Because by definition the eigenvalues of an operator $A$ are part of the point spectrum $\sigma_p(A)$. For self-adjoint operators, the continuous spectrum is the complement $\sigma_c(A)=\sigma(A) \setminus \sigma_p(A)$.

Therefore if, in any sense $Af=af$ for some $a \in \sigma_c(A)$, $f$ cannot be an eigenvector. For this reason it cannot belong to the Hilbert space.

As a matter of fact, the identity $Af=af$ where $a \in \sigma_c(A)$ holds in a different sense than the standard one, a distributional sense if the Hilbert space is $L^2(\mathbb R, d^nx)$.

Because by definition the eigenvalues of an operator $A$ are part of the point spectrum $\sigma_p(A)$. For self-adjoint operators, the continuous spectrum is the complement $\sigma_c(A)=\sigma(A) \setminus \sigma_p(A)$.

Therefore if, in any sense $Af=af$ for some $a \in \sigma_c(A)$, $f$ cannot be an eigenvector. For this reason it cannot belong to the Hilbert space.

As a matter of fact, the identity $Af=af$ where $a \in \sigma_c(A)$ holds in a different sense than the standard one, a distributional sense if the Hilbert space is $L^2(\mathbb R, d^nx)$.

It is worth noticing that the point spectrum, in spite of its name, may be a continuous set, all $\mathbb R$ for instance. In this case, however, the Hilbert space would not be separable. A famous theorem by Stone and von Neumann proves that the Hilbert space of a particle (irreducible representation of Weyl group) must be separable necessarily. For this reason Hilbert spaces of non-relativistic elementary systems in QM are separable and point spectra are at most countable.