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In [[real analysis]], a branch of [[mathematics]], the '''Darboux integral''' is constructed using '''Darboux sums''' and is one possible definition of the [[integral]] of a function.  Darboux integrals are equivalent to [[Riemann integral]]s, meaning that a function is Darboux-integrable if and only if it is Riemann-integrable, and the values of the two integrals, if they exist, are equal.  Darboux integrals have the advantage of being simpler to define than Riemann integrals.  Darboux integrals are named after their discoverer, [[Gaston Darboux]].
 
==Definition==
 
===Darboux sums===
A [[partition of an interval]] [''a'',''b''] is a finite sequence of values ''x''<sub>''i''</sub> such that
 
:<math>a = x_0 < x_1 < \cdots < x_n = b . \,\!</math>
 
Each interval [''x''<sub>''i''&minus;1</sub>,''x''<sub>''i''</sub>] is called a ''subinterval'' of the partition.  Let ƒ:[''a'',''b'']→'''R''' be a bounded function, and let
 
:<math>P = (x_0, \ldots, x_n) \,\!</math>
 
be a partition of [''a'',''b'']. Let
 
:<math>\begin{align}
M_i = \sup_{x\in[x_{i-1},x_{i}]} f(x) , \\
m_i = \inf_{x\in[x_{i-1},x_{i}]} f(x) .
\end{align}</math>
 
[[Image:Darboux.svg|thumb|right|Lower (green) and upper (green plus lavender) Darboux sums for four subintervals]]
 
The '''upper Darboux sum''' of ƒ with respect to ''P'' is
 
:<math>U_{f, P} = \sum_{i=1}^n (x_{i}-x_{i-1}) M_i . \,\!</math>
 
The '''lower Darboux sum''' of ƒ with respect to ''P'' is  
 
:<math>L_{f, P} = \sum_{i=1}^n (x_{i}-x_{i-1}) m_i . \,\!</math>
 
The lower and upper Darboux sums are sometimes called the lower and upper sums.
 
===Darboux integrals===
The '''upper Darboux integral''' of ƒ is  
 
:<math>U_f = \inf\{U_{f,P} \colon P \text{ is a partition of } [a,b]\} . \,\!</math>
 
The '''lower Darboux integral''' of ƒ is
 
:<math>L_f = \sup\{L_{f,P} \colon P \text{ is a partition of } [a,b]\} . \,\!</math>
 
In some literature an integral symbol with an underline and overline represent the lower and upper Darboux integrals respectively.
 
:<math>\begin{align}
L_f \equiv \underline{\int_{a}^{b}} f(x) \, dx &\quad U_f \equiv \overline{\int_{a}^{b}} f(x) \,dx
\end{align}</math>
 
And like Darboux sums they are sometimes simply called the lower and upper integrals.
 
If ''U''<sub>ƒ</sub>&nbsp;=&nbsp;''L''<sub>ƒ</sub>, then we call the common value the '''Darboux Integral'''.<ref>Wolfram MathWorld</ref> We also say that ƒ is ''Darboux-integrable'' or simply ''integrable'' and set
 
:<math>\int_a^b {f(t)\,dt} = U_f = L_f , \,\!</math>
 
An equivalent and sometimes useful criterion for the integrability of ''f'' is to show that for every ''ε > 0'' there exists a partition ''P''<sub>''ε''</sub> on  [''a'',''b''] such that<ref>Spivak 2008, chapter 13.</ref>
 
:<math> U_{f,P_\epsilon} - L_{f,P_\epsilon} < \epsilon </math>
 
==Properties==
*For any given partition, the upper Darboux sum is always greater than or equal to the lower Darboux sum. Furthermore, the lower Darboux sum is bounded below by the rectangle of width ''(b-a)'' and height ''inf(f)'' taken over [''a'',''b'']. Likewise, the upper sum is bounded above by the rectangle of width ''(b-a)'' and height ''sup(f)''.
 
