Finding the spectral family of $A^2$ $A$ is a bounded self-adjoint operator on a Hilbert space. The spectrum of $A$ is continuous (meaning, no point or residual spectrum). $\{E_\lambda\}$ is the spectral family of $A$.   
How can I show that the family $F_\lambda$ defined by    
\begin{cases}
0,  & \lambda<0 \\
E_\sqrt {\lambda} - E_{-{\sqrt {\lambda}}}, & \lambda\ge0
\end{cases}
is the spectral family of $A^2$?
I tried to show that the sum $\sum\lambda_k (F_{\lambda_{k+1}}-F_{\lambda _{k}}) $ converges to $A^2$ in operator norm, but it got me nowhere. And why do I need the assumption about the spectrum of $A$?
Any help would be appreciated.
 A: 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  $$A^2 = \int_{\mathbb{R}} \lambda^2 dE_\lambda = \int_{\mathbb{R}_+} \lambda \ dE_{\sqrt \lambda} - \int_{\mathbb{R}_+} \lambda \ dE_{-\sqrt \lambda}  = \int_{\mathbb{R}_+} \lambda \ d(E_{\sqrt \lambda}-E_{-\sqrt \lambda})  = \int_{\mathbb{R}} \lambda \ dF_\lambda$$
A: 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.
