Exercise 6.3

Question

Exercise 3: 
    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 $[-1,1]$.
    \begin{itemize}
        \item[(a)] Prove that $f\in\mathscr R(\beta_1)$ if and only if $f(0+)=f(0)$ and that then $\int f\,d\beta_1=f(0)$.
        \item[(b)] State and prove a similar result for $\beta_2$.
        \item[(c)] Prove that $f\in\mathscr R(\beta_3)$ if and only if $f$ is continuous at 0.
        \item[(d)] If $f$ is continuous at 0 prove that $\int f\,d\beta_1=\int f\,d\beta_2=\int f\,d\beta_3=f(0)$.
    \end{itemize}

ghostofgarborg · Accepted Answer

\def\where{\,|\,} \def\eps{\varepsilon} ( extit{Matt ``Frito'' Lundy}) \ (a) First suppose that $f(0+) = f(0)$ and let $\varepsilon > 0$ be given. Then there exists a $\delta^* >0$ such that $0 < x < \delta^*$ implies $\left| f(x) - f(0) ight| < \epsilon$. Let $\delta = \min\{1,\delta^*/2\}$ and form a partition $P = \{-1,0,\delta,1\}$. Then we have $$U(P,f,\beta_1) - L(P,f,\beta_1) = f(s) - f(t)$$ for some $s,t \in [0,\delta]$. But $$f(s) - f(t) \leq \left| f(s) - f(0) ight| + \left| f(0) - f(t) ight| $$ so we have $$U(P,f,\beta_1) - L(P,f,\beta_1) \leq \left| f(s) - f(0) ight| + \left| f(0) - f(t) ight| < 2\varepsilon,$$ which shows that $f \in \mathscr R (\beta_1)$. Now suppose that $f \in \mathscr R (\beta_1)$ and let $\varepsilon >0$ be given. Then there exists a partition $P$ for which $$U(P,f,\beta_1) - L(P,f,\beta_1) < \varepsilon.$$ Let $P^*$ be a refinement of $P$ that includes $0$. Then we have: $$U(P^*,f,\beta_1) - L(P^*,f,\beta_1) < \varepsilon.$$ But if $[0,\delta]$ is the subinterval of $P^*$ that contains $0$, then: $$U(P^*,f,\beta_1) - L(P^*,f,\beta_1) = f(s) - f(t)$$ for some $s,t\in[0,\delta]$ where $f(s) - f(t) \geq \left| f(x_1) - f(x_2) ight|$ for any $x_1,x_2 \in [0,\delta]$. So we have for any $0 < x < \delta$ $$\left| f(0) - f(x) ight| \leq f(s) - f(t) < \varepsilon,$$ which means that $f(0+) = f(0)$. As above, for any partition $P$ that contains $0$ we have: $$U(P,f,\beta_1) = M_l \quad L(P,f,\beta_1) = m_l$$ in the interval $[0,x_l]$ of $P$. Because $f$ is right-continuous at $0$, both $M_l$ and $m_l$ converge to $f(0)$ as $x_l o 0$, so $$\int f \ d\beta_1 = f(0).$$ (b) The statement is: $f\in \mathscr R (\beta_2)$ if and only if $f(0-) = f(0)$ and then $$\int f \ d\beta_2 = f(0).$$ The proof is similar to part (a). (c) Suppose that $f$ is continuous at $0$ and let $\varepsilon$ be given. Then there exists a $\delta^* > 0$ such that $\left| x ight| < 0$ implies $\left| f(x) - f(0) ight| < \varepsilon$. Let $\delta = \min\{1,\delta^*/2\}$, and $P = \{-1,-\delta,\delta,1\}$. Then $$U(P,f,\beta_3) - L(P,f,\beta_3) = f(s) - f(t)$$ where $s,t \in [-\delta,\delta]$. But $$f(s) - f(t) \leq \left| f(s) - f(0) ight| + \left| f(0) - f(t) ight| $$ so $$U(P,f,\beta_3) - L(P,f,\beta_3) < 2\varepsilon$$ and $f\in \mathscr R (\beta_3)$. Now suppose that $f\in \mathscr R (\beta_3)$ and let $\varepsilon > 0$ be given. There exists a partition $P$ such that $$U(P,f,\beta_3) - L(P,f,\beta_3) < \varepsilon.$$ Let $P^*$ be a refinement of $P$ that contains $0$ so that the partitions around $0$ are $[x_{l-1},0]$ and $[0,x_l]$, let $\delta = \min\{\left| x_l ight| ,\left| x_{l-1} ight| \}$ and let $P^{\#}$ be a refinement of $P^*$ that contains $\pm \delta$. Then $$U(P^{\#},f,\beta_3) - L(P^{\#},f,\beta_3) = \frac{1}{2} \left[ f(s) - f(t) + f(q) - f(r) ight] $$ where $s,t \in [0,\delta]$, $q,r \in [-\delta,0]$ $$f(s) - f(t) \geq \left| f(0) - f(x) ight| \quad ext{for any} \quad x \in [0,\delta]$$ $$f(q) - f(q) \geq \left| f(0) - f(x) ight| \quad ext{for any} \quad x \in [-\delta, 0].$$ So for any $x \in [-\delta,\delta]$ we have $$\left| f(0) - f(x) ight| < \epsilon$$ which shows that $f$ is continuous at $0$. (d) The result follows from parts (a) - (c) and the fact that if $f$ is continuous at $0$, then $f(0-) = f(0) = f(0+)$.

