, and $\mu_F$ is the Lebesgue-Stieltjes measure generated by $F$ in $\Bbb R$, so $\mu_F([a,b))=F(b)-F(a)$. Hence
$$ \mu_F(\Bbb R)=\mu_F\left(\bigcup_{k\in\Bbb Z}[k,k+1)\right)=\sum_{k\in\Bbb Z}\int_k^{k+1}f(t)\, dt=\int_{-\infty}^\infty f(t)\, dt<\infty$$
so $f\in\mathcal L_1(\Bbb R,\lambda_1,\overline{\Bbb R}^+)$, and by a previous result I know that if $f\in\mathcal L_1(J,E)$, for some interval $J\subset\Bbb R$ and $E$ some Banach space, then the function defined by
$$g(x):=\int_{\inf J}^x f(t)\, dt$$
is absolutely continuous, thus I can conclude that $F$ is also absolutely continuous. However I didnt used the fact that $\mu_F\ll\beta_1$, so it seems that something is wrong, where is my mistake?