Iterated Integrals and Fubini's Theorem
1. Iterated Integrals
For computational purposes, we want to be able to express integrals on \({\mathbb R}^n\) as iterated integrals of one variable.
Proposition
If \(f\) is Riemann integrable on the box \(R := [a_1,b_1] \times \tilde R\), then
\[ (L) \int_{a_1}^{b_1} f dx_1 \text{ and } (U) \int_{a_1}^{b_1} f dx_1 \]
are Riemann integrable on \(\tilde R\) and
\[{}\int_R f dx = \int_{\tilde R} \left[ (L) \int_{a_1}^{b_1} f dx_1 \right] d \tilde x{}\]
\[{}= \int_{\tilde R} \left[ (U) \int_{a_1}^{b_1} f dx_1 \right] d \tilde x.{}\]
Proof
Suppose
\[R := [a_1,b_1] \times \cdots \times [a_n,b_n].\]
Let
\[\tilde R := [a_2,b_2] \times \cdots \times [a_n,b_n].\]
(This is the same as \(R\) with first interval removed.) Let
\[{}I{}\]
\[{}:= \inf_{(x_1,\ldots,x_n) \in R} f(x_1,\ldots,x_n).{}\]
Then \(I \leq f(x_1,\ldots,x_n)\) for all \((x_1,\ldots,x_n) \in R\). So
\[{}(b_1 - a_1) I{}\]
\[{}\leq (L) \! \int_{a_1}^{b_1} f(x_1,x_2,\ldots,x_n) dx_1{}\]
\[{}\forall (x_2,\ldots,x_n) \in \tilde R.{}\]
Multiplying both sides by \(|\tilde R|\), using the fact that \(|R| = (b_1 - a_1) |\tilde R|\), and taking the infimum over all \((x_2,\ldots,x_n) \in \tilde R\) gives
\[{}|R| \inf_{x \in R} f(x){}\]
\[{}\leq |\tilde R| \inf_{x' \in \tilde R } (L) \int_{a_1}^{b_1} f(x_1,\tilde x) dx_1,{}\]
where \(\tilde x := (x_2,\ldots,x_n)\). Similar reasoning gives
\[{}|\tilde R| \sup_{x' \in \tilde R} (U) \int_{a_1}^{b_1} f(x_1,\tilde x) dx_1{}\]
\[{}\leq |R| \sup_{x \in R} f(x).{}\]
More generally, if \(\mathcal{P}_1 = \{I_1,\ldots,I_N\}\) is a partition of \([a_1,b_1]\) and \(\tilde R\) is any box in \({\mathbb R}^{n-1}\) such that \([a_1,b_1] \times \tilde R\) is in the domain of \(f\), then \(f(x_1,\ldots,x_n)\) is bounded below (at every point of \([a_1,b_1] \times \tilde R\)) by the function which equals \(\inf_{I_i \times \tilde R} f\) at all points in \(I_i \times \tilde R\) except possibly its boundary points. It's not difficult to see that the lower integral of such a function on \([a_1,b_1]\) would have to equal
\[ \sum_{i=1}^N |I_i| \inf_{I_i \times \tilde R} f \]
so it follows that
\[{}\sum_{i=1}^N |I_i| \inf_{I_i \times \tilde R} f{}\]
\[{}\leq (L) \int_{a_1}^{b_1} f(x_1,\tilde x) dx_1{}\]
for all \(\tilde x \in \tilde R\). Similarly
\[{}\sum_{i=1}^N |I_i| \sup_{I_i \times \tilde R} f{}\]
\[{}\geq (U) \int_{a_1}^{b_1} f(x_1,\tilde x) dx_1{}\]
for all \(\tilde x \in \tilde R\).
