But $e_{pTp}(-\infty, t)$ $\le$ $e_T(-\infty, t)$ does not imply that the range of $e_{pTp}(-\infty, t)$ is invariant under $T$.

Denis, Thanks for your answer. You guess is correct. I am sorry I should clarify my notations. If we write $T=\int_R \lambda d(e_{\lambda}) $, then $e_T(-\infty, t)$ means the projections corresponding to unique spectral decomposition of the operator $T$. Similarly, for $e_{pTp}(-\infty, t)$. And $e_{pTp}(-\infty, t)$ $\le$ $e_T(-\infty, t)$ means the range of the first projection is included in the range of the second projection.

Thanks for your answers. I am still not clear about that...