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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 is the link to my post on Math SE on Prob. 3 (a), 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 $[-1, 1]$, . . .

And, here is the link to my post here on Math SE on Prob. 3 (b), Chap. 6, in Baby Rudin:

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 . . .

My Attempt of Part (c):

Let us choose a real number $\varepsilon > 0$.

Suppose that $f$ is continuous at $0$. Then we can find a real number $\delta > 0$ such that $$ \lvert f(x) - f(0) \rvert < \frac{\varepsilon}{8} \tag{0} $$ for all real numbers $x \in [-1, 1]$ for which $\lvert x-0 \rvert < \delta$. Let us choose our $\delta \in (0, 1)$.

Let $n$ be a natural number such that $n > 3$. Let $P = \{ \ x_0, x_1, \ldots, x_n \ \}$ be a partition of $[-1, 1]$ such that $0$ is one of the points, say $x_j$ (for some $j \in \{ 1, \ldots, n-1 \ \}$, of $P$, and such that $\Delta x_i < \frac{\delta}{2}$ for each $i = 1, \ldots, n$.

Then as $x_{j-1} < x_j = 0 < x_{j+1}$, so $$ \begin{align} L \left( P, f, \beta_3 \right) &= \sum_{i=1}^n m_i \left[ \beta_3 \left( x_i \right) - \beta_3 \left( x_{i-1} \right) \right] \\ &= m_j \left[ \beta_3\left(x_j\right) - \beta_3\left(x_{j-1} \right) \right] + m_{j+1} \left[ \beta_3\left(x_{j+1} \right) - \beta_3\left(x_{j} \right) \right] \\ &= m_j \left( \frac{1}{2} - 0 \right) + m_{j+1} \left( 1 - \frac{1}{2} \right) \\ &= \frac{m_j + m_{j+1} }{2}, \tag{A} \end{align} $$ and similarly $$ \begin{align} U \left( P, f, \beta_3 \right) &= \sum_{i=1}^n M_i \left[ \beta_3 \left( x_i \right) - \beta_3 \left( x_{i-1} \right) \right] \\ &= M_j \left[ \beta_3\left(x_j\right) - \beta_3\left(x_{j-1} \right) \right] + M_{j+1} \left[ \beta_3\left(x_{j+1} \right) - \beta_3\left(x_{j} \right) \right] \\ &= M_j \left( \frac{1}{2} - 0 \right) + M_{j+1} \left( 1 - \frac{1}{2} \right) \\ &= \frac{M_j + M_{j+1} }{2}. \tag{B} \end{align} $$

Now as $\Delta x_i < \frac{\delta}{2}$ for each $i = 1, \ldots, n$ and as $x_j = 0$, so $$ -\frac{\delta}{2} < x_{j-1} < 0 < x_{j+1} < \frac{\delta}{2},$$ and therefore $$ \left[ x_{j-1}, x_{j+1} \right] \subset (-\delta, \delta), $$ and then from (0) we can conclude that $$ \lvert f(x) - f(0) \rvert < \frac{\varepsilon}{8} $$ for all $x \in \left[ x_{j-1}, x_{j+1} \right]$. That is, $$ f(0) - \frac{\varepsilon}{8} < f(x) < f(0) + \frac{\varepsilon}{8} $$ for all $x \in \left[ x_{j-1}, x_{j+1} \right]$. So we have the inequalities $$ f(0) - \frac{\varepsilon}{8} \leq m_j \leq M_j \leq f(0) + \frac{\varepsilon}{8}, $$ and $$ f(0) - \frac{\varepsilon}{8} \leq m_{j+1} \leq M_{j+1} \leq f(0) + \frac{\varepsilon}{8}. $$ By adding the last two chains of inequalities and then dividing the resulting chain out by $2$ we obtain $$ f(0) - \frac{ \varepsilon }{ 8 } \leq \frac{ m_j + m_{ j + 1 } }{2} \leq \frac{ M_j + M_{j+1 } }{ 2 } \leq f(0) + \frac{\varepsilon }{8}, $$ which together with (A) and (B) above implies that $$ f(0) - \frac{\varepsilon}{8} \leq L \left( P, f, \beta_3 \right) \leq U \left( P, f, \beta_3 \right) \leq f(0) + \frac{\varepsilon}{8}, \tag{1} $$ and this in turn implies that $$ U \left( P, f, \beta_3 \right) - L \left( P, f, \beta_3 \right) \leq \frac{\varepsilon}{4} < \varepsilon, $$ and this by virtue of Theorem 6.6 in Baby Rudin implies that $f \in \mathscr{R}\left(\beta_3\right)$ on $[-1, 1]$.

Is this part of the proof correct?

Conversely, suppose that $f \in \mathscr{R}\left(\beta_3\right)$ on $[-1, 1]$. Then by Theorem 6.6 in Baby Rudin, there exists a partition $Q$ of $[-1, 1]$ such that $$ U \left( Q, f, \beta_3 \right) - L \left( Q, f, \beta_3 \right) < \frac{\varepsilon}{2}. \tag{2} $$

Now let $P = \left\{ \ x_0, x_1, \ldots, x_n \ \right\}$ be a partition of $[-1, 1]$ such that $P \supset Q$ and such that $P$ contains $0$ as one of its points; let us suppose that $x_j = 0$ for some $j \in \{ \ 1, \ldots, n - 1 \ \}$.

