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**serious331** If the function $\displaystyle f : [a,b] \rightarrow R$ is bounded abd that $\displaystyle P = \{\{x_0\},.....,\{x_n\}\}$ is a partition of its domain [a,b]. For each index $\displaystyle i \geq 1$ we have:

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

and

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

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

$\displaystyle f(x)=\begin{cases} 0 & \mbox{if} \quad x\in\mathbb{Q} \\ 1 & \mbox{if}\quad x\notin\mathbb{Q}\end{cases}$

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:

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

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.