Iterated integrals and the Fubini theorem · Selected prerequisite proofs

L10.2.2: Full upper/lower compact-rectangle Fubini theorem, not just its continuous specialization.

Let

R \times S \subset \R^n \times \R^m

be a closed rectangle and

f \colon R \times S \to \R

be integrable. The functions

g \colon R \to \R

and

h \colon R \to \R

defined by

\begin{equation*} g(x) \coloneqq \underline{\int_S} f_x \qquad \text{and} \qquad h(x) \coloneqq \overline{\int_S} f_x \end{equation*}

are integrable on

R

and

\begin{equation*} \int_R g = \int_R h = \int_{R \times S} f . \end{equation*}

A partition of

R \times S

can be written as

(P,P') = (P_1,P_2,\ldots,P_n,P'_1,P'_2,\ldots,P'_m)\text{,}

where

P = (P_1,P_2,\ldots,P_n)

and

P' = (P'_1,P'_2,\ldots,P'_m)

are partitions of

R

and

S

respectively. Let

R_1,R_2,\ldots,R_N

be the subrectangles of

P

and

R'_1,R'_2,\ldots,R'_K

be the subrectangles of

P'\text{.}

The subrectangles of

(P,P')

are

R_i \times R'_j

where

1 \leq i \leq N

and

1 \leq j \leq K\text{.}

Let

\begin{equation*} m_{i,j} \coloneqq \inf_{(x,y) \in R_i \times R'_j} f(x,y) . \end{equation*}

Notice that

V(R_i \times R'_j) = V(R_i)V(R'_j)

and hence

\begin{equation*} L\bigl((P,P'),f\bigr) = \sum_{i=1}^N \sum_{j=1}^K m_{i,j} \, V(R_i \times R'_j) = \sum_{i=1}^N \left( \sum_{j=1}^K m_{i,j} \, V(R'_j) \right) V(R_i) . \end{equation*}

Define

\begin{equation*} m_j(x) \coloneqq \inf_{y \in R'_j} f(x,y) = \inf_{y \in R'_j} f_x(y) . \end{equation*}

For

x \in R_i\text{,}

we have

m_{i,j} \leq m_j(x)\text{,}

and therefore,

\begin{equation*} \sum_{j=1}^K m_{i,j} \, V(R'_j) \leq \sum_{j=1}^K m_j(x) \, V(R'_j) = L(P',f_x) \leq \underline{\int_S} f_x = g(x) . \end{equation*}

The inequality holds for all

x \in R_i\text{,}

and so

\begin{equation*} \sum_{j=1}^K m_{i,j} \, V(R'_j) \leq \inf_{x \in R_i} g(x) . \end{equation*}

We obtain

\begin{equation*} L\bigl((P,P'),f\bigr) \leq \sum_{j=1}^N \left( \inf_{x \in R_j} g(x) \right) V(R_j) = L(P,g) . \end{equation*}

Similarly,

U\bigl((P,P'),f\bigr) \geq U(P,h)\text{,}

and the proof of this inequality is left as an exercise. Putting the two inequalities together with the fact that

g(x) \leq h(x)

for all

x\text{,}

\begin{equation*} L\bigl((P,P'),f\bigr) \leq L(P,g) \leq U(P,g) \leq U(P,h) \leq U\bigl((P,P'),f\bigr) . \end{equation*}

Since

f

is integrable, it must be that

g

is integrable as

\begin{equation*} U(P,g) - L(P,g) \leq U\bigl((P,P'),f\bigr) - L\bigl((P,P'),f\bigr) , \end{equation*}

and we can make the right-hand side arbitrarily small. As for any partition we have

L\bigl((P,P'),f\bigr) \leq L(P,g) \leq U\bigl((P,P'),f\bigr)\text{,}

we have

\int_R g = \int_{R \times S} f\text{.}

Likewise,

\begin{equation*} L\bigl((P,P'),f\bigr) \leq L(P,g) \leq L(P,h) \leq U(P,h) \leq U\bigl((P,P'),f\bigr) , \end{equation*}

and hence

\begin{equation*} U(P,h) - L(P,h) \leq U\bigl((P,P'),f\bigr) - L\bigl((P,P'),f\bigr) . \end{equation*}

f

is integrable, so is

h\text{.}

Moreover,

L\bigl((P,P'),f\bigr) \leq L(P,h) \leq U\bigl((P,P'),f\bigr)

implies

\int_R h = \int_{R \times S} f\text{.}