:<math>(b-a)\inf_{x \in [a,b]} f(x) \leq L_{f,P} \leq U_{f,P} \leq (b-a)\sup_{x \in [a,b]} f(x)</math>
 
*The lower and upper Darboux integrals satisfy
 
:<math>\underline{\int_{a}^{b}} f(x) \, dx  \leq \overline{\int_{a}^{b}} f(x) \, dx </math>
 
*Given any ''c'' in (''a'',''b'')
 
:<math>\begin{align}
\underline{\int_{a}^{b}} f(x) \, dx  &= \underline{\int_{a}^{c}} f(x) \, dx +  \underline{\int_{c}^{b}} f(x) \, dx\\
\overline{\int_{a}^{b}} f(x) \, dx  &= \overline{\int_{a}^{c}} f(x) \, dx +  \overline{\int_{c}^{b}} f(x) \, dx
\end{align}</math>
 
*The lower and upper Darboux integrals are not necessarily linear. Suppose that ''g'':[''a'',''b'']→'''R''' is also a bounded function, then the upper and lower integrals satisfy the following inequalities.
 
:<math>\begin{align}
\underline{\int_{a}^{b}} f(x) \, dx + \underline{\int_{a}^{b}} g(x) \, dx &\leq \underline{\int_{a}^{b}} f(x) + g(x) \, dx\\
\overline{\int_{a}^{b}} f(x) \, dx + \overline{\int_{a}^{b}} g(x) \, dx &\geq \overline{\int_{a}^{b}} f(x) + g(x) \, dx
\end{align}</math>
 
*For a constant ''c'' ≥ 0 we have
 
:<math>\begin{align}
\underline{\int_{a}^{b}} cf(x) &= c\underline{\int_{a}^{b}} f(x)\\
\overline{\int_{a}^{b}} cf(x) &= c\overline{\int_{a}^{b}} f(x)
\end{align}</math>
 
*For a constant ''c'' ≤ 0 we have
 
:<math>\begin{align}
\underline{\int_{a}^{b}} cf(x) &= c\overline{\int_{a}^{b}} f(x)\\
\overline{\int_{a}^{b}} cf(x) &= c\underline{\int_{a}^{b}} f(x)
\end{align}</math>
 
*Consider the function ''F'':[''a'',''b'']→'''R''' defined as
 
:<math> F(x) = \underline{\int_{a}^{x}} f(x) \, dx </math>
then ''F'' is [[Lipschitz continuous]]. An identical result holds if ''F'' is defined using an upper Darboux integral.
 
==Examples==
 
===A Darboux-integrable function===
Suppose we want to show that the function ''f(x) = x'' is Darboux-integrable on the interval [0,1] and determine its value. To do this we partition [0,1] into ''n'' equally sized subintervals each of length ''1/n''. We denote a partition of ''n'' equally sized subintervals as ''P''<sub>''n''</sub>.
 
Now since ''f(x) = x'' is strictly increasing on [0,1], the infimum on any particular subinterval is given by its starting point. Likewise the supremum on any particular subinterval is given by its end point. The starting point of the ''k''th subinterval in ''P''<sub>''n''</sub> is ''(k-1)/n'' and the end point is ''k/n''. Thus the lower Darboux sum on a partition ''P''<sub>''n''</sub> is given by
 
:<math>\begin{align}
L_{f,P_n} &= \sum_{k = 1}^{n} f(x_{k-1})(x_{k} - x_{k-1})\\
        &= \sum_{k = 1}^{n} \frac{k-1}{n} \cdot \frac{1}{n}\\
        &= \frac{1}{n^2} \sum_{k = 1}^{n} [k-1]\\
        &= \frac{1}{n^2}\left[ \frac{(n-1)n}{2} \right]
\end{align}</math>
 
similarly, the upper Darboux sum is given by
 
:<math>\begin{align}
U_{f,P_n} &= \sum_{k = 1}^{n} f(x_{k})(x_{k} - x_{k-1})\\
        &= \sum_{k = 1}^{n} \frac{k}{n} \cdot \frac{1}{n}\\
        &= \frac{1}{n^2} \sum_{k = 1}^{n} k\\
        &= \frac{1}{n^2}\left[ \frac{(n+1)n}{2} \right]
\end{align}</math>
 
Since
:<math>\begin{align}
U_{f,P_n} - L_{f,P_n} &= \frac{1}{n}
\end{align}</math>
Thus for given any ''ε > 0'', we have that any partition ''P''<sub>''n''</sub> with ''n > 1/ε'' satisfies
:<math>\begin{align}
U_{f,P_n} - L_{f,P_n} &< \epsilon
\end{align}</math>
which shows that ''f'' is Darboux integrable. To find the value of the integral note that
 
:<math>\begin{align}
\int_{0}^{1}f(x) \, dx &= \lim_{n \to \infty} U_{f,P_n} =\lim_{n \to \infty} L_{f,P_n}  = \frac{1}{2}
\end{align}</math>
 
