Darboux integral

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In real analysis, a branch of mathematics, the Darboux integral is one possible definition of the integral of a function. Darboux integrals are equivalent to Riemann integrals, 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.

[edit] Definition

A partition of an interval [a,b] is a finite sequence of values xi such that

a = x_0 < x_1 < \cdots < x_n = b . \,\!

Each interval [xi−1,xi] is called a subinterval of the partition. A refinement of the partition

x_0,\ldots,x_n \,\!

is a partition

y_0, \ldots, y_m \,\!

such that for every i with

0 \le i \le n \,\!

there is an integer r(i) such that

x_{i} = y_{r(i)} . \,\!

In other words, to make a refinement, cut the subintervals into smaller pieces and do not remove any existing cuts.

Let ƒ:[a,b]→R be a bounded function, and let

P = (x_0, \ldots, x_n) \,\!

be a partition of [a,b]. Let

\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}
Lower (green) and upper (green plus lavender) Darboux sums for four subintervals
Lower (green) and upper (green plus lavender) Darboux sums for four subintervals

The upper Darboux sum of ƒ with respect to P is

U_{f, P} = \sum_{i=1}^n M_i (x_{i}-x_{i-1}) . \,\!

The lower Darboux sum of ƒ with respect to P is

L_{f, P} = \sum_{i=1}^n m_i (x_{i}-x_{i-1}) . \,\!

The upper Darboux integral of ƒ is

U_f = \inf\{U_{f,P} \colon P \text{ is a partition of } [a,b]\} . \,\!

The lower Darboux integral of ƒ is

L_f = \sup\{L_{f,P} \colon P \text{ is a partition of } [a,b]\} . \,\!

If Uƒ = Lƒ, then we say that ƒ is Darboux-integrable and set

\int_a^b {f(t)\,dt} = U_f = L_f , \,\!

the common value of the upper and lower Darboux integrals.

[edit] Facts about the Darboux integral

When passing to a refinement, the lower sum increases and the upper sum decreases.
When passing to a refinement, the lower sum increases and the upper sum decreases.

If

P' = (y_0,\ldots,y_m) \,\!

is a refinement of

P = (x_0,\ldots,x_n) , \,\!

then

U_{f, P} \ge U_{f, P'} \,\!

and

L_{f, P} \le L_{f, P'} . \,\!

If P1, P2 are two partitions of the same interval (one need not be a refinement of the other), then

L_{f, P_1} \le U_{f, P_2} . \,\!.

It follows that

L_f \le U_f . \,\!

Riemann sums always lie between the corresponding lower and upper Darboux sums. Formally, if

P = (x_0,\ldots,x_n) \,\!

and

T = (t_1,\ldots,t_n) \,\!

together make a tagged partition

x_0 \le t_1 \le x_1\le \cdots \le x_{n-1} \le t_n \le x_n \,\!

(as in the definition of the Riemann integral), and if the Riemann sum of ƒ corresponding to P and T is R, then

L_{f, P} \le R \le U_{f, P}.\,\!

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.

[edit] See also

v • d • e
Integrals