Fubini’s Theorem

Section 6.3 Fubini’s Theorem

An integral is rarely evaluated using the limit of Riemann sum. Fubini’s Theorem gives a mechanism to evaluate the integral of a function defined on a rectangle using iterated integrals in one variable. When the integrand is a continuous function on a rectangle, both the statement and proof of the theorem is straightforward. For the general case of a Riemann integrable function on a rectangle, the formulation and proof require some modification.

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Suppose that $R_{1}\subset \bbR^{n}$ and $R_{2}\subset \bbR^{m}$ are two rectangles in $\bbR^{n}$ and $\bbR^{m}$ respectively, and $f(\bx, \by)$ is a Riemann integrable function on $R:=R_{1}\times R_{2}\text{.}$ Then for any fixed $\bx\in R_{1}\text{,}$ we can consider the integrability of the function $\by\mapsto f(\bx, \by)$ for $\by \in R_{2}\text{.}$ We can also reverse the role between $\bx$ and $\by\text{.}$ To indicate the difference of the integration with respect to the variables, we write

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\begin{equation*} \int_{R} f(\bx, \by)\; dA \text{ or } \int_{R} f(\bx, \by)\; d\bx\, d\by \text{ for } \int_{R} f(\bx, \by), \end{equation*}

and

\begin{equation*} \int_{R_{2}}f(\bx, \by)\, d\by \text{ for the integral of $\by\mapsto f(\bx, \by)$ over $\by \in R_{2}$} \end{equation*}

when the integral exists. Likewise the upper and lower integrals

\begin{equation*} \upint_{R_{2}}f(\bx, \by)\, d\by \text{ and } \lowint_{\;R_{2}}f(\bx, \by)\, d\by \end{equation*}

make natural sense.

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Theorem 6.3.1. Fubini’s Theorem for Continuous Functions.

Suppose that $R_{1}\subset \bbR^{n}$ and $R_{2}\subset \bbR^{m}$ are two rectangles in $\bbR^{n}$ and $\bbR^{m}$ respectively, and $f(\bx, \by)$ is a continuous function on $R:=R_{1}\times R_{2}\text{.}$ Then $\int_{R_{2}} f(\bx, \by)\, d\by$ is a continuous function of $\bx \in R_{1}\text{,}$ and

\begin{equation} \int_{R} f(\bx, \by)\; dA = \int_{R_{1}} \left( \int_{R_{2}} f(\bx, \by)\, d\by \right)\, d\bx.\tag{6.3.1} \end{equation}

Likewise, $\int_{R_{1}} f(\bx, \by)\, d\bx$ is a continuous function of $\by \in R_{2}\text{,}$ and

\begin{equation} \int_{R} f(\bx, \by)\; dA = \int_{R_{2}} \left( \int_{R_{1}} f(\bx, \by)\, d\bx \right)\, d\by.\tag{6.3.2} \end{equation}

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Proof.

The key property used here is the uniform continuity of $f$ on $R\text{:}$ both sides of (6.3.1) and (6.3.2) can be approximated by any Riemann sum with respect to a partition whose size is sufficiently small. More specifically, for any $\epsilon >0\text{,}$ there exists $\delta >0$ such that for any rectangle $S_{1}$ of $R_{1}$ and $S_{2}$ of $R_{2}$ whose side length is less than $\delta\text{,}$ we have

\begin{equation*} \osc (f(\bx, \cdot), S_{2}) \lt \epsilon \text{ for any $\bx \in R_{1}$,} \end{equation*}

\begin{equation*} \osc (f(\cdot, \by), S_{1}) \lt \epsilon \text{ for any $\by \in R_{2}$,} \end{equation*}

and

\begin{equation*} \osc (f, S_{1}\times S_{2}) \lt \epsilon. \end{equation*}

It follows that for any partition $\cP_{1}$ of $R_{1}$ and $\cP_{2}$ of $R_{2}$ whenever $\lambda (\cP_{1}), \lambda (\cP_{2}) \lt \delta\text{,}$

