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Let $f(x) = \begin{cases} x &\mbox{if } x\in [0,1]\bigcap\mathbb{Q} \\ -x & \mbox{if } x\in [0,1]\bigcap\mathbb{Q}^c. \end{cases}$

I want to show that $f:[0,1]$ is not integrable.

My professor said that my result is not rigorous enough and should incorporate more Darboux Sum ideas? I keep looking at this and I feel it makes sense. So where did I go wrong and how do I make it more rigorous?

Here's my proof: By definition of Riemann integral we know f is integrable if the lower integral of $f$ on $[a,b]$ is the same as the upper integral of $f$ on $[a,b]$. So for them to not be integrable the lower and upper integral must not be equal.

Let $P=\{x_0,\ldots,x_1\}$ be a partition of $[0,1]$. Since the rationals and irrationals are dense in R for each index i greater than one such that $m_i=0$ and $M_i=1$ the collection of lower darboux sums consist of $0$. By definition of supremum the lower integral is $0$. On the other hand the collection of upper Darboux sums consist of $1$ and by infimum the upper integral is $1$. Since $0\neq1$ we have our result.

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  • $\begingroup$ Where do $m_i=0$ and $M_i=1$ come from? if $m_i:=\inf_{x\in [x_i,x_{i+1}]}f(x)$ then $m_i=-x_{i+1}$... $\endgroup$ – Milly Nov 20 '14 at 2:20
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Let $P=\{x_0,\cdots,x_n\}$ a partition. On each $[x_{i-1},x_{i}]$ you get $$m_i:=\inf_{[x_{i-1},x_{i}]} f(x)=-x_i,\qquad M_i:=\sup_{[x_{i-1},x_{i}]} f(x)=x_i$$ since there exist both rationals and irrationals are dense. Thus lower/upper Darboux sums are $$L_P:=-\sum x_i(x_i-x_{i-1}),\qquad U_P:=\sum x_i(x_i-x_{i-1}).$$ Let for any partition $Q=\{x_0,\cdots ,x_N\}$, let $x_q$ the smallest element such that $x_q\geq 1/2$. Thus $$L_Q:=-\sum x_i(x_i-x_{i-1}) \leq0,\qquad U_Q:=\sum x_i(x_i-x_{i-1})\geq\frac12\sum_{i\geq q} (x_i-x_{i-1}).$$ Letting the finess of $Q$ going to 0 gives $L_Q\leq 0$, $U_Q\geq 1/4$.

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