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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?

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

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  • $\begingroup$ 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 ...."? $\endgroup$
    – stefano
    Jun 16 '20 at 21:12
  • $\begingroup$ $I_A$ 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 $\endgroup$ Jun 16 '20 at 23:13
  • $\begingroup$ 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? $\endgroup$
    – stefano
    Jun 16 '20 at 23:46
  • $\begingroup$ $\mu$ is Lebesgue measure. Riemann integral and integral w.r.t $\mu$ coincide. @stefano $\endgroup$ Jun 16 '20 at 23:55

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