Whether the spectral projection $E^{T}(0,\frac1n)\downarrow 0$ as $n\rightarrow \infty$ in the strong operator topology?

Question

Let $T$ be a self-joint operator on a Hilbert $H$. Whether the spectral projection $E^{T}(0,\frac1n)\downarrow 0$ as $n\rightarrow \infty$ in the strong operator topology?

My proof is:

Recall that spectral measure is countably additive. Since $E^T(0,1] =\vee_{n=1}^\infty E^T(\frac{1}{1+n},\frac1n]$, we have $E^T(\frac{1}{1+m},1]= \vee_{n=1}^m E^T(\frac{1}{1+n},\frac1n] \uparrow_m E^T(0,1]$. Hence, $E^T(0,\frac{1}{1+m}]\downarrow 1$.

Is it correct?