Products of measures

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1 Products of measurable spaces

1.1 The product of \(\sigma\)-algebras

Let \((X,\mathcal{F}_{X})\), \((Y,\mathcal{F}_{Y})\) be measurable spaces. Then the product of measurable spaces is the set \(X \times Y\) together with the sigma-algebra

\[ \mathcal{F}_{X} \otimes \mathcal{F}_{Y} := \sigma \left(\left\{ A \times B \; : \; A \in \mathcal{F}_{X}, B \in \mathcal{F}_{Y}\right\} \right) \]

That is, this is the Cartesian product of \(X\) and \(Y\) together with the minimal sigma-algebra on the rectangles \(A \times B\).

Let \(E \subset X \times Y\). Let \(a \in X\). The vertical section (cut) of the set \(E\) at the point \(a\) is the set

\[ [E]_{a} := \left\{ y \in Y \; : \; (a, y) \in E\right\} \]

The set \([E]^b\), \(b \in Y\), is defined analogously.

The vertical section \([E]_a\) and the horizontal section \([E]^b\) of the set \(E \subset X \times Y\) — both are measurable.

Lemma 1 (Sections of measurable sets are measurable) If \(E \in \mathcal{F}_{X} \otimes \mathcal{F}_{Y}\), then \([E]_{a} \in \mathcal{F}_{Y}\) for every \(a \in X\), and \([E]^{b} \in \mathcal{F}_{X}\) for every \(b \in X\).

Sections of a function are defined analogously. If \(f: X \times Y \to \mathbb {R}\) is a function and \(a \in X\), then \([f]_{a}: Y \to \mathbb {R}\) is defined by

\[ [f]_{a}(y) := f(a, y), \quad y \in Y \]

Similarly for \([f]^b\).

Lemma 2 (Sections of measurable functions are measurable) Let \(f: X \times Y \to \mathbb {R}\) be an \(\mathcal{F}_{X} \otimes \mathcal{F}_{Y}\)-measurable function. Then for every point \(a \in X\) the section \([f]_a\) is an \(\mathcal{F}_Y\)-measurable function.

Similarly for \([f]^b\).

ProblemProblem 1

Suppose \((X, \mathcal{F}_{X})\) and \((Y, \mathcal{F}_{Y})\) are measurable spaces. Prove that every rectangle in \(\mathcal{F}_{X} \otimes \mathcal{F}_{Y}\) can be obtained only as the Cartesian product of elements of \(\mathcal{F}_{X}\) and \(\mathcal{F}_{Y}\). That is, if \(A\) is a nonempty subset of \(X\) and \(B\) is a nonempty subset of \(Y\) such that \(A \times B \in \mathcal{F}_{X} \otimes \mathcal{F}_{Y}\), then necessarily \(A \in \mathcal{F}_{X}\) and \(B \in \mathcal{F}_{Y}\).

ProblemProblem 2

Suppose \((X, \mathcal{F}_{X})\) is a measurable space. Prove that if \(E \in \mathcal{F}_{X} \otimes \mathcal{F}_{X}\), then \[ \left\{ x \in X:(x, x) \in E\right\} \in \mathcal{F}_{X} . \]

1.2 The product of measures

The following result will let us introduce the product of \(\sigma\)-finite measures.

Lemma 3 (The measure of a section is a measurable function) Let \((X, \mathcal{F}_{X}, \mu )\), \((Y, \mathcal{F}_{Y}, \nu )\) be two spaces with \(\sigma\)-finite measures. Let \(E \in \mathcal{F}_{X} \otimes \mathcal{F}_{Y}\). Then the function \[ x \mapsto \nu \left([E]_{x}\right) \] is \(\mathcal{F}_{X}\)-measurable. Similarly for \(y \mapsto \mu \left([E]^{y}\right)\).

