Can someone check this exercise please? Thank you.
Suppose that $F$ is a Banach space and $A\in\mathcal L(E,F)$. Then show that for $f\in\mathcal S(I,E)$ that $$Af:=(x\mapsto A(f(x)))\in\mathcal S(I,F)$$ and that $\int_I Af=A\int_I f$.
About the notation: the Banach space $\mathcal S(I,F)$ is the space of jump-continuous functions from $I$ to $F$, that is, functions that have at most jump discontinuities.
The $\mathcal L(E,F)$ is the vector space of bounded linear operators from $E$ to $F$, and we knows that $\int_I\in\mathcal L(\mathcal S(I,E),E)$ or $\int_I\in\mathcal L(\mathcal S(I,F),F)$ where $\int$ represent the integral of Riemann.
And $I$ represent an interval in $\Bbb R$ with more than one point, and $E$ and $F$ are Banach spaces.
My work:
Observe that $Af:=(x\mapsto Af(x))$ is a composition of a continuous function and a jump-continuous function, hence $Af$ is continuous for each sub-interval where $f$ is continuous, then $Af$ is jump-continuous.
We define a set of partitions over $I:=[\alpha,\beta]$ by
$$I_n:=\left\{\alpha+k\cdot\frac{\beta-\alpha}n:k=0,\ldots,n\right\}$$
for some $n\in\Bbb N$. Then
$$\int_\alpha^\beta f=\lim_{n\to\infty}\sum_{k=0}^nf(\zeta_k)\frac{\beta-\alpha}n,\quad \zeta_k\in\left[\alpha+k\cdot\frac{\beta-\alpha}n,\alpha+(k+1)\cdot\frac{\beta-\alpha}n\right)$$
Then if we define $x_n:=\sum_{k=0}^nf(\zeta_k)\frac{\beta-\alpha}n$, and because $A$ is continuous and linear, we have that
$$\lim_n Ax_n=A(\lim_n x_n)\implies A\int_\alpha^\beta f=\int_\alpha^\beta Af$$