Amit Awasthi · Answer

\begin{proof}[Proof of $(a)$] Let our partition $P$ include the point $x_{j-1} = 0$ and put $\delta = x_j - x_{j-1}$. By Definition 6.1, \begin{align*} U(P,f,\beta_1) &= \sup_{x \in [0,\delta]} f(x)\ L(P,f,\beta_1) &= \inf_{x \in [0,\delta]} f(x) \end{align*} since $\Delta\beta_1 = 1$ on the interval $[0,\delta]$, and 0 otherwise. \par If we assume $f \in \mathscr{R}(\beta_1)$, then given $\epsilon > 0$, we can find a refinement $P^*$ with a value of $x_j = \delta$ such that \begin{equation*} \sup_{x \in [0,\delta]} f(x) - \inf_{x \in [0,\delta]} f(x) < \epsilon, \end{equation*} by Theorem 6.6. Since $|f(0) - f(\delta)| \le \sup f(x) - \inf f(x)$ over this interval, \begin{equation*} |f(0) - f(\delta)| < \epsilon, \end{equation*} and so $f(0+) = f(0)$ since $\epsilon$ is arbitrary. \par If we assume $f(0+) = f(0)$, then given $\epsilon > 0$, we can find a $\delta > 0$ such that \begin{equation*} |f(x) - f(0)| < \epsilon, \end{equation*} for $0 < x < \delta$, so $f$ is continuous at $x = 0$, and therefore $\int_0^\delta f d\beta_1$ exists. Since $\int f d\beta_1 = 0$ for any other subinterval of $\mathbb{R}$, $f \in \mathscr{R}(\beta_1)$. \par Now evaluate $\int f d\beta_1$. Since $f \in \mathscr{R}(\beta_1)$, \begin{equation*} \int f d\beta_1 = \sup_{x \in [0,\delta]} f(x) = \inf_{x \in [0,\delta]} f(x), \end{equation*} and since $f(0+) = f(0)$, \begin{equation*} \int f d\beta_1 = f(0).\qedhere \end{equation*} \end{proof} \begin{claim}[$b$] $f \in \mathscr{R}(\beta_2)$ if and only if $f(0-) = f(0)$ and \begin{equation*} \int f d\beta_2 = f(0). \end{equation*} \end{claim} \begin{proof}[Proof of $(b)$] Let our partition $P$ include the point $x_{j+1} = 0$ and put $\delta = x_j - x_{j+1}$. By Definition 6.1, \begin{align*} U(P,f,\beta_2) &= \sup_{x \in [\delta,0]} f(x)\ L(P,f,\beta_2) &= \inf_{x \in [\delta,0]} f(x) \end{align*} since $\Delta\beta_2 = 1$ on the interval $[\delta,0]$, and 0 otherwise. \par If we assume $f \in \mathscr{R}(\beta_2)$, then given $\epsilon > 0$, we can find a refinement $P^*$ with a value of $x_j = \delta$ such that \begin{equation*} \sup_{x \in [\delta,0]} f(x) - \inf_{x \in [\delta,0]} f(x) < \epsilon, \end{equation*} by Theorem 6.6. Since $|f(0) - f(\delta)| \le \sup f(x) - \inf f(x)$ over this interval, \begin{equation*} |f(0) - f(\delta)| < \epsilon, \end{equation*} and so $f(0-) = f(0)$ since $\epsilon$ is arbitrary. \par If we assume $f(0-) = f(0)$, then given $\epsilon > 0$, we can find a $\delta < 0$ such that \begin{equation*} |f(\delta) - f(0)| < \epsilon, \end{equation*} so $f$ is continuous at $x = 0$, and therefore $\int_\delta^0 f d\beta_2$ exists. Since $\int f d\beta_2 = 0$ for any other subinterval of $\mathbb{R}$, $f \in \mathscr{R}(\beta_2)$. \par Now evaluate $\int f d\beta_2$. Since $f \in \mathscr{R}(\beta_2)$, \begin{equation*} \int f d\beta_1 = \sup_{x \in [\delta,0]} f(x) = \inf_{x \in [\delta,0]} f(x), \end{equation*} and since $f(0-) = f(0)$, \begin{equation*} \int f d\beta_1 = f(0).\qedhere \end{equation*} \end{proof}

Exercise 6.3

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Exercise 6.3

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