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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.2012-10-08
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    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.2012-10-08
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    @Chris, you are right, thanks.2012-10-08

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