Theorem 1.5.8 in "Positive Definite Matrices" written by R. Bhatia is that the function $f(A)=A^r$ is convex on $L_+(\mathbb{H})$ for $1\leq r\leq 2$.
I'm struggling to understand some part of the proof : From the integral formula $$ t^r = \int_0^\infty \frac{t^2}{\lambda+t} \;d\mu(\lambda) $$ the author conclude that for positive oprators $A$ $$ A^r = \int_0^\infty A^2(\lambda +A)^{-1} \;d\mu(\lambda). $$
However, I'm not familiar with forms like this; integrad contains operators.
My attempt to understand this is considering the spectral measure $E_A$ of $A$. From $A^r = \int t^r\; dE_A$, we have $$ A^r = \int t^r\;dE_A(t) = \int \int_0^\infty \frac{t^2}{\lambda+t}\; d\mu(\lambda)dE_A(t). $$ But I don't know the Fubini theorem holds for this case, i.e., $$ \int_0^\infty \int \frac{t^2}{\lambda+t}\; dE_A(t)d\mu(\lambda) = \int \int_0^\infty \frac{t^2}{\lambda+t}\; d\mu(\lambda)dE_A(t). $$ How can I understand this?