Exercise 6.1

Question

Exercise 1: 
    Suppose $\alpha$ increase on $[a,b]$, $a\le x_0\le b$, $\alpha$ is continuous at $x_0$, $f(x_0)=1$, and $f(x)=0$ if $x
eq x_0$. Prove that $f\in\mathscr R(\alpha)$
    and that $\int f\,d\alpha=0$.

ghostofgarborg · Accepted Answer

\def\where{\,|\,} \def\eps{\varepsilon} ( extit{Matt ``Frito'' Lundy}) \ ( extit{Note: I should probably consider the cases where $x \pm d otin [a,b]$ in the solutions to 1 and 2 below}) \ First note that for any partition $P$ of $[a,b]$, $L(P,f,\alpha) = 0$, so we have $$\underline{\int_a^b} f \ d\alpha = 0$$ and $0 \leq U(P,f,\alpha)$. Let $\varepsilon > 0$ be given. $\alpha$ is continuous at $x_0$ means there exists a $\delta > 0$ such that $\left| x - x_0 ight| < \delta$ and $a \leq x \leq b$ implies $\left| \alpha(x) - \alpha(x_0) ight| < \varepsilon$. Let $P = \{a, x_0 - \delta, x_0 + \delta, b\}$. Then we have $$U(P,f,\alpha) = \alpha(x_0 + \delta) - \alpha(x_0 - \delta) < 2\varepsilon$$ and $$0 \leq U(P,f,\alpha) < 2\varepsilon.$$ Because $\varepsilon$ was arbitrary, we have $$\overline{\int_a^b} f \ d\alpha = 0$$ so that $f \in \mathscr{R}(\alpha)$ and $$\int_a^b f \ d\alpha = 0.$$

Amit Awasthi · Answer

\begin{proof}[Proof that $f \in \mathscr{R}(\alpha)$]
      Since $0 \le f \le 1$ on $[a,b]$, and $\alpha$ is continuous at the one discontinuity of $f$ (at $x_0$), $f \in \mathscr{R}(\alpha)$ by Theorem 6.10.
    \end{proof}
    \begin{proof}[Proof that $\int f d\alpha = 0$]
      Let our partition $P = {x_1,\ldots,x_n}$ be such that $x_0 \in (x_{j-1},x_j)$ for some $j \le n$, and every $x_i$ is distinct. Then, by Definition 6.2,
      \begin{align*}
        L(P,f,\alpha) &= \sum_{i=1}^n m_i \Delta \alpha_i = 0,
      \end{align*}
      where $m_i$ is as defined by Definition 6.1, since the $\inf$ over any interval of $f$ is 0. Since $f \in \mathscr{R}(\alpha)$ by the proof above, $\int f d\alpha = L(P,f,\alpha) = 0$ by Definition 6.2.
    \end{proof}

Exercise 6.1

Answers

Comments

Comments

Exercise 6.1

Answers

Comments

Comments

Add answer