Suppose \((X, \mathcal{F}_{X}, \mu )\) and \((Y, \mathcal{F}_{Y}, \nu )\) are measure spaces, and \(f: X \times Y \rightarrow \mathbb {R}\) is a function. Then

\[ \int _{X} \int _{Y} \; f(x, y) d \nu (y) d \mu (x) \quad \text{ means } \quad \int _{X}\left(\int _{Y} f(x, y) d \nu (y)\right) d \mu (x) . \]

In other words, to compute \(\int_{X} \int_{Y} f(x, y) d \nu (y) d \mu (x)\), first (temporarily) fix \(x \in X\) and compute the integral \(\int_{Y} [f]_x(y) \; d \nu (y)\) of the section \([f]_x\) (if this integral makes sense). Then compute the integral with respect to \(\mu\) of the function

\[ x \mapsto \int _{Y} [f]_x(y) d \nu (y) \]

(if this integral makes sense).

ExampleExample 1

Compute \[ \int _{[0,4]}\int _{[0,4]}f(x,y)\; d\lambda (x) \; d\lambda (y), \qquad \int _{[0,4]}\int _{[0,4]}f(x,y)\; d\lambda (y) \; d\lambda (x) \] where \(f(x,y) = x^2 + y\).

SolutionSolution

\([f]_x(y)\) is a continuous function on \([0,4]\) (as a function of \(y\)) for every \(x\). Hence it is Riemann integrable in the proper sense, and the Riemann integral equals the Lebesgue integral, so we may freely pass to the Riemann integral. Similarly for \([f]^y\). In that case we obtain \[ \begin{align} \int _{[0,4]} \int _{[0,4]}\left(x^2+y\right) d \lambda (y) d \lambda (x) & =\int _{[0,4]}\left(4 x^2+8\right) d \lambda (x) =\frac{352}{3} \\ \int _{[0,4]} \int _{[0,4]}\left(x^2+y\right) d \lambda (x) d \lambda (y) & =\int _{[0,4]}\left(\frac{64}{3}+4 y\right) d \lambda (y) =\frac{352}{3} \end{align} \] The two iterated integrals in this example turned out to be equal to \(\frac{352}{3}\), even though at the intermediate stage of the computation they look different. As we shall see in the next section, this equality of integrals under a change of the order of integration is not a coincidence.

AnswerAnswer

In both cases, \(\frac{352}{3}.\)

Definition 1 (The product of two measures: \(\mu \times \nu\)) Suppose \((X, \mathcal{F}_{X}, \mu )\) and \((Y, \mathcal{F}_{Y}, \nu )\) are spaces with \(\sigma\)-finite measures. For \(E \in \mathcal{F}_{X} \otimes \mathcal{F}_{Y}\) define the value of the measure \((\mu \times \nu )(E)\) as the iterated integral of the indicator: \[ (\mu \times \nu )(E) := \int _{X} \int _{Y} I_E \; d \nu \, d \mu \]

As we shall prove later (see Tonelli’s theorem), it does not matter in which order we integrate in the definition above.

The definition of \((\mu \times \nu )(E)\) given above makes sense because the inner integral equals \(\nu \left([E]_x\right)\), and above we stated that sections of measurable sets are measurable (i.e. we may “plug” \([E]_x\) into \(\nu\)). Furthermore, the outer integral makes sense because above we showed that \(x \mapsto \nu ([E]_x)\) is a measurable function.

Lemma 4 (The product of two sigma-finite measures is a measure) Suppose \((X, \mathcal{F}_{X}, \mu )\) and \((Y, \mathcal{F}_{Y}, \nu )\) are spaces with \(\sigma\)-finite measures. Then \(\mu \times \nu\) is a measure on \((X \times Y, \mathcal{F}_{X} \otimes \mathcal{F}_{Y})\).

We restrict the definition above to \(\sigma\)-finite measures because the main results we will obtain fail without \(\sigma\)-finiteness (see Example Example 3 in the next section).

ExampleExample 2

Prove that under the conditions above, the measure of a rectangle is the product of the measures of its sides.