\begin{equation*} U(f(\bx, \by), \cP_{2})-L(f(\bx, \by), \cP_{2}) \lt \epsilon |R_{2}| \text{ for any $\bx \in R_{1}$,} \end{equation*}

\begin{equation*} U(f(\bx, \by), \cP_{1})-L(f(\bx, \by), \cP_{1}) \lt \epsilon |R_{1}| \text{ for any $\by \in R_{2}$,} \end{equation*}

and

\begin{equation*} U(f(\bx, \by), \cP_{1}\times \cP_{2})-L(f(\bx, \by), \cP_{1}\times \cP_{2}) \lt \epsilon |R_{1}| |R_{2}|. \end{equation*}

Since $\int_{R_{1}\times R_{2}} f(\bx, \by)\, dA$ is sandwiched between $U(f(\bx, \by), \cP_{1}\times \cP_{2})$ and $L(f(\bx, \by), \cP_{1}\times \cP_{2})\text{,}$ it follows that

\begin{equation*} \left\vert \int_{R_{1}\times R_{2}} f(\bx, \by)\, dA - U(f(\bx, \by), \cP_{1}\times \cP_{2}) \right \vert \lt \epsilon |R_{1}| |R_{2}|. \end{equation*}

Likewise, $\int_{R_{2}} f(\bx, \by)\, d\by$ is sandwiched between $U(f(\bx, \by), \cP_{2})$ and $L(f(\bx, \by), \cP_{2})\text{,}$ it follows that

\begin{equation*} \left\vert \int_{R_{2}} f(\bx, \by)\, d\by - U(f(\bx, \by), \cP_{2}) \right \vert \lt \epsilon |R_{2}| \text{ for any $\bx \in R_{1}$,} \end{equation*}

therefore for any sampling $\{\bx_{\alpha}\}$ of points corresponding to $\cP_{1}$ we have

\begin{equation*} \left\vert \sum_{S_{\alpha}\in \cP_1} \left(\int_{R_{2}} f(\bx_{\alpha}, \by)\, d\by \right) |S_{\alpha}|- \sum_{S_{\alpha}\in \cP_1} U(f(\bx_{\alpha}, \by), \cP_{2}) |S_{\alpha}| \right \vert \lt\epsilon |R_{1}| |R_{2}|. \end{equation*}

Observe that $\sum_{S_{\alpha}\in \cP_1} U(f(\bx_{\alpha}, \by), \cP_{2}) |S_{\alpha}| $ is sandwiched between $U(f(\bx, \by), \cP_{1}\times \cP_{2})$ and $L(f(\bx, \by), \cP_{1}\times \cP_{2})\text{,}$ so

\begin{equation*} \left\vert \sum_{S_{\alpha}\in \cP_1} U(f(\bx_{\alpha}, \by), \cP_{2}) |S_{\alpha}| - \int_{R_{1}\times R_{2}} f(\bx, \by)\, dA \right\vert \lt \epsilon |R_{1}| |R_{2}|. \end{equation*}

It then follows that

\begin{equation*} \left\vert \sum_{S_{\alpha}\in \cP_1} \left(\int_{R_{2}} f(\bx_{\alpha}, \by)\, d\by \right) |S_{\alpha}|- \int_{R_{1}\times R_{2}} f(\bx, \by)\, dA \right\vert \lt 2 \epsilon |R_{1}| |R_{2}|. \end{equation*}

But $\sum_{S_{\alpha}\in \cP_1} \left(\int_{R_{2}} f(\bx_{\alpha}, \by)\, d\by \right) |S_{\alpha}|$ is a Riemann sum of the integral $\int_{R_{1}} \left( \int_{R_{2}} f(\bx, \by)\, d\by \right) \, d\bx$ only subject to $\lambda (\cP_{1}) \lt \delta\text{,}$ it follows that

\begin{equation} \left\vert \int_{R_{1}} \left( \int_{R_{2}} f(\bx, \by)\, d\by \right) \, d\bx- \int_{R_{1}\times R_{2}} f(\bx, \by)\, dA \right\vert \le 2 \epsilon |R_{1}| |R_{2}|.\tag{6.3.3} \end{equation}

Since $\epsilon >0 $ is arbitrary, it follows from (6.3.3) that (6.3.1) holds. (6.3.2) can be proved in a similar way.