As $P$ is a refinement of $Q$, so by Theorem 4.4 in Baby Rudin we have $$ L \left( Q, f, \beta_3 \right) \leq L \left( P, f, \beta_3 \right) \leq U \left( P, f, \beta_3 \right) \leq U \left( Q, f, \beta_3 \right), $$ and so $$ U \left( P, f, \beta_3 \right) - L \left( P, f, \beta_3 \right) \leq U \left( Q, f, \beta_3 \right) - L \left( Q, f, \beta_3 \right), $$
which together with (2) implies that $$ U \left( P, f, \beta_3 \right) - L \left( P, f, \beta_3 \right) < \frac{\varepsilon}{2}. \tag{3} $$

Now using (A) and (B) in (3) we obtain $$ \frac{ M_j + M_{j+1} }{2} - \frac{ m_j + m_{j+1} }{2} < \frac{\varepsilon}{2}, $$ which is the same as $$ \frac{M_j - m_j}{2} + \frac{M_{j+1} - m_{j+1} }{2} < \frac{\varepsilon}{2}. \tag{4} $$ But as $M_j - m_j \geq 0$ and $M_{j+1} - m_{j+1} \geq 0$, so (4) implies that $$ \frac{M_j - m_j}{2} < \frac{\varepsilon}{2} \qquad \mbox{ and } \qquad \frac{M_{j+1} - m_{j+1} }{2} < \frac{\varepsilon}{2}; $$ therefore $$ M_j - m_j < \varepsilon \qquad \mbox{ and } \qquad M_{j+1} - m_{j+1} < \varepsilon. \tag{5} $$

Now let $\delta$ be any real number such that $$ 0 < \delta < \min \left\{ \ \Delta x_j, \ \Delta x_{j+1} \ \right\} = \min \left\{ \ x_j - x_{j-1}, \ x_{j+1} - x_j \ \right\} = \min \left\{ \ - x_{j-1} , \ x_{j+1} \ \right\}. $$ Then we know that our choice of $\delta$ implies that $$ x_{j-1} < -\delta < 0 < \delta < x_{j+1}. \tag{6} $$

Now if $x$ is any real number which satisfies $$ \lvert x-0 \rvert < \delta,$$ then that $x$ also satisfies $$ - \delta < x < \delta, $$ and hence (6) implies that that real number $x$ also satisfies $$ x_{j-1} < x < x_{j+1}, $$ and therefore such an $x$ belongs to $ \left[ x_{j-1}, x_{j+1} \right]$. But $$ \left[ x_{j-1}, x_{j+1} \right] = \left[ x_{j-1}, x_j \right] \cup \left[ x_j, x_{j+1} \right]. $$ Thus any real number $x$ which satisfies $\lvert x-0 \rvert < \delta$ must belong to at least one of the intervals $\left[ x_{j-1}, x_j \right]$ and $\left[x_j, x_{j+1} \right]$, and so we must have $$ m_j \leq f(x) \leq M_j \qquad \mbox{ or } \qquad m_{j+1} \leq f(x) \leq M_{j+1} \tag{7} $$ for any real number $x$ such that $\lvert x-0 \rvert < \delta$.

Since both the inequalities in (7) hold for $x_j = 0$, therefore we can conclude from (7) that $$ \lvert f(x) - f(0) \rvert \leq \max \left\{ \ M_j - m_j, \ M_{j+1} - m_{j+1} \ \right\} \tag{8} $$ for any real number $x$ such that $\lvert x-0 \rvert < \delta$.

From (5) and (8) we can conclude that $$ \lvert f(x) - f(0) \rvert < \varepsilon $$ for any real number $x$ such that $\lvert x-0 \rvert < \delta$. Thus it follows that $f$ is continuous at $0$.

Is this proof of the converse part correct?

Thus we have shown that $f \in \mathscr{R}\left(\beta_3\right)$ on $[-1, 1]$ if and only if $f$ is continuous at $0$. In that case we also have $$ L \left( Q, f, \beta_3 \right) \leq \int_{-1}^1 f \ \mathrm{d} \beta_3 \leq U \left( Q, f, \beta_3 \right) $$ for any partition $Q$ of $[-1, 1]$; so (1) implies that $$ f(0) - \frac{\varepsilon}{8} \leq \int_{-1}^1 f \ \mathrm{d} \beta_3 \leq f(0) + \frac{\varepsilon}{8}, $$ which in turn implies that $$ \left\lvert \int_{-1}^1 f \ \mathrm{d} \beta_3 \ - \ f(0) \right\rvert < \varepsilon $$ for any real number $\varepsilon > 0$. Hence $$ \int_{-1}^1 f \ \mathrm{d} \beta_3 = f(0). $$

Is this the correct conclusion about $\int_{-1}^1 f \ \mathrm{d} \beta_3 $, and if so, then is it derived using the correct procedure?

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    $\begingroup$ The same down-vote!! What's so offensive about my post, I wonder? $\endgroup$ – Saaqib Mahmood Jul 27 '17 at 14:28

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