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could any one Give an example of a non-monotonic function on $[0,1]$ with infinitely many points of discontinuity such that the function is bounded & Riemann integrable on $[0,1].$?

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    $\begingroup$ Let $C$ be the Cantor set, and let $f=\chi_{C}$ be its indicator function. It is discontinuous at every point of $C$, which in particular shows that the set of discontinuities of $f$ is uncountable. But it is an easy exercise that it is Riemann integrable on $[0,1]$ with the integral value 0. $\endgroup$ – Sangchul Lee Aug 25 '12 at 15:31
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    $\begingroup$ Dirichlet's function is one. $\endgroup$ – David Mitra Aug 25 '12 at 16:58
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    $\begingroup$ Perhaps the simplest example is $f:[0,1]\rightarrow\Bbb R$ defined by $f(x)=\cases{1,& $x=1/n$\cr 0,& \text{otherwise}}$. $\endgroup$ – David Mitra Aug 25 '12 at 17:18
  • $\begingroup$ @DavidMitra please provide proof of it $\endgroup$ – user1942348 Mar 4 '18 at 11:58
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Try $f=\sum\limits_{n=1}^{+\infty}(-1)^na_n\mathbf 1_{[0,x_n]}$ with $(a_n)_{n\geqslant1}$ and $(x_n)_{n\geqslant1}$ decreasing to $0$.

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If by 'non-monotonic' , you simply mean a function which is not monotonic on $[0,1]$ (i.e. you are not asking a function which is not monotone on any sub-interval of $[0,1]$ , of any positive length), then simply consider :

$A:=\{\frac{1}{n}:n\in \Bbb N\}$ and the function

$f(x):= x $ if $x \in A^c \cap [0,\frac{1}{2}]$

$:= (x-\frac{1}{2}) $ if $ x \in A^c \cap [\frac{1}{2},1]$

$:= (1+\frac{1}{n}) $ if $ x =\frac{1}{n} $

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  • $\begingroup$ Why not just look at $\chi_A?$ $\endgroup$ – zhw. May 24 '18 at 16:57
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Thomae's function $T(x)$ (see https://en.wikipedia.org/wiki/Thomae%27s_function) fits the bill nicely. It is bounded on $[0,1],$ continuous at each irrational, discontinuous at each rational, and is Riemann integrable on $[0,1].$

It also fails to be monotonic on every subinterval of $[0,1]$ of positive length. To see this,suppose $0\le a<b\le 1.$ Then $[a,b]$ contains a rational $r$ in its interior, at which $T(r)>0.$ But $[a,r)$ contains an irrational $x$ and $(r,b]$ contains an irrational $y.$ Thus, as we move from left to right through $x,r,y$ we have the values of $T$ equal to $0,T(r),0.$ It follows that $T$ is not monotonic on $[a,b].$

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