SolutionSolution

Suppose \((X, \mathcal{F}_{X}, \mu )\) and \((Y, \mathcal{F}_{Y}, \nu )\) are spaces with a \(\sigma\)-finite measure. If \(A \in \mathcal{F}_{X}\) and \(B \in \mathcal{F}_{Y}\), then \[ \begin{aligned} (\mu \times \nu )(A \times B) & =\int _{X} \underbrace{\left(\int _{Y} I_{A \times B}(x, y) d \nu (y)\right)}_{=\nu ([A \times B]_x) = \nu (B) \cdot I_{A}(x)} d \mu (x) \\ & =\int _{X} \nu (B) I_A(x) d \mu (x) = \nu (B) \cdot \int _{X} I_A(x) d \mu (x) =\\ & =\nu (B) \cdot \mu (A) \end{aligned} \]

ProblemProblem 3

Suppose \(\mu\) and \(\nu\) are \(\sigma\)-finite measures. Prove that \(\mu \times \nu\) is \(\sigma\)-finite.

2 Multiple integrals

2.1 Equality of multiple and iterated integrals: Tonelli’s and Fubini’s theorems

The conclusion of Tonelli’s theorem and of Fubini’s theorem is (almost) the same, but the hypotheses differ slightly: in Tonelli’s theorem \(f\) is nonnegative, whereas in Fubini’s theorem \(f\) has arbitrary sign but is integrable.

NoteTonelli’s theorem and Fubini’s theorem

Let

• \((X, \mathcal{F}_{X}, \mu)\) and \((Y, \mathcal{F}_{Y}, \nu)\) be spaces with \(\sigma\)-finite measures,
• \(f: X\times Y \to \bar{\mathbb{R}}\) be an \(\mathcal{F}_{X} \otimes \mathcal{F}_{Y}\)-measurable function.

Tonelli’s theorem	Fubini’s theorem
Let \(f\) be nonnegative: \(f: X \times Y \to [0,+\infty]\)	\(f\) is integrable: \(\int_{X \times Y} \|f\| \, d(\mu \times \nu) < \infty\)

Then:

• The function \(x \mapsto \int_Y f(x,y)\,d\nu(y)\) is \(\mathcal{F}_{X}\)-measurable. Similarly for \(y \mapsto \int_X f(x,y)\,d\mu(x)\).
• The multiple integral equals the iterated integrals:

\[ \int_{X \times Y} f\,d(\mu \times \nu) = \int_Y \left(\int_X f(x,y)\,d\mu(x)\right)d\nu(y) = \int_X \left(\int_Y f(x,y)\,d\nu(y)\right)d\mu(x) \]

---

P.S. Fubini’s theorem additionally asserts that for \(\mu\)-almost all \(x \in X\) the section \([f]_x\) is \(\nu\)-integrable. Similarly for \([f]^y\).

ExampleExample 3

Show that Tonelli’s theorem can fail without the hypothesis of \(\sigma\)-finiteness.

SolutionSolution

• Let the 1st space be the space \(([0,1], \mathscr {B}\left([0,1]\right), \lambda )\), where \(\lambda\) is Lebesgue measure on \([0,1]\).

• Let the 2nd space be the space \(([0,1], \mathscr {B}\left([0,1]\right), \# )\), where \(\#\) is counting measure. This space is not sigma-finite.

• We showed earlier in a problem that \(\mathscr {B}\left([0,1]\right) \otimes \mathscr {B}\left([0,1]\right)\) is the Borel sigma-algebra on the square \([0,1]^2\). Hence the diagonal \(D = \left\{ (x,x) \; : \; x \in [0,1]\right\}\) lies in the product of these sigma-algebras: the diagonal is a closed set, and hence Borel.

• Consider the indicator function of the diagonal \(f = I_{D}: [0,1]^2 \to \overline{\mathbb {R}}\). It is measurable, being the indicator of a measurable set. It is nonnegative. Hence the hypotheses of Tonelli’s theorem are satisfied.

