operator theory – On the dimension of the range of the resolution of the identity


I want to prove the following: Let $A,B$ be bounded self-adjoint operators in a complex-Hilbert space and $E_A(lambda)$, $E_B(lambda)$ its corresponding spectral resolutions, i.e.,
$$A=int_{(m_A,M_A)}t;dE_A(t)qquadtext{and}qquad B=int_{(m_B,M_B)}t;dE_B(t).$$
If $Ageq B$ (in the sense of positive operators) then $mathrm{dim}(mathrm{rg}E_A(lambda))leq mathrm{dim}(mathrm{rg}E_B(lambda))$ for all $lambdainmathbb{R}$.

I think for each $lambda$ we can define the linear operator $T_lambda:mathrm{rg}(E_B(lambda))tomathrm{rg}(E_A(lambda)),;xmapsto E_A(lambda)x$ and prove that this opeator is surjective but I do not know how to do it.

Can someone give me an idea? I will be grateful.