Iterated Integrals — MTH 429H

We now turn to defining integration over domains in {\mathbb{R}^{d}}. We will take a very practical approach to integration at first. Rather than trying to make a general theory, we will define the integral as an interated integral. That is we will define the integral of a continuous function {f\in C(I_{d})} by integrating “one variable at a time:”

\displaystyle  \int_{I_{d}}f(\mathbf{x})d\mathbf{x}:=\int_{a_{1}}^{b_{1}}\left(\int_{a_{2}}^{b_{2}}\cdots\left(\int_{a_{d}}^{b_{d}}f(x_{1},\ldots,x_{d})dx_{d}\right)\cdots dx_{2}\right)dx_{1},

where {I_{d}} is a d-cell,

\displaystyle  I_{d}=[a_{1},b_{1}]\times\cdots\times[a_{d},b_{d}].

(So a {1}-cell in {\mathbb{R}} is a compact interval, a {2}-cell in {\mathbb{R}^{d}} is a rectangle, and a {d}-cell in {\mathbb{R}^{d}} is the Cartesian product of {d-1}-cell in {\mathbb{R}^{d-1}} with a closed interval.)

A PDF version of this lecture is available here.

In more detail, if {f\in C(I_{d})} then for fixed {x_{1},\ldots,x_{d-1}} the function

\displaystyle  \phi(t)=f(x_{1},\ldots,x_{d-1},t)

is continuous on {[a_{d},b_{d}]}. Thus {\phi} is integrable and it makes sense to define:

\displaystyle  f_{d-1}(x_{1,}\ldots,x_{d-1}):=\int_{a_{d}}^{b_{d}}\phi(t)dt=\int_{a_{d}}^{b_{d}}f(x_{1,}\ldots,x_{d-1},x_{d})dx_{d}

Since {f} is uniformly continuous (why?), it is an easy exercise to show that {f_{d-1}} is continuous on the {d-1}-cell {I_{d-1}=[a_{1},b_{1}]\times\cdots\times[a_{d-1},b_{d-1}]}. Iterating the procedure we obtain for each {j=1,\ldots,d} a function {f_{j}\in C\left(I_{j}\right)} where {I_{j}=[a_{1},b_{1}]\times\cdots\times[a_{j},a_{j}]} with the following properties

  1. {f_{d}=f}.
  2. {f_{j}\left(x_{1,}\ldots,x_{j}\right)=\int_{a_{j+1}}^{b_{j+1}}f_{j+1}\left(x_{1},\ldots,x_{j},t\right)dt} for each {j=1,\ldots,d-1}.

Then we define

\displaystyle  \int_{I_{d}}f(\mathbf{x})d\mathbf{x}:=\int_{a_{1}}^{b_{1}}f_{1}(t)dt.

Remark 1 If we simply assume that {f\left(x_{1},\ldots,x_{d}\right)} is a Riemann integrable function of {x_{d}} for each fixed {x_{1},\ldots,x_{d-1}} then {f_{d-1}\left(x_{1},\ldots,x_{d-1}\right)} may not be a Riemann integrable function of {x_{d-1}}. This is why we restrict our attention to continuous funcitons. Ultimately, using the Lebesgue theory of integration, it is possible to show that the iterated integral makes sense for all Lebesgue integral functions (which includes all Riemann integrable functions) and gives the same result as a more geometrically defined integral.

Recall that the support of a function {f} is the smallest closed set containing all points at which {f} does not vanish,

\displaystyle  \mathrm{supp}f:=\overline{\left\{ x\ :\ f(x)\neq0\right\} }.

The set of all compactly supported continuous functions on {\mathbb{R}^{d}} is denoted {C_{c}\left(\mathbb{R}^{d}\right)}. Given {f\in C_{c}\left(\mathbb{R}^{d}\right)} we have {\mathrm{supp}f\subset[-M,M]^{d}} for sufficiently large {M} by the Heine-Borel theorem. We define

\displaystyle  \int_{\mathbb{R}^{d}}f\left(\mathbf{x}\right)d\mathbf{x}:=\int_{[-M,M]^{d}}f\left(\mathbf{x}\right)d\mathbf{x}.

To see that this definition makes sense we need the following

Theorem 2 Let {f} be a continuous function on {\mathbb{R}^{d}}. If {I_{d}} and {J_{d}} are {d}-cells such that {f\left(\mathbf{x}\right)=0} whenever {\mathbf{x}\in I_{d}\bigtriangleup J_{d}} then

\displaystyle  \int_{I_{d}}f\left(\mathbf{x}\right)d\mathbf{x}=\int_{J_{d}}f\left(\mathbf{x}\right)d\mathbf{x}.

Remark 3 Recall that the symmetric difference {A\bigtriangleup B} of two sets {A} and {B} is

\displaystyle  A\triangle B=\left(A\setminus B\right)\cup\left(B\setminus A\right)=\left(A\cup B\right)\cap\left(A\cap B\right)^{c}.

Proof: First suppose {J_{d}\subset I_{d}}. So

\displaystyle  \begin{array}{rcl}  I_{d} & = & [\alpha_{1},\delta_{1}]\times[\alpha_{2},\delta_{2}]\times\cdots\times[\alpha_{d},\delta_{d}]\\ J_{d} & = & [\beta_{1},\gamma_{1}]\times[\beta_{2},\gamma_{2}]\times\cdots\times[\beta_{d},\gamma_{d}] \end{array}

where

\displaystyle  \alpha_{j}\le\beta_{j}\le\gamma_{j}\le\delta_{j},\quad j=1,\ldots,d.