Now take the two inequalities immediately above, take an infimum and supremum, respectively, over \(\tilde x \in \tilde R\) and sum over all \(\tilde R\) in some grid partition of \([a_2,b_2] \times \cdots \times [a_n,b_n]\). The result is this: if \(\mathcal G\) is a grid on a rectangle \(R \subset {\mathbb R}^n\) and if \(\tilde{\mathcal G}\) is the same grid with the first direction removed,
\[{}{\mathcal L}(f,\mathcal G) \leq {\mathcal L} \left((L) \int_{a_1}^{b_1} f dx_1,\tilde {\mathcal G}\right){}\]
\[{}\leq {\mathcal U} \left((U) \int_{a_1}^{b_1} f dx_1,\tilde {\mathcal G}\right) \leq {\mathcal U}(f,\mathcal G){}\]
By monotonicity for lower and upper sums,
\[{}{\mathcal L} \left((L) \int_{a_1}^{b_1} f dx_1,\tilde {\mathcal G}\right){}\]
\[{}\leq {\mathcal L} \left((U) \int_{a_1}^{b_1} f dx_1,\tilde {\mathcal G}\right){}\]
\[{}\leq {\mathcal U} \left((U) \int_{a_1}^{b_1} f dx_1,\tilde {\mathcal G}\right){}\]
and
\[{}{\mathcal L} \left((L) \int_{a_1}^{b_1} f dx_1,\tilde {\mathcal G}\right){}\]
\[{}\leq {\mathcal U} \left((L) \int_{a_1}^{b_1} f dx_1,\tilde {\mathcal G}\right){}\]
\[{}\leq {\mathcal U} \left((U) \int_{a_1}^{b_1} f dx_1,\tilde {\mathcal G}\right).{}\]
Letting the grid \(\mathcal G\) become sufficiently fine gives the desired conclusion because the upper and lower integrals of \((U) \int_{a_1}^{b_1} f dx_1\) and \((L) \int_{a_1}^{b_1} f dx_1\) are necessarily both bounded below and above by \({\mathcal L}(f,\mathcal G)\) and \({\mathcal U}(f,\mathcal G)\), respectively, which can be made as close to equal as desired.
2. Fubini's Theorem
Theorem (Fubini's Theorem, Induction Version)
If \(R = [a_1,b_1] \times \tilde R\) is a box, \(f\) is Riemann integrable on \(R\), and if
\[ \int_{a_1}^{b_1} f(x_1,x_2,\ldots,x_n) dx_1 \]
exists for all \(\tilde x := (x_2,\ldots,x_n) \in \tilde R\), then
\[ \int_R f(x) dx = \int_{\tilde R} \left[ \int_{a_1}^{b_1} f(x_1,\tilde x) dx_1 \right] d \tilde x. \]
Proof
The proof is immediate: existence of the integral \(\int_{a_1}^{b_1} f(x_1,\tilde x) dx_1\) for each \(\tilde x\) guarantees that one-dimensional upper and lower integrals are always equal (and equal to the 1D integral itself).
Exercise
Use Fubini's Theorem to compute
\[ \int_R x ~ dA \]
where \(R \subset {\mathbb R}^2\) is the region
\[ \left\{(x,y) \in {\mathbb R}^2 \ : \ x \geq 0 \text{ and } x^2 + y^2 \leq 1 \right\}.\]
Hint
Recall how integrals over Jordan regions are defined in terms of multiplication by the characteristic function; this reduces the problem to integrating over a box.
Warning (Iterated Integrability Does Not Imply Integrability)
There exists a countable dense subset of \([0,1]^2\) which intersects each horizontal or vertical line in either zero or one points. The indicator function of this set is not Riemann integrable on \([0,1]^2\) (because upper sums are always \(1\) and lower sums always \(0\)), but the iterated integrals exist trivially.
Proof
Let \(Q_0 := [0,1]^2\). Let \(Q_1,\ldots,Q_4\) be the squares obtained by subdividing \(Q_0\) into \(4\) congruent parts. Then let
\[{}Q_5,Q_6,Q_7,Q_8,\ldots{}\]
\[{},Q_{17},Q_{18},Q_{19},Q_{20}{}\]
be the subdivisions of \(Q_1,\ldots,Q_4\), respectively, and continue this process forever. Let \(\{x_n\}_{n=0}^\infty\) be a sequence of points in \([0,1]^2\) chosen so that \(x_n \in Q_n\) for each \(n\) and \(x_n\) does not have its first or second coordinates in common with any of \(x_0,\ldots,x_{n-1}\) (which is possible because for any \(n\), there are only finitely many values of each coordinate which are prohibited and uncountably many to choose from). Let \(f(x) = 1\) if \(x\) is a term of the sequence and \(0\) otherwise. Then \(f\) is \(1\) on a countable dense subset of \([0,1]^2\) and zero elsewhere, but along any horizontal or vertical line, there can be no more than one point where \(f\) is nonzero. Thus the iterated integrals of \(f\) exist and are zero, but \(f\) itself is not integrable.