  Note that

  \[ I_{D}(x,y) = \begin{cases} 1, & x = y \\ 0, & x \neq y \end{cases} \]

• Let us compute the first iterated integral: for fixed \(y\), the function \([f]^y(x)\) equals \(1\) at only one point, namely \(y = x\). Hence it is equal to zero Lebesgue-almost everywhere, and \(\int_{[0,1]} [f]^y(x) \; d\lambda (x) = 0\) for all \(y \in [0,1]\). Hence,

  \[ \int _{Y} \left(\int _{X} f(x,y) \; d\lambda (x)\right) \; d\# (y) = \int _{Y} 0 \; d\# (y) = 0 \]

• Let us compute the second iterated integral: for fixed \(x\), the function \([f]_x(y)\) equals \(0\) everywhere except at the point \(y = x\), where it equals \(1\). The integral with respect to counting measure is simply the sum of the values of the function at all nonzero points. Hence \(\int_{[0,1]} [f]_x(y) \; d\# (y) = 1\) for all \(x \in [0,1]\), and

  \[ \int _{X} \left(\int _{Y} f(x,y) \; d\# (y)\right) \; d\lambda (x) = \int _{X} 1 \; d\lambda (x) = 1 \]

The two iterated integrals turned out to be different.

ProblemProblem 4

1. Let \(\lambda\) denote Lebesgue measure on \([0,1]\). Show that

\[ \int _{[0,1]} \int _{[0,1]} \frac{x^{2}-y^{2}}{\left(x^{2}+y^{2}\right)^{2}} d \lambda (y) d \lambda (x) = \frac{\pi }{4} \]

and

\[ \int _{[0,1]} \int _{[0,1]} \frac{x^{2}-y^{2}}{\left(x^{2}+y^{2}\right)^{2}} d \lambda (x) d \lambda (y) = -\frac{\pi }{4} \]

2. Explain why the previous part does not contradict either Tonelli’s theorem or Fubini’s theorem.

HintHint

Enum-item(1)

Fix \(x \in [0,1]\). Polar substitution: \(\theta = \arctan \frac{y}{x}\), \(y = x\tan \theta\), \(dy = \frac{x}{\cos^2 \theta } d\theta\), \(x^2+y^2 = \frac{x^2}{\cos^2 \theta }\). Then

\[ \begin{align} \int _0^1 \frac{x^2-y^2}{(x^2+y^2)^2} dy & = \int _{0}^{\arctan (1/x)}\frac{x^2 - x^2\tan ^2\theta }{\left(\frac{x^2}{\cos ^2\theta }\right)^2} \cdot \frac{x}{\cos ^2 \theta } \; d\theta = \frac{1}{x} \cdot \int _0^{\arctan (1/x)} \cos ^2\theta - \sin ^2\theta \; d\theta = \\ & = \int _0^{\arctan (1/x)} \cos 2\theta \; d\theta = \frac{1}{2x} \sin (2\arctan (1/x)) \end{align} \]

where \(\sin (2\arctan 1/x) = \frac{2x}{1+x^2}\).

Enum-item(2)

On which part of the square \([0,1]^2\) is the integrand nonnegative? On which part is it negative?

2.2 The area under a graph

Suppose \(X\) is a set (of arbitrary nature, in general) and \(f: X \to [0, +\infty ]\) is a function. Then the region under the graph of the function \(f\), denoted \(U_f\), is defined as follows:

\[ U_f=\{ (x, t) \in X \times (0, +\infty ): 0<t<f(x)\} . \]

We call \(U_f\) the region under the graph of \(f\) even when \(X\) is not a subset of \(\mathbb {R}\). The term “area” in the next paragraph should remind you of the ordinary notion of area, although what we are actually dealing with is the measure of \(U_f\) in the product space.

The first equality in the result below can be viewed as a reminder of Riemann’s concept of the integral as the area under a graph (though now in a far more general context, with arbitrary \(\sigma\)-finite measures). The second equality in the result below can be viewed as a strengthening of Lebesgue’s concept of computing the area under a graph by integrating in the direction perpendicular to Riemann’s direction.

