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Show that there is no riemann integrable function $f$ on $[0,1]$ such that $f=\chi_{C}$ a.e. (almost everywhere), where $C$ is the fat cantor set.

Proof:

Would it suffice to show that $\chi_C$ is not riemann integrable??? Or am I missing something? Any suggestions?

Attempt:

Let $x$ be an element of $C$ then since $C$ contains no intervals if we define $C_n = (x-\frac{1}{n}, x+\frac{1}{n})$ then there exists some $x_n$ in $C_n$ such that $x_n\neq x$ and $x_n\not\in C$. Then $\forall \delta>0,\exists n$ such that $\delta>\frac{1}{n}.$ So then $$\left|x_n-x\right|<\frac{1}{n}<\delta.$$ But, if we take $\epsilon=1/2$ we get $$|f(x_n)-f(x)|=1>\epsilon$$ so $f$ is discontinuous at every $x\in C$, thus it cannot be Riemann integrable.

Would this work??

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@leo that is not true as written. A Riemann integrable function is discontinuous on a set of measure zero, but that does not mean that there is a continuous function which equals it almost everywhere. Any function with a jump discontinuity is a counterexample. –  Chris Janjigian Oct 8 '12 at 2:14
    
And no. It no suffices. Let $\{r_n\}$ an enumeration of the rationals in $[0,1]$. Let $f_n=\chi_{\{r_1,\ldots,r_n\}}$ then $f_n$ is Riemann integrable in $[0,1]$ and $$f_n=\chi_{\Bbb Q\cap [0,1]},$$ a.e., but $\chi_{\Bbb Q\cap [0,1]}$ is not Riemann integrable. –  leo Oct 8 '12 at 2:20
    
@Chris, you are right, thanks. –  leo Oct 8 '12 at 2:25
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2 Answers

A Riemann integrable function has the property that its set of discontinuities has Lebesgue measure zero. By contrast, the fat Cantor set is nowhere dense and of positive measure, so its indicator function has the property that in a neighborhood of every point of the fat Cantor set, the function takes on values of 1 and 0. In particular the fat Cantor set is contained in the set of discontinuities of $\chi_C$.

Changing the values on any measure zero subset will not change the size of the set of discontinuities in this case. This follows from the fact that, up to removing a set of measure zero, $C$ will still be contained in the discontinuities set (it matters that $C$ is nowhere dense for this part). To see this, let $f = \chi_C$ $a.e.$ and let $M$ be the Lebesgue measure zero set on which we have changed the value of the function. Take $x \in C\cap M^c$ and observe that since $C$ is nowhere dense and since $M$ cannot contain an interval, there is a point of $C^c\cap M^c$ in any neighborhood of $x$ and so $x$ is a discontinuity point of $f$. But $\mu(C \cap M^c) = \mu(C)$ so $f$ cannot be Riemann integrable.

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Attempt: Let $x$ be an element of $C$ then since $C$ contains no intervals if we define $C_n = (x-\frac{1}{n}, x+\frac{1}{n})$ then there exists some $x_n$ in $C_n$ such that $x_n\neq x$ and $x_n\not\in C$. Then $\forall \delta>0,\exists n$ such that $\delta>\frac{1}{n}.$ So then $$\left|x_n-x\right|<\frac{1}{n}<\delta.$$ But, if we take $\epsilon=1/2$ we get $$|f(x_n)-f(x)|=1>\epsilon$$ so $f$ is discontinuous at every $x\in C$, thus it cannot be Riemann integrable. Would this work?? –  Brandon Oct 8 '12 at 2:24
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@commenter the difference pops up when you try to rewrite the function on a set of measure zero (call it $M$). You need to argue that every point of $C\cap M^c$ is still a point of discontinuity (this the first point where a difference pops up: $C$ has positive measure so this set is nonempty and of positive measure). Since $C$ is nowhere dense, the argument still works that we can find a point in a neighborhood of any point of $C$ for which $f(x)=0$. –  Chris Janjigian Oct 8 '12 at 2:29
    
@Brandon no. You need to show that for any function equal to $\chi_C$ a.e. that function has a set of discontinuities of positive Lebesgue measure. See my previous comment for what I mean by that. –  Chris Janjigian Oct 8 '12 at 2:30
    
@ Chris: But, if correct, what I did above would show that the set of discontinuities of $f$ is exactly the fat cantor set $C$, and since $C$ has positive measure, wouldn't that conclude the proof? How else would I procede to show that? –  Brandon Oct 8 '12 at 2:32
    
@Chris: Now I agree with you, thanks for clarifying. I removed my obsolete comment (it addressed an earlier version of the answer anyway). May I suggest that you move this argument from a comment to the main body of the answer to make it more visible (maybe you could add "and has positive measure" in the last parentheses at the end)? –  commenter Oct 8 '12 at 2:43
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Yes it would suffice to show $\chi_C$ is not riemann integrable. Then proceed by contradiction.

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It's not immediately obvious that showing $\chi_C$ is not integrable by itself is sufficient without more work. The problem asks to show an entire class of functions is not integrable. –  Potato Oct 8 '12 at 2:08
    
@Potato Why it suffices? There are non Riemann integrable functions which are equal a.e. to a Riemann integrable function –  leo Oct 8 '12 at 2:33
    
Indeed, but one must say that. –  Potato Oct 8 '12 at 2:44
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