Suppose $f$ is a real function on $(0, 1]$ and $f \in \mathscr{R}$ on $[c,1]$ for every $c>0$. Define $\int_0^1 f(x)dx=\lim_{c\to 0} \int_c^1 f(x)dx$ if this limit exists (and is finite).

(a) If $f \in \mathscr{R}$ on $[0,1]$, show that this definition of the integral agrees with the old one.

(b) Construct a function $f$ such that the above limit exists, although it fails to exist with $|f|$ in place of $f$.

This is Problem 7 of Chapter 6 in Principles of Mathematical Analysis by Rudin. For (a), I can prove the equation is correct but I am not sure what does 'definition agrees' mean? For (b), I have no idea.

Thank you in advance.

  • 1
    $\begingroup$ For (b) suppose, for $n\in \Bbb N,$ that $\int_{1/(n+1)}^{1/n}f(x)dx=(-1)^n/n .$ Suppose that when $x\in [1/(n+1),1/n]$ then $f(x)\leq 0$ if $n$ is odd, while $f(x)\geq 0$ if $n$ is even. $\endgroup$ – DanielWainfleet Nov 17 '18 at 10:26
  • $\begingroup$ The above answer is more intuitive. $\endgroup$ – Tengerye Nov 20 '18 at 9:02

$b$) Using the well known integral $$ \int_{1}^\infty \frac{\sin x}{x}\mathrm dx $$ which converges conditionally, we reflect reflect everything to near $0$ by sending $$ x\to 1/x $$ and find $$ \int_{1}^\infty\frac{\sin x}{x}\mathrm dx= \int_0^1\frac{\sin(1/y)}{y}\mathrm dy $$

For part $a$, you must prove using the definition of the Riemann integral that for a function which is integrable on $[0,1]$, the number $$ \int_0^1 f(x)\mathrm dx $$ is the same as the number $$ \lim_{c\to 0^+}\int_c^1f(x)\mathrm dx $$

  • $\begingroup$ I think, it should be $\int_0^1\frac{\sin(1/y)}{1/y}\mathrm dy$ in third display. Awesome answer by the way. I had seen a discrete construction before for this, and now know a well-known function can also serve the purpose. $\endgroup$ – Silent Dec 12 '18 at 2:29
  • $\begingroup$ @Silent thank you. Did you forget the derivative term in the substitution? $\endgroup$ – qbert Dec 12 '18 at 16:47
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    $\begingroup$ Thank you very much, for pointing that out. It is integration by substitution! $\endgroup$ – Silent Dec 13 '18 at 3:10

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