Let $A$ be a self-adjoint operator on a Hilbert space $\mathscr{H}$ with dense domain $D(A)$. From the continuous functional calculus, for every continuous function $f \in C(\sigma(A))$ we can associate a bounded operator $f(A)$ on $\mathscr{H}$. Moreover, if $\psi \in \mathscr{H}$, the functional:
$$\omega(f) := \langle \psi, f(A)\psi\rangle$$
is a positive linear functional, so by Riesz Representation Theorem there is an isomorphism between $\mathscr{H}$ and the set of regular Borel measures on $\sigma(A)$, denoted by $M(\sigma(A))$, such that each $\psi \in \mathscr{H}$ is uniquely associated to a measure $\mu_{\psi}$, with
$$\langle \psi, f(A)\psi\rangle = \int_{\sigma(A)}f(\lambda)d\mu_{\psi}(\lambda).$$
I now that every Borel measure $\mu$ has a decomposition $\mu = \mu_{pp}+\mu_{ac}+\mu_{sc}$, which is a consequence of the Lebesgue Decomposition Theorem. Here $\mu_{pp}$ is a pure point measure, $\mu_{ac}$ is absolutely continuous with respect to the usual Borel measure on $\mathbb{R}$ and $\mu_{sc}$ is singular with respect to the latter.
Hence, there is an identification $\psi \to \mu_{\psi} = \mu_{\psi,pp}+\mu_{\psi,ac}+\mu_{\psi,sc}$. This shows that $\mathscr{H}$ can be written as $\mathscr{H}_{pp}+\mathscr{H}_{ac}+\mathscr{H}_{sc}$, with $\mathscr{H}_{pp} := \{\psi \in \mathscr{H}: \mbox{$\mu_{\psi}$ is pure point}\}$, $\mathscr{H}_{ac} := \{\psi \in \mathscr{H}: \mbox{$\mu_{\psi}$ is absolutely continuous with respect to the usual Borel measure}\}$ and $\mathscr{H}_{sc} := \{\psi \in \mathscr{H}: \mbox{$\mu_{\psi}$ is singular with respect to the the usual Borel measure}\}$.
My question is: why this sum is actually a direct sum? That is, how to prove further that:
$$\mathscr{H} = \mathscr{H}_{pp}\oplus \mathscr{H}_{ac}\oplus \mathscr{H}_{sc}?$$
Best Answer
This is easier to see from the Borel functional calculus. In particular, for every Borel set $B$ we have a bounded, nonnegative definite operator $1_B(A)$, with the properties that:
$1_{\mathbb{R}}(A) = I$, $1_\emptyset(A)=0$
$1_B(A) 1_C(A) = 1_{B \cap C}(A)$
If $B_1, B_2, \dots$ are pairwise disjoint, with $B = \bigcup_n B_n$, then $1_{B}(A) = \sum_n 1_{B_n}(A)$, with the sum converging strongly.
Now the Lebesgue decomposition can be restated as saying that for any measure $\mu = \mu_\psi$ we can partition $\mathbb{R}$ into three disjoint Borel sets $B, C, D$, such that the restrictions of $\mu$ to these three sets are respectively pure point, absolutely continuous, and singular continuous. These restricted measures are the spectral measures associated to the vectors $\psi_{pp} = 1_B(A) \psi$, $\psi_{ac} = 1_C(A)\psi$, and $ \psi_{sc} = 1_D(A)\psi$.
Then we have $$\psi_{pp} + \psi_{ac} + \psi_{sc} = (1_B(A) + 1_C(A) + 1_D(A)) \psi = 1_\mathbb{R}(A) \psi = \psi$$ so that $\psi$ decomposes as a sum of three vectors in the appropriate spaces. Moreover, since the operator $1_B(A)$ is self adjoint, we have $$\langle \psi_{pp} , \psi_{ac} \rangle = \langle 1_B(A) \psi, 1_C(A) \psi \rangle = \langle \psi, 1_B(A) 1_C(A) \psi \rangle = \langle \psi, 1_{B \cap C}(A) \psi \rangle = 0$$ because $B$ and $C$ are disjoint. The other orthogonality relations follow similarly.
To see the spaces are closed:
Note that a measure $\mu$ is pure point if and only if it is supported on a countable set $B$. When $\mu = \mu_{\psi}$ this says that $\psi = 1_B(A) \psi$. Now if $\psi_n$ is a sequence of such vectors converging to some $\psi$, with corresponding countable sets $B_n$, let $B = \bigcup_n B_n$ which is also countable. Verify that $1_B(A) \psi_n = \psi_n$ for all $n$, and now $1_B(A)$ is a bounded operator so $1_B(A) \psi = \psi$.
A measure $\mu$ is absolutely continuous iff $\mu(A) = 0$ whenever $m(A) = 0$, where $m$ is the Lebesgue measure. When $\mu = \mu_\psi$ this simply says $1_A(B) \psi = 0$. So if $\psi_n \to \psi$ is a sequence of such, let $A$ be an arbitrary Lebesgue null set, note $1_A(B) \psi_n = 0$ for all $n$, and since $1_A(B)$ is bounded we have $1_A(B) \psi = 0$ as well.
A measure $\mu$ is singular continuous iff $\mu(B) = 0$ for every countable $B$ and it is supported on a set of Lebesgue measure zero. The proof for this part follows similarly to the previous two.