1

Here is the link to my earlier post here on Math Stack Exchange on Theorem 6.15 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Theorem 6.15 in Baby Rudin: If $a<s<b$, $f$ is bounded on $[a,b]$, $f$ is continuous at $s$, and $\alpha(x)=I(x-s)$, then . . .

Now here is Definition 6.23 in Baby Rudin:

Let $f_1, \ldots, f_k$ be real functions on $[a, b]$, and let $\mathbf{f} = \left( f_1, \ldots, f_k \right)$ be the corresponding mapping of $[a, b]$ into $\mathbb{R}^k$. If $\alpha$ increases monotonically on $[a, b]$, to say that $\mathbf{f} \in \mathscr{R}(\alpha)$ means that $f_j \in \mathscr{R}(\alpha)$ for $j = 1, \ldots, k$. If this is the case, we define $$ \int_a^b \mathbf{f} \ \mathrm{d} \alpha = \left( \int_a^b f_1 \ \mathrm{d} \alpha , \ldots, \int_a^b f_k \ \mathrm{d} \alpha \right). $$ In other words, $\int \mathbf{f} \ \mathrm{d} \alpha$ is the point in $\mathbb{R}^k$ whose $j$th coordinate is $\int f_j \ \mathrm{d} \alpha$.

Now my questions is, Is Theorem 6.15 in Baby Rudin not valid for vector-valued functions?

My feeling is that it certainly is. Here is my proposed generalization:

Theorem 6.15 for vector-valued functions:

If $a < s < b$, and if $\mathbf{f}$ is a mapping of $[a, b]$ into $\mathbb{R}^k$ such that $\mathbf{f}$ is bounded on $[a, b]$, $\mathbf{f}$ is continuous at $s$, and $\alpha(x) = I(x-s)$, then $$ \int_a^b \mathbf{f} \ \mathrm{d} \alpha = \mathbf{f}(s). $$

And, here is my proof:

Let $\mathbf{f} \colon= \left( f_1, \ldots, f_k \right)$, where each $f_j$ is a real function on $[a, b]$. As $\mathbf{f}$ is bounded, so is each $f_j$; furthermore, as $\mathbf{f}$ is continuous at $s$, so is each $f_j$. Therefore $f_j \in \mathscr{R}(\alpha)$ on $[a, b]$, for each $j = 1, \ldots, k$. So $\mathbf{f} \in \mathscr{R}(\alpha)$ on $[a, b]$ also. Then we obtain $$ \begin{align} \int_a^b \mathbf{f} \ \mathrm{d} \alpha &= \left( \int_a^b f_1 \ \mathrm{d} \alpha , \ldots, \int_a^b f_k \ \mathrm{d} \alpha \right) \qquad \mbox{ [ by Definition 6.23 in Baby Rudin ] } \\ &= \left( f_1(s), \ldots, f_k(s) \right) \qquad \mbox{ [ by Theorem 6.15 in Baby Rudin applied to each $f_j$ ] } \\ &= \mathbf{f}(s) \qquad \mbox{ [ using the definition of $\mathbf{f}$ ] }, \end{align} $$ as required.

Is there any problem with this theorem or its proof?

Saaqib Mahmood
  • 26,762

1 Answers1

1

Your theorem is well-written and the question you ask is well-posed. There is no problem with your proof either; all is well.

For future reference, if you intend to post questions of this nature whose answers may be as succinct as mine here, you may consider adding to your original post where your doubts lie. Since you would likely not ask if you were completely confident in what you have written down, it helps us to know what you think needs more attention.

Alex Ortiz
  • 24,844