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 $y=1+x^{2}$).