Exercise 6.7

Question

Exercise 7: 
    Suppose $f$ is a real function on $(0,1]$ and $f\in\mathscr R$ on $[c,1]$ for every $c>0$. Define $$\int_0^1f(x)\,dx=\lim_{c\rightarrow 0}\int_c^1f(x)\,dx$$ if this limit exists
    (and is finite).
    \begin{itemize}
        \item[(a)] If $f\in\mathscr R$ on $[0,1]$, show that this definition of the integral agrees with the old one.
        \item[(b)] Construct a function $f$ such that the above limit exists, although it fails to exist with $|f|$ in place of $f$.
    \end{itemize}

analambanomenos · Accepted Answer

(a) By Theorems 6.12(c) and 6.20, $$\lim_{cightarrow0}\int_c^1f(x)\,dx=\int_0^1f(x)\,dx-\lim_{cightarrow0}\int_0^cf(x)\,dx=\int_0^1f(x)\,dx.$$

(b) Let 
    $$f(x) = 
    \begin{cases}
        0 & x=0  \
        -\displaystyle\frac{2^{2n-1}}{2n-1} & 2^{-(2n-1)}<x\le2^{-(2n-2)},\;n=1,2,\ldots \
        \displaystyle\frac{2^{2n}}{2n} & 2^{-2n}<x\le2^{-(2n-1)},\;n=1,2,\ldots
    \end{cases} $$
    Then for any positive integer $N$, $$\int_{2^{-N}}^1f(x)\,dx=\sum_{n=1}^N\frac{(-1)^n}{n}$$ which converges as $Nightarrow\infty$ by Theorem 3.43. However,
    $$\int_{2^{-N}}^1\big|f(x)\big|\,dx=\sum_{n=1}^N\frac{1}{n}$$ fails to converge as $Nightarrow\infty$ by Theorem 3.28.

Exercise 6.7

Answers

Comments

Exercise 6.7

Answers

Comments

Add answer