# Prob. 3 (d), Chap. 6, in Baby Rudin: If $\beta_j(x)=0$ if $x<0$ and $\beta_j(x)=1$ for $x>0$ and if $f$ is . . .

Here is Prob. 3, Chap. 6, in the book Principles of Mathematical Analysis by Walter Rudin,3rd edition:

Define three functions $\beta_1$, $\beta_2$, $\beta_3$ as follows: $\beta_j(x) = 0$ if $x < 0$, $\beta_j(x)=1$ if $x>0$ for $j=1, 2, 3$; and $\beta_1(0)=0$, $\beta_2(0) = 1$, $\beta_3(0) = \frac{1}{2}$. Let $f$ be a bounded function on $[ -1, 1]$.

(a) Prove that $f \in \mathscr{R}\left(\beta_1\right)$ if and only if $f(0+) = f(0)$ and that then $$\int f \ \mathrm{d} \beta_1 = f(0).$$

(b) State and prove a similar result for $\beta_2$.

(c) Prove that $f \in \mathscr{R}\left(\beta_3\right)$ if and only if $f$ is continuous at $0$.

(d) If $f$ is continuous at $0$ prove that $$\int f \ \mathrm{d}\beta_1 = \int f \ \mathrm{d}\beta_2 = \int f \ \mathrm{d}\beta_3 = f(0).$$

Here are the links to my earlier posts here on Math SE on Probs. 3 (a), (b), and (c), Chap. 6, in Baby Rudin, 3rd edition:

Prob. 3 (a), Chap. 6, in Baby Rudin: If $\beta_j(x)=0$ for $x<0$ and $\beta_j(x)=1$ for $x>0$ and $f$ is bounded on $[-1, 1]$, . . .

Prob. 3 (b), Chap. 6, in Baby Rudin: If $\beta_j(x)=0$ if $x<0$ and $\beta_j(x)=1$ if $x>0$, then . . .

Prob. 3 (c), Chap. 6, in Baby Rudin: If $\beta_j(x)=0$ if $x<0$ and $\beta_j(x)=1$ if $x>0$ for $j=1, 2, 3$, and if . . .

Here I'll attempt a solution to Prob. 3 (d).

My Attempt of Part (d):

If $f$ is continuous at $0$, then $f \in \mathscr{R} \left( \beta_3 \right)$ on $[-1, 1]$ (by Prob. 3 (c), Chap. 6, in Baby Rudin, 3rd edition), and (as in my post on Pob. 3 (c), Chap. 6, in Baby Rudin, 3rd edition) we also have $$\int_{-1}^1 f \ \mathrm{d} \beta_3 = f(0). \tag{1}$$

Moreover, if $f$ is continuous at $0$, then we also have $$\lim_{x \to 0} f(x) = f(0),$$ which implies that $$f(0+) = f(0) = f(0-).$$ So by Probs. 3 (a) and (b), Chap. 6, in Baby Rudin we can conclude that $f \in \mathscr{R}\left(\beta_1\right)$ and $f \in \mathscr{R}\left(\beta_2\right)$ on $[-1, 1]$, and that $$\int_{-1}^1 f \ \mathrm{d} \beta_1 = f(0) = \int_{-1}^1 f \ \mathrm{d} \beta_2. \tag{2}$$

From (1) and (2) we have our desired result.

Are there any problems with my proof?

• I'm sorry for the confusion, but this post (in its entirety) is not at all a duplicate of any of my earlier posts. The confusion is because I've included in all four of my posts the entire Prob. 3, Chap. 6, in Baby Rudin, 3rd edition. However, I've attempted only a singe part of that problem in any given post. So no two of the posts are duplicates of each other. – Saaqib Mahmood Jul 27 '17 at 15:24
• @amWhy please read through the two posts carefully and you'll see that these aren't duplicates of each other. Or, please read my comment above. – Saaqib Mahmood Jul 27 '17 at 15:26
• As I said in a very recent comment, this sight isn't a whiteboard so progress with three/four separate versions of the same thing. You are always free to edit and update your original question. – Namaste Jul 27 '17 at 15:33
• @amWhy but that will make that particular post too lengthy, won't it? Imagine how lengthy my post on Prob. 3(a), Chap. 6, in Baby Rudin would get if I edited / updated it to include my attempts of the Parts (b), (c), and (d) also! Please have a look at all the four posts one by one, and you'll then have an idea! – Saaqib Mahmood Jul 27 '17 at 15:41
• this site isn't meant to be an editorial site. You can not post a question (submit a proof), then fine tune it and resubmit, and then fine tune and "re-resubmit" it, .... It's quite the nerve to post and repost very lengthy proofs or questions, with the presumption that users are obliged to read each and every update, when you should have taken the time to consider and reconsider and then edit the original post. – Namaste Jul 27 '17 at 15:41