Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The following is an exercise from Stein's Real Analysis (ex. 10 Chapter 1). I know it should be easy but I am somewhat confused at this point; it mostly consists of providing the Cantor-like construction for continuous functions on the interval $[0,1]$ whose pointwise limit is not Riemann integrable.

So, let $C'$ be a closed set so that at the $k$th stage of the construction one removes $2^{k-1}$ centrally situated open intervals each of length $l^{k}$ with $l_{1}+\ldots+2^{k-1}l_{k}<1$; in particular, we know that the measure of $C'$ is strictly positive. Now, let $F_{1}$ denote a piece-wise linear and continuous function on $[0,1]$ with $F_{1}=1$ in the complement of the first interval removed in the consutrction of $C'$, $F_{1}=0$ at the center of this interval, and $0 \leq F_{1}(x) \leq 1$ for all $x$. Similarly, construct $F_{2}=1$ in the complement of the intervals in stage two of the construction of $C'$, with $F_{2}=0$ at the center of these intervals, and $0 \leq F_{2} \leq 1$, and so on, and let $f_{n}=F_{1}\cdot \ldots F_{n}$.

Now, obviously $f_{n}(x)$ converges to a limit say $f(x)$ since it is decreasing and bounded and $f(x)=1$ if $x \in C'$; so in order to show that $f$ is discontinuous at every point of $C'$, one should show that there is a sequence of points $x_{n}$ so that $x_{n} \rightarrow x$ and $f(x_{n})=0$; I can't see this, so any help is welcomed, thanks a lot!

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Take a point $c\in C'$ and any open interval $I$ containing $c$. Then there is an open interval $D\subseteq I $ that was removed in the construction of $C'$. Indeed, since $C'$ has no isolated points, there is a point $y\in C'\cap I$ distinct from $x$. Between $x$ and $y$, there is an open interval removed from the construction of $C'$, which we take to be our $D$.

Now, by the definition of the $f_n$, there is a point $d\in D$ (namely the center of $D$) such that $f(d)=0$.

To recap: given $x\in C'$ and any open interval $I$ containing $x$, there is a point $d\in I$ with $f(d)=0$. As $f(x)=1$, this implies that $f$ is not continuous at $x$.

share|improve this answer
    
Heh, nice. Thanks. Feel bad I didn't see it –  Anna Feb 13 '12 at 0:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.