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I know that if $f:[a,b]\to[m,M]$ is Riemann integrable and $g:[m,M]\to\mathbb{R}$ is continuous, then $g\circ f$ is also integrable on $[a,b]$. I'm trying to think about the following 3 cases:

1) $f$ the same, $g$ is Riemann integrable but not continuous on $[m,M]$, such that the statement is false.

2) $f:[a,b]\to\mathbb{R}$ is continuous and $g:[m,M]\to\mathbb{R}$ is bounded and Riemann integrable, then what about $g\circ f$?

3) $f$ is Riemann integrable on $[a,b]$ but not continuous, $g$ same as in 2), what about $g\circ f$?

Thank you.

[EDIT]: Let $f$ be continuous on $[0,1]$, vanish on cantor set C and positive otherwise. What kind of riemann integrable bounded real $g$ will make $g\circ f$ not riemann integrable?

mez
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  • For 1): let $g(x)=1$ for $0<x\le 1$ and $g(0)=0$; let $f(x)=0$ if $x$ is irrational, and $f(x)=1/m$ if $x=n/m$ with $n$, $m$ integers in lowest terms. – David Mitra Feb 17 '13 at 17:04

1 Answers1

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Here is an example to think about: Let $f\colon [0,1]\to\mathbb{R}$ be a continuous, nonnegative function which vanishes on a fat Cantor set $C$ and is positive outside $C$. Let $g$ be the characteristic function of $(0,\infty)$. Then $g\circ f$ is discontinuous everywhere on $C$, and hence not Riemann integrable.

Harald Hanche-Olsen
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  • Can you say more about characteristic function of $(0,\infty)$, what do you mean? Also doesn't characteristic functions take complex values? – mez Feb 20 '13 at 16:07
  • The characteristic function of a set $E$, often denoted $\chi_E$, takes the value $1$ for points in $E$ and $0$ for points outside $E$. You may be thinking of the characteristic function in statistics, which is the Fourier transform of a probability distribution. To avoid this confusion, some people say indicator function where I wrote characteristic function. – Harald Hanche-Olsen Feb 20 '13 at 16:39
  • Let $C$ be cantor set, and $f$ be continuous, 0 on $C$ and elsewhere positive. Why is $(g\circ f)(x)$ nowhere continuous? Isn't it continuous on for example $(1/3,2/3)$? – mez Feb 20 '13 at 16:56
  • It is discontinuous everywhere on $C$, not everywhere on $[0,1]$. And this is so because $C$ has empty interior. Also, for my example $C$ needs to be a fat Cantor set, not the usual Cantor set. The difference is in how much you remove at each stage in the construction: If you do it right, $C$ will have positive measure (this is the “fatness”), and that is why $g\circ f$ is not Riemann integrable. – Harald Hanche-Olsen Feb 20 '13 at 19:06