Darboux Sums, Upper/Lower integral

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May 3rd 2010, 04:24 PM
serious331

Darboux Sums, Upper/Lower integral

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

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

and

$M_i \equiv \mbox{ sup} \{f(x) | x \in [x_{i-1}, x_i] \}$
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Then for a function $f: [0,1] \rightarrow R$ defined by:

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

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

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

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

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

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

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

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

then how does

$\int_{*a}^b f = 0$ and $\int_a^{*b} f \geq \frac{1}{2}$ 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)?
May 3rd 2010, 04:43 PM
Drexel28

Quote:

Originally Posted by serious331

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

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

and

$M_i=\sup \left\{f(x) | x in [x_{i-1}, x_i]\right\}$
-----------------------------------------------------
Then for a function $f: [0,1] \rightarrow R$ defined by:

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

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

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

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

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

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

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.

I'm not sure what you mean, but the upper sums are always one and the lower zero.

Quote:

-----------------------------
So, if we have another function

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

then how does
$\int_{*a^b} f = 0$ and $\int_{a^*b} f \geq \frac{1}{2}$ in this case??? Following the function 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)?

I actually really like this problem, give it some more thought but here's the idea

Spoiler:

So, let us denote $\mathcal{P}_n=\left\{0,\frac{1}{n},\frac{2}{n},\cd ots,1\right\}$ .

So, we cut $[0,1]$ into subintervals of the form $\left[\tfrac{k}{n},\tfrac{k+1}{n}\right]$ . And so clearly $\sup_{x\in[\frac{k}{n},\frac{k+1}{n}]}f(x)=\frac{k+1}{n}$ . And so, we get $U\left(\mathcal{P}_n,f\right)=\sum_{j=0}^{n}\mathc al{M}_j\Delta x_j=\sum_{j=0}^{n}\frac{j+1}{n}\cdot\frac{1}{n}=\f rac{1}{n^2}\sum_{j=0}^{n}j+1=\frac{(n+1)(n+2)}{2n^ 2}\geqslant\frac{1}{2}$

So?
May 3rd 2010, 04:58 PM
serious331

Quote:

Originally Posted by Drexel28

I'm not sure what you mean, but the upper sums are always one and the lower zero.

I actually really like this problem, give it some more thought but here's the idea

Spoiler:

So, let us denote $\mathcal{P}_n=\left\{0,\frac{1}{n},\frac{2}{n},\cd ots,1\right\}$ .

So, we cut $[0,1]$ into subintervals of the form $\left[\tfrac{k}{n},\tfrac{k+1}{n}\right]$ . And so clearly $\sup_{x\in[\frac{k}{n},\frac{k+1}{n}]}f(x)=\frac{k+1}{n}$ . And so, we get $U\left(\mathcal{P}_n,f\right)=\sum_{j=0}^{n}\mathc al{M}_j\Delta x_j=\sum_{j=0}^{n}\frac{j+1}{n}\cdot\frac{1}{n}=\f rac{1}{n^2}\sum_{j=0}^{n}j+1=\frac{(n+1)(n+2)}{2n^ 2}\geqslant\frac{1}{2}$

So?

then,
$U\left(\mathcal{P}_n,f\right) \geq \frac{1}{2}$ proves that the upper integral of f on [a,b] is greater than or equal to 1/2.

and are we using $\inf_{x\in[\frac{k}{n},\frac{k+1}{n}]}f(x)=\frac{k}{n}$ to prove that the lower integral of f on [a,b] = 0?

Thank you so much for the help!
May 3rd 2010, 05:00 PM
Drexel28

Quote:

Originally Posted by serious331

then,
$U\left(\mathcal{P}_n,f\right) \geq \frac{1}{2}$ proves that the upper integral of f on [a,b] is greater than or equal to 1/2.

Well, it proves that $\overline{\int_0^1}f(x)\text{ }dx\geqslant\frac{1}{2}$ , for a general interval I leave you to generalize.