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Exercise 6.3.2.

Verify that $\sum_{S_{\alpha}\in \cP_{1}} U(f(\bx_{\alpha}, \by), \cP_{2}) |S_{\alpha}| $ is sandwiched between $U(f(\bx, \by), \cP_{1}\times \cP_{2})$ and $L(f(\bx, \by), \cP_{1}\times \cP_{2})\text{.}$

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Exercise 6.3.3.

Verify (6.3.3). What’s the reason by replacing $\lt$ by $\le\text{?}$

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To obtain a similar theorem for a general Riemann integrable function $f$ on $R_{1}\times R_{2}\text{,}$ the main issue is that the integrability of $f$ on $R_{1}\times R_{2}$ does not necessarily guarantee that the iterated integrals $\int_{R_{2}} f(\bx, \by)\, d\by\text{,}$ and respectively $\int_{R_{1}} f(\bx, \by)\, d\bx\text{,}$ are defined for every $\bx\in R_{1}\text{,}$ and respectively for every $\by\in R_{2}\text{.}$ One can see this through the simple example on $[0, 1]\times [0, 1]$

\begin{equation*} f(x, y)= \begin{cases} 1-\frac 1q \amp x=\frac pq, p, q \text{ co-prime, and $y\in \bbQ$}, \\ 1 \amp \text{elsewhere.} \end{cases} \end{equation*}

This $f(x,y)$ fails to be Riemann integrable in $y\in [0, 1]$ for every $x\in \bbQ\text{.}$ Yet, both $\upint_{0}^{1} f(x, y)\, dy$ and $\lowint_{\;0}^{\;1} f(x, y)\, dy$ are well defined for every $x\in [0, 1]\text{.}$ It turns out that we can formulate and prove a Fubini theorem using either the upper or the lower integral as a replacement in the iterated integral.

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Exercise 6.3.4.

Verify that the function $f(x, y)$ above is Riemann integrable on $[0, 1]\times [0, 1]\text{.}$ Also evaluate $\upint_{0}^{1} f(x, y)\, dy$ and $\lowint_{\;0}^{\;1} f(x, y)\, dy$ for each $x \in [0, 1]$ and determine their integrability in $x \in [0, 1]\text{.}$

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Here is a typical application of the Fubini’s Theorem.

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Example 6.3.5.

Evaluate $\int_{A} e^{-x^{3}}y\, dx\, dy$ over the triangular region $A=\{(x, y): 0\le x \le 1, 0\le y \le x\}\text{.}$

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First the domain of integration is not a rectangle. We can treat this integral as the integral of $e^{-x^{3}}y \chi_{A}(x, y)$ over the square $R=\{(x, y): 0\le x \le 1, 0\le y \le 1\}\text{.}$

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We can try to evaluate the integral $\int_{R} e^{-x^{3}}y \chi_{A}(x, y)$ via

\begin{equation*} \int_{0}^{1} \left( \int_{0}^{1} e^{-x^{3}}y \chi_{A}(x, y) \, dx \right)\, dy =\int_{0}^{1} \left( \int_{x}^{1} e^{-x^{3}}y\, dx \right)\, dy, \end{equation*}

but the integral $\int_{x}^{1} e^{-x^{3}}y\, dx$ is not so easy to evaluate.

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We then try to the alternate way of iterated integrals

\begin{align*} \int_{0}^{1} \left( \int_{0}^{1} e^{-x^{3}}y \chi_{A}(x, y) \, dy \right)\, dx =\amp \int_{0}^{1} \left( \int_{0}^{x} e^{-x^{3}}y \, dy \right)\, dx \\ =\amp \int_{0}^{1} \left(\frac 12 x^{2} e^{-x^{3}} \right)\, dx\\ =\amp \frac 16 \left(1-e^{-1}\right). \end{align*}

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Exercise 6.3.6.

