Darboux Sums, Upper/Lower integral

If the function $\displaystyle f : [a,b] \rightarrow R$ is bounded and that $\displaystyle P = \{\{x_0\},.....,\{x_n\}\}$ is a partition of its domain [a,b]. For each index $\displaystyle i \geq 1$ we have:

$\displaystyle {m_i} \equiv \mbox{inf} \{f(x) | x \in [x_{i-1}, x_i] \} $

and

$\displaystyle M_i \equiv \mbox{ sup} \{f(x) | x \in [x_{i-1}, x_i] \} $

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Then for a function $\displaystyle f: [0,1] \rightarrow R$ defined by:

$\displaystyle f(x)=\left\{\begin{array}{cc}0,&\mbox{ if the point x in [0,1] is rational}\\1, & \mbox{ if the point x in [0,1] is irrational}\end{array}\right.$

in this case, we suppose $\displaystyle P = \{{x_0},....., {x_n}\}$ a partition of its domain [0,1]. since the rationals and irrationals are dense in R, it follows that for each index $\displaystyle i \geq 1$, if $\displaystyle m_i$ & $\displaystyle M_i$ are defined as above, then $\displaystyle m_i = 0$ and $\displaystyle M_i = 1$.

Therefore, the collection of the lower Darboux sums consists of the single number 0,

$\displaystyle \int_{*a}^b f = 0$

and the upper collection of darboux sums consists of the single number 1, therefore:

$\displaystyle \int_a^{*b} f = 1$

note: the * sign at the bottom and top denotes the lower integral of f on [a,b] and the higher integral of f on [a,b] respectively.

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So, if we have another function

$\displaystyle f(x)=\left\{\begin{array}{cc}x,&\mbox{ if the point x in [0,1] is rational}\\0, & \mbox{ if the point x in [0,1] is irrational}\end{array}\right.$

then how does

$\displaystyle \int_{*a}^b f = 0$ and $\displaystyle \int_a^{*b} f \geq \frac{1}{2}$ in this case???

Following the example above, isn't the collection of lower darboux sum supposed to be x here, and how is it shown that the upper integral of f is greater than (1/2)?