Show that $\mu$ is absolutely continuous w.r.t. $\mathcal{L}$ and find $\frac{d\mu} {d\mathcal{L}}$

Question

Let $\mu$ be the unique Borel measure on $\mathbb{R}$ satisfying $\mu((a,b])=\arctan b-\arctan a$. Show that for any $\mu$-measurable subset $E$ of $\mathbb{R}$, $\mathcal{L}(E)=0$ implies $\mu(E)=0$.And find $\frac{d\mu} {d\mathcal{L}}$

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Attempt: It suffices to find an integrable function $f$ such that for any $\mu$-measurable set $E$, $\mu(E)=\int_Efd\mathcal{L}$. For $E=(a,b]$ we get $f(x)=\frac{1}{1+x^2}$. But how can we guarantee that for arbitrary $\mu$-measurable set $E$, we have $\mu(E)=\int_Efd\mathcal{L}$ ?Thanks!

Answer 1

Just define a measure (check that this is a measure!) $\nu(E)=\int_E 1/(1+x^2)\, d\mathcal{L}$.

As you observed yourself, we have $\mu(E)=\nu(E)=\arctan(b)-\arctan(a)$ if $E=(a,b]$.

The exercise now tells you that $\mu$ is the unique measure with this property. Hence, $\mu=\nu$.

The uniqueness can also be shown (e.g.) as an application of Dynkin's $\pi$-$\lambda$ theorem. But you probably know some Theorem which tells you the uniqueness, otherwise the exercise would be phrased differently.