Evaluate the integral $\int_{R}\frac{x y^{2}}{1+x^{2}+y^{2}}\text{,}$ where $R=\{(x, y): -1\le x, y \le 1\}\text{.}$

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Theorem 6.3.7. Fubini’s Theorem for Riemann Integrable Functions.

Suppose that $R_{1}\subset \bbR^{n}$ and $R_{2}\subset \bbR^{m}$ are two rectangles in $\bbR^{n}$ and $\bbR^{m}$ respectively, and $f(\bx, \by)$ is a Riemann integrable function on $R:=R_{1}\times R_{2}\text{.}$ Then both $\upint_{R_{2}} f(\bx, \by)\, d\by$ and $\lowint_{\;R_{2}} f(\bx, \by)\, d\by$ are Riemann integrable of $\bx \in R_{1}\text{,}$ and

\begin{align} \int_{R} f(\bx, \by)\; dA \amp = \int_{R_{1}} \left( \upint_{R_{2}} f(\bx, \by)\, d\by \right)\, d\bx\notag\\ \amp = \int_{R_{1}} \left( \lowint_{\;R_{2}} f(\bx, \by)\, d\by \right)\, d\bx.\tag{6.3.4} \end{align}

Likewise, both $\upint_{R_{1}} f(\bx, \by)\, d\bx$ and $\lowint_{\;R_{1}} f(\bx, \by)\, d\bx$ are Riemann integrable of $\by \in R_{2}\text{,}$ and

\begin{equation*} \int_{R} f(\bx, \by)\; dA = \int_{R_{2}} \left( \upint_{R_{1}} f(\bx, \by)\, d\bx \right)\, d\by = \int_{R_{2}} \left( \lowint_{\;R_{1}} f(\bx, \by)\, d\bx \right)\, d\by. \end{equation*}

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Proof.

We will focus on the first equality in (6.3.4). It relies on the observation that the lower and upper sums of $\upint_{R_{2}} f(\bx, \by)\, d\by$ for any partition $\cP_{1}$ of $R_{1}$ and $\cP_{2}$ of $R_{2}$ are sandwiched between the lower sum $L(f(\bx, \by), \cP_{1}\times \cP_{2})$ and the upper sum $U(f(\bx, \by), \cP_{1}\times \cP_{2})\text{,}$ and these two sums approach $\int_{R} f(\bx, \by)\; dA$ when the partition size goes to zero.

\begin{align*} \amp L(f(\bx, \by), \cP_{1}\times \cP_{2})\le L(\upint_{R_{2}} f(\bx, \by)\, d\by, \cP_{1}) \le U( \upint_{R_{2}} f(\bx, \by)\, d\by, \cP_{1}) \\ \le \amp U(f(\bx, \by), \cP_{1}\times \cP_{2}). \end{align*}

This follows by noting that

\begin{equation*} \sum_{S_{2,\beta}\in \cP_{2}} m_{S_{2,\beta}}(f(\bx, \by)) |S_{2,\beta}| \le \lowint_{\;R_{2}} f(\bx, \by)\, d\by \end{equation*}

and

\begin{equation*} \upint_{R_{2}} f(\bx, \by)\, d\by \le \sum_{S_{2,\beta}\in \cP_{2}} M_{S_{2,\beta}}(f(\bx, \by)) |S_{2,\beta}| \end{equation*}