Since {f\left(x_{1},\ldots,x_{d}\right)} vanishes whenever {\gamma_{j}\le x_{j}\le\delta_{j}} or {\alpha_{j}\le x_{j}\le\beta_{j}} it is an easy exercise to see that according to the above definitions {\int_{J_{d}}f=\int_{I_{d}}f}.

Now for a general pair of {d}-cells, {J_{d}\cap I_{d}} is a {d}-cell and so if {f} vanishes on {I_{d}\bigtriangleup J_{d}} we have

\displaystyle  \int_{I_{d}}f=\int_{I_{d}\cap J_{d}}f=\int_{J_{d}}f.

\Box

Permuation invariance of the integral.

The above definition of the integral is very practical but it has one annoying defect: In principle the integral might depend on what order we choose to present the variables {x_{1},\ldots,x_{d}}. Put differently we can ask whether permuting the coordinates changes the integral?

Definition 4 Given a permutation {\sigma\in\mathcal{P}_{d}} we define the associated permutation map of {\mathbb{R}^{d}} denoted {P_{\sigma}} as follows

\displaystyle  P_{\sigma}\left(x_{1},\ldots,x_{d}\right)=\left(x_{\sigma(1)},\ldots,x_{\sigma(d)}\right).

Remark 5 Note that {P_{\sigma}} is a linear map. The matrix of {P_{\sigma}} consists of {0}‘s and {1}‘s, with exactly one {1} in each row and in each column;

\displaystyle  \left[P_{\sigma}\right]_{i,j}=\begin{cases} 1 & \mbox{ if }j=\sigma(i),\\ 0 & \mbox{ otherwise.} \end{cases}

The question of whether the integral depends on the order of the variables can be phrased as asking whether, with the above definition,

\displaystyle  \int_{I_{k}}f(\mathbf{x})d\mathbf{x}=\int_{P_{\sigma}(I_{k})}f(P_{\sigma}^{-1}\mathbf{x})d\mathbf{x} \ \ \ \ \ (1)

for arbitrary {d}-permutation?

Theorem 6 Let {I_{d}} be a {d}-cell, let {f\in C(I_{d})} and let {P_{\sigma}} be a permutation map. Then (1) holds.

Proof: We first prove this result in a special case. Suppose {f} is a product function, that is

\displaystyle  f(\mathbf{x})=h_{1}(x_{1})\cdots h_{k}(x_{d})=\prod_{i=1}^{d}h_{i}(x_{i}),

so {f} is a product of functions of a single variable. Then the left and right sides of (1) are both equal to

\displaystyle  \prod_{i=1}^{d}\int_{a_{_{i}}}^{b_{i}}h_{i}(t)dt.

Since both sides of (1) are linear in the function {f}, it follows that (1) holds whenever {f} is a linear combination of product functions, i.e. for

\displaystyle  f(\mathbf{x})=\sum_{m=1}^{n}\prod_{i=1}^{d}h_{i,m}(x_{i})

with {h_{i,m}\in C([a_{i},b_{i}])} for {i=1,\ldots,d} and {m=1,\ldots,n}. Let us denote by {\mathcal{A}} the collection of all such linear combinations of product functions,

\displaystyle  \mathcal{A}:=\left\{ \sum_{m=1}^{n}\prod_{i=1}^{d}h_{i,m}(x_{i})\ :\ n\in\mathbb{N}\text{ and }h_{i,m}\in C([a_{i},b_{i}])\text{ for }i=1,\ldots,d\text{ and }m=1,\ldots,n.\right\}  \ \ \ \ \ (2)

It is easy to see that {\mathcal{A}} is an algebra. The words “Stone-Weierstrass” should pop into your head now! Indeed the algebra {\mathcal{A}} is nowhere vanishing because the constant functions are in {\mathcal{A}}. Likewise it separates points since if {\mathbf{y}\neq\mathbf{z}} then {y_{i}\neq z_{i}} for some {i=1,\ldots,d} so the function

\displaystyle  f(\mathbf{x})=x_{i}-y_{i},

which is in {\mathcal{A},} satisfies {f(\mathbf{y})=0} while {f(\mathbf{z})=z_{i}-y_{i}\neq0}. So {\mathcal{A}} is dense in {C\left(I_{d}\right)} by Stone-Weierstrass!

Suppose we want to integrate {f\in C(I_{d})} in two different orders. Then for any {\epsilon>0} we can find {h\in\mathcal{A}} such that {\left|f(\mathbf{x})-h(\mathbf{x})\right|<\epsilon} for all {\mathbf{x}\in I_{k}}. But then

\displaystyle  \begin{array}{rl}  \left|\int_{I_{d}}f(\mathbf{x})d\mathbf{x}-\int_{P_{\sigma}(I_{d})}f(P_{\sigma}^{-1}\mathbf{x})d\mathbf{x}\right| \le  &   \int_{I_{d}}\left|f(\mathbf{x})-h(\mathbf{x})\right| d\mathbf{x} \\ & +\left|\int_{I_{d}}h(\mathbf{x})d\mathbf{x}-\int_{P_{\sigma}(I_{d})}h(P_{\sigma}^{-1}\mathbf{x})d\mathbf{x}\right|\\ & +\int_{P_{\sigma}(I_{d})}\left|f(P_{\sigma}^{-1}\mathbf{x})-h(P_{\sigma}^{-1}\mathbf{x})\right|d\mathbf{x}\\ < & 2\epsilon{\displaystyle \prod_{i=1}^{d}}(b_{i}-a_{i}).  \end{array}

Since {\epsilon} is arbitrary, we see that (1) holds\Box

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One thought on “Iterated Integrals — MTH 429H

  1. Pingback: Change of Variables Formula — MTH 429H | Jeffrey Schenker

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