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Prove $f(x) = \begin{cases} 2+x, & x \in \Bbb Q \\ 1, & x \notin \Bbb Q \end{cases}$ is not integrable on $[0,1]$

My attempt:

Let $P=\{t_0,...,t_n\}$ be a partition of $[0,1], I_i=[t_{i-1},t_i], m_i=\inf_{I_i}f, M_i=\sup_{I_i}f$

Since there exists at least one irrational number in $I_i$ we have that $m_i=1$

Since there exists at least one rational number in $I_i$ we have that $M_i=2+t_i$ or $M_i \approx 2+t_i$, where $t_i \in \Bbb Q$

$\implies L(f,P)=1-0=1 \ne \displaystyle{\sum _{i=1}^{n}} {M_i(t_i-t_{i-1})=U(f,P)}$, because $M_i \ne m_i$ $\forall i$

Is this correct? If not, then what can be improved? Thanks in advance.

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You could replace $$M_i \approx 2+t_i $$ by

$$M_i\ge 2$$ which is more precise.

then $$M_i-m_i\ge 2-1$$ and $$U (f,P)-L (f,P)=\sum_{i=0}^{n-1}(M_i-m_i)(t_{i+1}-t_i)\geq 1$$

if we take $\epsilon=\frac {1}{2} $ then for all partition $P $ of $[0,1] $

$$U (f,P)-L (f,P)>\epsilon $$ $f $ is not integrable at $[0,1] $.

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