\begin{align*} L(f(\bx, \by), \cP_{1}\times \cP_{2})=\amp \sum_{S_{1,\alpha}\in \cP_{1}}\sum_{S_{2,\beta}\in \cP_{2}} m_{S_{1,\alpha}\times S_{2,\beta}}(f(\bx, \by)) |S_{1,\alpha}| |S_{2,\beta}| \\ \le \amp \sum_{S_{1,\alpha}\in \cP_{1}}\sum_{S_{2,\beta}\in \cP_{2}} m_{S_{1,\alpha}}(m_{S_{2,\beta}}(f(\bx, \by)) |S_{1,\alpha}| |S_{2,\beta}| \\ \le \amp \sum_{S_{1,\alpha}\in \cP_{1}} m_{S_{1,\alpha}} ( \lowint_{\;R_{2}} f(\bx, \by)\, d\by) |S_{1,\alpha}|\\ \le \amp \sum_{S_{1,\alpha}\in \cP_{1}} m_{S_{1,\alpha}} ( \upint_{R_{2}} f(\bx, \by)\, d\by) |S_{1,\alpha}|\\ \le \amp \sum_{S_{1,\alpha}\in \cP_{1}} M_{S_{1,\alpha}} ( \upint_{R_{2}} f(\bx, \by)\, d\by) |S_{1,\alpha}|\\ \le \amp \sum_{S_{1,\alpha}\in \cP_{1}} \sum_{S_{2,\beta}\in \cP_{2}} M_{S_{1,\alpha}} ( M_{S_{2,\beta}}(f(\bx, \by)) |S_{1,\alpha}| |S_{2,\beta}|\\ \le \amp U(f(\bx, \by), \cP_{1}\times \cP_{2}). \end{align*}

Since

\begin{equation*} \sup_{\cP_{1}\times \cP_{2} }L(f(\bx, \by), \cP_{1}\times \cP_{2}) =\inf_{\cP_{1}\times \cP_{2} }U(f(\bx, \by), \cP_{1}\times \cP_{2}) = \int_{R} f(\bx, \by)\; dA \end{equation*}

it follows that

\begin{equation*} \sup_{\cP_{1}} L(\upint_{R_{2}} f(\bx, \by)\, d\by, \cP_{1}) = \inf_{\cP_{1}} U( \upint_{R_{2}} f(\bx, \by)\, d\by, \cP_{1}) \end{equation*}

proving that $\upint_{R_{2}} f(\bx, \by)\, d\by$ is Riemann integrable over $\bx \in R_{1}$ with

\begin{equation*} \int_{R_{1}} \left(\upint_{R_{2}} f(\bx, \by)\, d\by \right)\, d\bx = \int_{R} f(\bx, \by)\; dA. \end{equation*}

The rest of the equalities can be proved in a similar way.

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Corollary 6.3.8. A Riemann integrable function $f(\bx, \by)$ of $(\bx, \by)$ is Riemann integrable in $\by$ except perhaps for $\bx$ on a set of measure $0$.

In the setting of the above theorem, $\upint_{R_{2}} f(\bx, \by)\, d\by= \lowint_{\;R_{2}} f(\bx, \by)\, d\by$ except perhaps on a set of measure $0$ of $R_{1}\text{.}$

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Proof.

It follows from (6.3.4) that

\begin{equation*} \int_{R_{1}} \left( \upint_{R_{2}} f(\bx, \by)\, d\by - \lowint_{\;R_{2}} f(\bx, \by)\, d\by\right)\, d\bx =0. \end{equation*}

Since $\upint_{R_{2}} f(\bx, \by)\, d\by - \lowint_{\;R_{2}} f(\bx, \by)\, d\by\ge 0$ for every $\bx\in R_{1}\text{,}$ the conclusion follows from the following exercise.

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Exercise 6.3.9.

Suppose that $g(\bx)\ge 0$ on $R_{1}$ and $\upint_{R_{1}} g(\bx )=0\text{.}$ Then $g(\bx)=0$ except perhaps on a set of measure $0$ of $R_{1}\text{.}$

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Exercise 6.3.10.

Prove the rest of the equalities in Theorem 6.3.7.

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Math412 Notes Select Topics in Mathematical Analysis of Functions of One or Several Variables

Section 6.3 Fubini’s Theorem

Theorem 6.3.1. Fubini’s Theorem for Continuous Functions.

Proof.

Exercise 6.3.2.

Exercise 6.3.3.

Exercise 6.3.4.

Example 6.3.5.

Exercise 6.3.6.

Theorem 6.3.7. Fubini’s Theorem for Riemann Integrable Functions.

Proof.

Corollary 6.3.8. A Riemann integrable function \(f(\bx, \by)\) of \((\bx, \by)\) is Riemann integrable in \(\by\) except perhaps for \(\bx\) on a set of measure \(0\).

Proof.

Exercise 6.3.9.

Exercise 6.3.10.