Lemma 5 (The area under the graph of a function equals the integral) Suppose \((X, \mathcal{F}_{X}, \mu )\) is a space with a \(\sigma\)-finite measure, and \(f: X \to [0, +\infty ]\) is a function measurable with respect to \(\mathcal{F}_{X}\). Then the region under the graph is measurable (\(U_f \in \mathcal{F}_{X} \otimes \mathscr {B}\left([0,+\infty )\right)\)) and \[ (\mu \times \lambda )\left(U_f\right)=\int _{X} f d \mu = \int _{0}^{+\infty } \mu (f > t) \; d t \]

It follows from Markov’s inequality that if \(f\) and \(\mu\) are as in the result above, then

\[ \mu (\{ x \in X: f(x)>t\} ) \leq \frac{\int _{X} f d \mu }{t} \]

for all \(t>0\). Thus, if \(\int_{X} f d \mu <\infty\), the result above should be regarded as somewhat stronger than Markov’s inequality (since \(\int_{(0, \infty )} \frac{1}{t} d \lambda (t)=\infty\)).

3 Lebesgue integration in \(\mathbb {R}^{n}\)

3.1 Lebesgue measure and the Borel sigma-algebra in \(\mathbb {R}^{n}\)

Recall that a set in \(\mathbb {R}^{n}\) is called open if every point in it belongs together with some open ball around it. The Borel sigma-algebra in \(\mathbb {R}^{n}\) is the minimal sigma-algebra containing all open sets.

Lemma 6 Let \(m,n\in \mathbb {N}\).

• If \(G_1\), \(G_2\) are open subsets of \(\mathbb {R}^{m}\), \(\mathbb {R}^{n}\), then \(G_1 \times G_2\) is an open subset of \(\mathbb {R}^{m + n}\);

• \(\mathscr {B}\left(\mathbb {R}^{m}\right) \otimes \mathscr {B}\left(\mathbb {R}^{n}\right) = \mathscr {B}\left(\mathbb {R}^{m +n}\right)\);

• \(\mathcal{M}(\lambda_m) \otimes \mathcal{M}(\lambda_n) \subsetneq \mathcal{M}(\lambda_{m+n})\);

• If \(\lambda_{m}\), \(\lambda_n\) are Lebesgue measures on \((\mathbb {R}^{m}, \mathscr {B}\left(\mathbb {R}^{m}\right))\) and \((\mathbb {R}^{n}, \mathscr {B}\left(\mathbb {R}^{n}\right))\), then

  \[ \lambda _{m+n} = \lambda _m \times \lambda _n \]

  where \(\lambda_{m + n}\) is Lebesgue measure on \((\mathbb {R}^{m+n}, \mathscr {B}\left(\mathbb {R}^{m+n}\right))\).

ProblemProblem 5

Let \(X = Y = \mathbb {R}\). Prove that

1. \(\mathscr {B}\left(\mathbb {R}\right) \otimes \mathscr {B}\left(\mathbb {R}\right) = \mathscr {B}\left(\mathbb {R}^{2}\right)\);

2. \(\mathcal{M}(\lambda ) \otimes \mathcal{M}(\lambda ) \subsetneq \mathcal{M}(\lambda_2)\), where \(\lambda_2\) is Lebesgue measure in \(\mathbb {R}^{2}\), and \(\mathcal{M}(\lambda_2)\) is the sigma-algebra of Lebesgue-measurable subsets of \(\mathbb {R}^{2}\)?

ProblemProblem 6

Show that there exists a set \(E \subseteq \mathbb {R}^{2}\) such that the sections \([E]_{a}\) and \([E]^{a}\) are open subsets of \(\mathbb {R}\) for every \(a \in \mathbb {R}\), but \(E \notin \mathscr {B}\left(\mathbb {R}^{2}\right)\).

ProblemProblem 7

Suppose \(G_{1}\) is a nonempty subset of \(\mathbb {R}^{m}\) and \(G_{2}\) is a nonempty subset of \(\mathbb {R}^{n}\). Prove that \(G_{1} \times G_{2}\) is an open subset of \(\mathbb {R}^{m} \times \mathbb {R}^{n}\) if and only if \(G_{1}\) is an open subset of \(\mathbb {R}^{m}\) and \(G_{2}\) is an open subset of \(\mathbb {R}^{n}\).

