Computing an explicit Radon-Nikodym derivative

Question

Q/ let $\lambda$ be the Lebesgue measure and $\delta_0$ be the Dirac measure at 0. Show that $\lambda$ is abs cts wrt $\lambda+\delta_0$ (have done this part) and find the R-N derivative $\frac{d\lambda}{d(\delta_0+\lambda)}$

So I understand that I have to find a real valued function f st $\forall\; [a,b]\subset\mathbb{R}$ we have;

$\lambda([a,b])=b-a=\int_a^b f\;d(\delta_0+\lambda)$

However I'm really quite confused by the $d(\delta_0+\lambda)$ at the end of the integral seeing as I cannot turn it into a Riemann integral, for intervals $[a,b]$ which do not contain 0 its clear that $f=1$ since $\int_a^b \;d(\delta_0+\lambda)=(\delta_0+\lambda)([a,b])=b-a$ however when 0 is contained in the interval I can't really see what fits.

Thanks

Answer 1

$$f(x) := \begin{cases} 1 & x \neq 0 \\ 0 & x=0 \end{cases}$$ does the job. Indeed: For $a<b$, we have

$$\begin{align*} \int_a^b f(x) \, d(\delta_0+\lambda)(x) &= \int_a^b f(x) \, d\delta_0(x) + \int_a^b f(x) d\lambda(x) \\ &= f(0) + \int_a^b (1-1_{\{0\}}(x)) \, d\lambda(x) \\ &= 0 + (b-a). \end{align*}$$

Here, we have used that $$\int_a^b 1_{\{0\}}(x) \, d\lambda(x)=0$$ since $\lambda(\{0\})=0$.