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Let $f,g:\mathbb{R}\to \mathbb{R}$ be given by \begin{align} f(x)=\frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}, \qquad x \in \mathbb{R},\\ g(x) = \begin{cases} 0 , \qquad x \in (-\infty,0)\\ e^{-x}, \qquad x \in [0,\infty). \end{cases} \end{align} Let $\mu, \nu: \mathcal{B}(\mathbb{R}) \to [0,1]$ be probability measures defined by \begin{align} \mu(B) = \int_B f(x)\ dx, \qquad \nu(B)= \int_B g(x)\ dx, \qquad B \in \mathcal{B}(\mathbb{R}). \end{align} By the Radon-Nikodym theorem there exists unique measures $\mu_a$ and $\mu_s$ on $\mathcal{B}(\mathbb{R})$, such that $\mu = \mu_a + \mu_s$, $\mu_a \ll \nu$ and $\mu_s \perp \nu$. Now, I want to provide $\mu_a$ and $\mu_s$ explicitly.

I know that $v(F)=0$ should imply that $\mu_a(F)=0$ for $F \in \mathcal{B}(\mathbb{R})$ and that $\mu_s(N)=0=\nu(N^c)$ for a $N \in \mathcal{B}(\mathbb{R})$. However I do not know how to make $\mu_a$ and $\mu_s$ explicitly.

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    Consider splitting $f=f_1+f_2$ where $f_1=f\chi_{\{x<0\}}$, $f_2=f\chi_{\{x\geq 0\}}$. Then note that $f_1$ is singular w.r.t. $\nu$, $f_2$ is abs. cont. w.r.t. $\nu$. – 2017-01-31
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    On $x \ge 0$, you want some $h$ such that $\mu_a A = \int_A h d \nu = \int_A h \cdot g dm$, so pick $m = {f \over g}$. – 2017-01-31

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Consider splitting $f=f_1+f_2$ where $f_1=f\chi_{\{x<0\}}$, $f_2=f\chi_{\{x\geq 0\}}$. Define $$ \mu_i(B):=\int_B f_i(x)dx,i=1,2. $$ Then note that $\mu_1$ is singular w.r.t. $\nu$, $\mu_2$ is abs. cont. w.r.t. $\nu$.