ProblemProblem 8

Suppose \(E\) is a subset of \(\mathbb {R}^{m} \times \mathbb {R}^{n}\) and \[ A = \left\{ x \in \mathbb {R}^{m}:(x, y) \in E \text{ for some } y \in \mathbb {R}^{n}\right\} \]

1. Prove that if \(E\) is an open subset of \(\mathbb {R}^{m} \times \mathbb {R}^{n}\), then \(A\) is an open subset of \(\mathbb {R}^{m}\).

2. Prove or give a counterexample: if \(E\) is a closed subset of \(\mathbb {R}^{m} \times \mathbb {R}^{n}\), then \(A\) is a closed subset of \(\mathbb {R}^{m}\).

3.2 The volume of the unit ball in \(\mathbb {R}^n\)

Lemma 7 (The scaling measure) Suppose \(t>0\). If \(E \in \mathscr {B}\left(\mathbb {R}^{n}\right)\), then \(t E \in \mathscr {B}\left(\mathbb {R}^{n}\right)\) and \(\lambda_n(t E)=t^n \lambda_n(E)\).

Denote the open unit ball in \(\mathbb {R}^n\) by:

\[ \mathbf{B}_n := \left\{ \left(x, \ldots , x_n\right) \in \mathbb {R}^n: x^2+\cdots +x_n^2<1\right\} . \]

The open unit ball \(\mathbf{B}_n\) is open in \(\mathbb {R}^n\) (you should convince yourself of this) and therefore belongs to the family of Borel sets \(\mathscr {B}\left(\mathbb {R}^{n}\right)\).

Theorem 1 (The volume of the unit ball in \(\mathbb {R}^n\)) The volume of the \(n\)-dimensional unit ball equals \[ \lambda _n\left(\mathbf{B}_n\right) = \frac{\pi ^{n / 2}}{\Gamma \left(\frac{n}{2}+1\right)}= \begin{cases} \dfrac {\pi ^{n / 2}}{(n / 2)!} & \text{ if } n \text{ is even, } \\ \dfrac {2^{(n+1) / 2} \pi ^{(n-1) / 2}}{1 \cdot 3 \cdot 5 \cdots \cdot n} & \text{ if } n \text{ is odd. } \end{cases} \]

ProblemProblem 9

Prove the formula for the volume of the unit ball.

HintHint

Prove that \(\lambda_n\left(\mathbf{B}_n\right) = \frac{2\pi }{n}\lambda_{n-2}\left(\mathbf{B}_{n-2}\right)\) for \(n \geq 3\). Use a change to polar coordinates in the integral.

ProblemProblem 10

1. Prove that \(\lambda_{n}\left(\mathbf{B}_{n}\right) \to 0\) as \(n \to +\infty\).

2. Find the value of \(n\) that maximizes \(\lambda_{n}\left(\mathbf{B}_{n}\right)\).

3.3 The curse of dimensionality

ExampleExample 4

er

SolutionSolution

Let \(X_1, \ldots , X_n\), \(Y_1, \ldots , Y_n\), \(\xi_{i} = X_{i} - Y_{i}\). The central moments of \(\xi_i\)

\[ \begin{align} \mu _1 = 0, \quad \mu _2 = 2\int _0^1x^2(1-x)dx = 2 \left(\frac{1}{3} - \frac{1}{4}\right) = \frac{1}{6} \\ \mu _3 = 0, \quad \mu _4 = 2\int _{0}^1x^4(1-x)dx = 2 \left(\frac{1}{5} - \frac{1}{6}\right) = \frac{1}{15} \end{align} \]

Then

\[ \mathbb {E}\left[\xi _i^2\right] = \frac{1}{6}, \quad \operatorname {Var}\left[\xi _i^2\right] = \mathbb {E}\left[\xi _i^4\right] - \left(\mathbb {E}\left[\xi _i^2\right]\right)^2 = \frac{1}{15} - \frac{1}{36} = \frac{12 - 5}{180} = \frac{7}{180} \]

Then

\[ D^2 = \xi _1^2 + \ldots + \xi _n^2 \approx N\left(na, n\sigma ^2\right) \]

\(a = \frac{1}{6}\), \(\sigma^2 = \frac{7}{180}\).

