# An application of Radon-Nicodym theorem

Consider $$M$$ be the $$\sigma -$$algebra of Lebesgue measurable sets and $$\mu$$ the Lebesgue measure. Denote by $$P$$ the set of $$p-$$measurable sets, that is the sets $$A\in \mathcal{P}\left( \mathbb{R} \right)$$ such that the set $$\begin{equation*} \left\{ x\in \left[ 0,1\right] \mid 1+x^{2}\in A\right\} \end{equation*}$$ is $$M_{\left[ 0,1\right] \text{ }}-$$measurable. Prove

(a) the function $$\begin{equation*} \nu :A\rightarrow \left[ 0,\infty \right] , \quad \nu \left( A\right) =\mu \left( \left\{ x\in \left[ 0,1\right] \mid 1+x^{2}\in A\right\} \right) \end{equation*}$$ is a complete measure and

(b) show that $$\int_{\left[ 0,\infty \right) }\sqrt{x}d\nu \left( x\right) =\int_{\left[ 0,1% \right] }\sqrt{x^{2}+1}d\mu \left( x\right)$$ and compute $$\int_{\left[ 0,\infty \right) }\sqrt{x}d\nu \left( x\right) .$$ Let's say that for the part (a) is an easy verification of definitions: the image of the empty set is $$0$$, $$\nu$$ is countable additive and for all $$Z$$ in A with $$\nu(Z)=0$$, every subset of $$Z$$ lies in $$A$$. For the second part, I don't know how to apply the Radon-Nycodim theorem to prove the equality from b. I think that the last integral can be computed using the equality above and the fact that the RHS integral is equal with Riemann integral of the function $$\sqrt{x^2+1}$$ on $$[0,1]$$. Am I right?

By definition $$\int f d\nu =\int f(1+x^{2})d\mu$$ when $$f=I_A$$. This implies that the equation holds for simple functions and then for all non-negative Borel measurable functions, In particular it holds for $$f(x)=\sqrt x$$. Thus $$\int \sqrt x d\nu =\int \sqrt {1+x^{2}} d\mu$$. (Evaluation of $$\int \sqrt {1+x^{2}} d\mu$$ is standard: You can write it as $$x\sqrt {1+x^{2}}|_0^{1}-\frac 1 2\int \frac x {\sqrt {1+x^{2}}} d\mu (x)$$ and then make the substitution $$y=1+x^{2}$$).
• I am a bit confused. There is no need of the Radon-Nycodym theorem here or it is hidden somewhere in your argument? And by $I_A$ you denoted the characteristic function of $A$ or something else? Do you have a reference book for "by definition ...."? Jun 16 '20 at 21:12
• $I_A$ is the characteristic function of $A$. There is no need for RNT. $\int I_A g(x) d\mu(x)=\int_A g(x)d\mu(x)$. This is the definition of $\int_A g(x)d\mu(x)$. You are just over thinking and the enitire argument is very elementary not requiring any theorem. @stefano Jun 16 '20 at 23:13
• Probably is an easy computation, but I don't understand what was used in the following remark: "you can write it as $x\sqrt {1+x^{2}}|_0^{1}-\frac 1 2\int \frac x {\sqrt {1+x^{2}}} d\mu (x)$ ". Why is $-\frac 1 2\int \frac x {\sqrt {1+x^{2}}} d\mu (x)$ and not $-\int \frac 2x^2 {\sqrt {1+x^{2}}} d\mu (x)$ ? Or is not the same as integration by parts in the Riemann integral? Jun 16 '20 at 23:46
• $\mu$ is Lebesgue measure. Riemann integral and integral w.r.t $\mu$ coincide. @stefano Jun 16 '20 at 23:55