===An unintegrable function===
Suppose we have the function ''f'':[0,1]→'''R''' defined as
:<math>\begin{align}
f(x) &=
\begin{cases}
0, & \text{if }x\text{ is rational} \\
1, & \text{if }x\text{ is irrational}
\end{cases}
\end{align}</math>
Since the rational and irrational numbers are both [[dense subset]]s of '''R''', it follows that ''f'' takes on the value of 0 and 1 on every subinterval of any partition. Thus for any partition ''P'' we have
:<math>\begin{align}
L_{f,P} &=\sum_{k = 1}^{n}(x_{k} - x_{k-1})\inf_{x \in [x_{k-1},x_{k}]}f = 0\\
U_{f,P} &=\sum_{k = 1}^{n}(x_{k} - x_{k-1}) \sup_{x \in [x_{k-1},x_{k}]}f = 1
\end{align}</math>
from which we can see that the lower and upper Darboux integrals are unequal.
 
==Facts about the Darboux integral==
[[Image:Darboux refinement.svg|250px|thumb|right|When passing to a refinement, the lower sum increases and the upper sum decreases.]]
A ''refinement'' of the partition
 
:<math>x_0,\ldots,x_n  \,\!</math>
 
is a partition
 
:<math>y_0, \ldots, y_m \,\!</math>
 
such that for every ''i'' with
 
:<math>0 \le i \le n \,\!</math>
 
there is an integer ''r''(''i'') such that
 
:<math> x_{i} = y_{r(i)} . \,\!</math>
 
In other words, to make a refinement, cut the subintervals into smaller pieces and do not remove any existing cuts.  If
 
:<math>P' = (y_0,\ldots,y_m) \,\!</math>
 
is a refinement of
 
:<math>P = (x_0,\ldots,x_n) , \,\!</math>
 
then
 
:<math>U_{f, P} \ge U_{f, P'} \,\!</math>
 
and
 
:<math>L_{f, P} \le L_{f, P'} . \,\!</math>
 
If ''P''<sub>1</sub>, ''P''<sub>2</sub> are two partitions of the same interval (one need not be a refinement of the other), then
 
:<math>L_{f, P_1} \le U_{f, P_2} . \,\!</math>. 
 
It follows that
 
:<math>L_f \le U_f . \,\!</math>
 
Riemann sums always lie between the corresponding lower and upper Darboux sums.  Formally, if
 
:<math>P = (x_0,\ldots,x_n) \,\!</math>
 
and
 
:<math>T = (t_1,\ldots,t_n) \,\!</math>
 
together make a tagged partition
 
:<math> x_0 \le t_1 \le x_1\le \cdots \le x_{n-1} \le t_n \le x_n \,\!</math>
 
(as in the definition of the [[Riemann integral]]), and if the Riemann sum of ƒ corresponding to ''P'' and ''T'' is ''R'', then
 
:<math>L_{f, P} \le R \le U_{f, P}.\,\!</math>
 
From the previous fact, Riemann integrals are at least as strong as Darboux integrals: If the Darboux integral exists, then the upper and lower Darboux sums corresponding to a sufficiently fine partition will be close to the value of the integral, so any Riemann sum over the same partition will also be close to the value of the integral.  It is not hard to see that there is a tagged partition that comes arbitrarily close to the value of the upper Darboux integral or lower Darboux integral, and consequently, if the Riemann integral exists, then the Darboux integral must exist as well.
 
==See also==
* [[Regulated integral]]
* [[Lebesgue integration]]
 
==Notes==
{{Reflist}}
 
==References==
* {{cite web
| url = http://mathworld.wolfram.com/DarbouxIntegral.html
| title = Darboux Integral
| accessdate = 2013-01-08
| work=[[Wolfram MathWorld]]
}}
* [http://www.encyclopediaofmath.org/index.php/Darboux_integral ''Darboux integral'' at Encyclopaedia of Mathematics]
* {{springer|title=Darboux sum|id=p/d030160}}
* {{Citation
| last = Spivak
| first = Michael
| title = Calculus
| publisher = Publish or Perish
| year = 2008
| edition = 4
| isbn = 978-0914098911
}}
 
{{Integral}}
 
{{DEFAULTSORT:Darboux Integral}}
[[Category:Definitions of mathematical integration]]

Latest revision as of 17:13, 2 December 2014

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