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Let $\mu$ be Lebesgue's measure on $[0,2]$ and $\nu$ - Lebesgue's measure on $[1, 3]$. Determine Lebesgue's decomposition $\nu_a + \nu_s$ of measure $\nu$ with respect to $\mu$.

I want to make sure that I understand this correctly. If I let $\nu_a$ be Lebesgue's measure on $[1, 2]$ and $\nu_s$ - Lebesgue's measure on $(2, 3]$, would it be the correct answer?

$\nu_a$ is absolutely continuous with respect to $\mu$, since it measures subsets of $[0,2]$ and they're both one-dimensional Lebesgue's measures. $\nu_a$ and $\nu_s$ are singular, since they're concentrated on disjoint sets. Same for $\nu_s$ and $\mu$.

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    Almost: you need $\nu_s$ and $\mu$ to be singular (which they are), not $\nu_s$ and $\nu_a$. – 2017-01-25
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    @UmbertoP. but if $\nu_a$ is absolutely continuous with respect to $\mu$, and $\nu_s$ is singular with $\mu$, doesn't it also mean that $\nu_a$ and $\nu_s$ are also singular? – 2017-01-25

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Yes. We let \begin{align*} \nu = \nu_0 + \nu_1 \end{align*} where $\nu_0$ is Lebesgue on $[1,2]$ and $\nu_1$ is Lebesgue on $(2,3]$.

If $\mu(E) = 0$, then $\nu_0(E) = 0$. So $\nu_0 \ll \mu$.

Since $\mu = 0$ on $(2,3]$ and $\nu_1 = 0$ on $[0,2]$, we also have $\nu_1 \perp \mu$.

It follows that $\nu = \nu_0 + \nu_1$ is the unique decomposition as given by Lebesgue's Decomposition Theorem.