Let $X$ and $Y$ be topological vector spaces such that $X^{\ast}$ and $Y^{\ast}$ respectively separate points on the aforementioned spaces. Let $\mu$ be a Borel measure on $Q$, a compact Hausdorff space and let $T\in\mathcal{B}(X,Y)$ and $f:Q\to X$ be continuous. I want to prove that $$T\int_{Q} f d\mu=\int_{Q}(Tf)d\mu.$$
I said: let $y:=\int_{Q}fd\mu$ whose uniqueness follows from the fact that $X^{\ast}$ separates points on $X$. Moreover, if $\Lambda\in Y^{\ast}$, then $\Lambda T\in X^{\ast}$. Then, from the definition of the Pettis integral, we get: $$(\Lambda T)(y)=(\Lambda T)\left(\int_{Q}f d\mu\right)=\int_{Q}(\Lambda T)(f)d\mu,$$ for every $\Lambda\in Y^{\ast}$. Thus $$T\int_{Q}fd\mu=\int_{Q}(Tf)d\mu.$$ However, I am unsure about the final step. Furthermore, we have not established uniqueness of $y$. Certainly, this holds if $X$ is a Fréchet space, or if we were working on a Banach algebra.
Edit: Please feel free to comment that the existence of $y$ cannot be proven under the current assumptions placed on $f$.