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 $[0,1].$?
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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0@DavidMitra please provide proof of it – 2018-03-04
3 Answers
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 $0$.
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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0Why not just look at $\chi_A?$ – 2018-05-24
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 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 $[a,b].$