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Here is the link to my earlier post here on Math Stack Exchange on Theorem 6.16 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Theorem 6.16 in Baby Rudin: $\int_a^b f d \alpha = \sum_{n=1}^\infty c_n f\left(s_n\right)$

Now my question is, Is this theorem valid for vector-valued functions too?

My feeling is that the answer to this question is in the affirmative. Am I right?

Here is my Theorem 6.16 in Baby Rudin for vector-valued functions:

Suppose $\sum c_n$ is a convergent series of non-negative real numbers, $\left( s_n \right)$ is a sequence of distinct points in $[a, b]$, and $$ \alpha(x) = \sum_{n=1}^\infty c_n I\left(x-s_n \right) $$ for all $x \in [a, b]$. Let $\mathbf{f}$ be a continuous mapping of $[a, b]$ into $\mathbb{R}^k$. Then $\mathbf{f} \in \mathscr{R}(\alpha)$ on $[a, b]$, and $$ \int_a^b \mathbf{f} \ \mathrm{d} \alpha = \sum_{n=1}^\infty c_n \mathbf{f} \left( s_n \right). $$

And, here is my proof:

Let $\mathbf{f} \colon= \left( f_1, \ldots, f_k \right)$, where each $f_j$ is a real function defined on $[a, b]$. As $\mathbf{f}$ is continuous on $[a, b]$, so is each $f_j$; therefore each $f_j \in \mathscr{R}(\alpha)$ on $[a, b]$, which implies that $\mathbf{f} \in \mathscr{R}(\alpha)$ on $[a, b]$ also, and $$ \begin{align} \int_a^b \mathbf{f} \ \mathrm{d} \alpha &= \left( \int_a^b f_1 \ \mathrm{d} \alpha, \ldots, \int_a^b f_k \ \mathrm{d} \alpha \right) \qquad \mbox{ [ by Definition 6.23 in Baby Rudin ] } \\ &= \left( \sum_{n=1}^\infty c_n f_1 \left( s_n \right) , \ldots , \sum_{n=1}^\infty c_n f_k \left( s_n \right) \right) \\ & \qquad \qquad \mbox{ [ by Theorem 6.16 in Baby Rudin applied to each $f_j$ ] } \\ &= \end{align} $$

What next?

Can we go from here to our desired answer?

Or, is this result not valid for vector-valued functions?

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    $\begingroup$ You have it. Your last vector in your proof is precisely $\sum c_n\mathbf f(s_n)$. $\endgroup$ – Ted Shifrin Jul 25 '17 at 20:08

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