Finding the spectral family of $A^2$

By the spectral theorem $A^2 = \int_{\mathbb{R}} \lambda^2 dE_\lambda$ you want to show that $A^2 = \int_{\mathbb{R}} \lambda \ dF_\lambda$.

Now, $\int_{\mathbb{R}} \lambda^2 \ dE_{\lambda} = \int_{\mathbb{R}_-} \lambda ^2\ dE_{\lambda} +\int_{\mathbb{R}_+} \lambda^2 \ dE_{\lambda} $ (notice that $0\ \in \mathbb{R}_- \cap \mathbb{R}_+ $ is $dE$- negligible otherwise we would have point spectrum).

Then by a change of variables ($\phi=x\mapsto x^2$),

$\int_{\mathbb{R}_+} \lambda^2 \ dE_{\lambda} = \int_{\mathbb{R}_+} \lambda \ dE_{\sqrt \lambda} $ and $\int_{\mathbb{R}_-} \lambda^2 \ dE_{\lambda} = - \int_{\mathbb{R}_+} \lambda \ dE_{-\sqrt \lambda} $ where the (outermost) minus sign is due to the fact that $\phi|_{(-\infty,0]}^{[0,+\infty)}$ is NOT an orientation preserving diffeomorphism.

It follows that $$A^2 = \int_{\mathbb{R}} \lambda^2 dE_\lambda = \int_{\mathbb{R}_+} \lambda \ dE_{\sqrt \lambda} - \int_{\mathbb{R}_+} \lambda \ dE_{-\sqrt \lambda} = \int_{\mathbb{R}_+} \lambda \ d(E_{\sqrt \lambda}-E_{-\sqrt \lambda}) = \int_{\mathbb{R}} \lambda \ dF_\lambda$$

If $\mu$ is a Borel measure on $\mathbb{R}$ with no atoms, then $m(\lambda)=\mu(-\infty,\lambda]$ is continuous, and $$ \int_{-\infty}^{\infty}\lambda^2 dm(\lambda)=\int_{0}^{\infty}\lambda d(m(\sqrt{\lambda})-m(-\sqrt{\lambda})). $$ However, there are problems in the case that $m$ has atoms, if you want a normalization such that $m(\lambda)$ and $m(\sqrt{\lambda})-m(-\sqrt{\lambda})$ are both continuous from the right (or left.) So, to avoid renormalization issues, the problem was stated in such a way that the spectral measures have no atoms.