We have

\[ \mathbb {P}\left(D \leq t\right) = \mathbb {P}\left(D^2 \leq t^2\right) = \mathbb {P}\left(\frac{D^2 - an}{\sqrt{n}\sigma } \leq \frac{t^2 - an}{\sqrt{n}\sigma }\right) = \Phi \left(\frac{t^2 - an}{\sqrt{n}\sigma }\right) \]

Density:

\[ f\left(\frac{t^2 - an}{\sqrt{n}\sigma }\right) \cdot \frac{2}{\sqrt{n}\sigma }t \]

We have

\[ \begin{align} \mathbb {E}\left[D\right] \approx \sqrt{na} \\ \operatorname {Var}\left[D\right] = \mathbb {E}\left[D^2\right] - \mathbb {E}\left[D\right]^2 = na - \end{align} \]

3.4 Equality of mixed partial derivatives

Definition 2 Suppose \(G\) is an open subset of \(\mathbb {R}^2\) and \(f: G \rightarrow \mathbb {R}\) is a function. For \((x, y) \in G\) the partial derivatives \(\left(D_1 f\right)(x, y)\) and \(\left(D_2 f\right)(x, y)\) are defined as follows: \[ \left(D_1 f\right)(x, y)=\lim _{t \rightarrow 0} \frac{f(x+t, y)-f(x, y)}{t} \] and \[ \left(D_2 f\right)(x, y)=\lim _{t \rightarrow 0} \frac{f(x, y+t)-f(x, y)}{t} \] if these limits exist.

Using the notation for the section of a function (see 5.7), we can write the definitions of \(D_1\) and \(D_2\) as follows:

\[ \left(D_1 f\right)(x, y)=\left([f]^y\right)^{\prime }(x) \quad \text{ and } \quad \left(D_2 f\right)(x, y)=\left([f]_x\right)^{\prime }(y) . \]

ExampleExample 5

Examples of partial derivatives of \(x^y\)

SolutionSolution

Let \(G=\left\{ (x, y) \in \mathbb {R}^2: x>0\right\}\) and define \(f: G \rightarrow \mathbb {R}\) by \(f(x, y)=x^y\). Then \[ \left(D_1 f\right)(x, y)=y x^{y-1} \quad \text{ and } \quad \left(D_2 f\right)(x, y)=x^y \ln x, \] (check this yourself). Taking partial derivatives of these partial derivatives, we have \[ \left(D_2\left(D_1 f\right)\right)(x, y)=x^{y-1}+y x^{y-1} \ln x \] and \[ \left(D_1\left(D_2 f\right)\right)(x, y)=x^{y-1}+y x^{y-1} \ln x, \] (check this yourself). We obtain \(D_1\left(D_2 f\right)=D_2\left(D_1 f\right)\) as functions on \(G\).

In the example above, the two mixed partial derivatives turn out to be equal, even though the intermediate results look completely different. The following result shows that the behavior in the example above is typical rather than a coincidence.

Some proofs of the result below do not use Fubini’s theorem. However, Fubini’s theorem leads to an elegant proof below.

The integrals appearing in the proof below make sense because continuous real-valued functions on \(\mathbb {R}^2\) are measurable (since for a continuous function the preimage of every open set is open), and also because continuous real-valued functions on closed bounded subsets of \(\mathbb {R}^2\) are bounded.

Although the continuity hypotheses in the result below can be weakened somewhat, they cannot be dropped entirely, as shown in Exercise 14 of this section.

Theorem 2 (Equality of mixed partial derivatives) Suppose \(G\) is an open subset of \(\mathbb {R}^2\), and \(f: G \rightarrow \mathbb {R}\) is a function such that \(D_1 f\), \(D_2 f, D_1\left(D_2 f\right)\), and \(D_2\left(D_1 f\right)\) exist and are continuous functions on \(G\). Then \[ D_1\left(D_2 f\right)=D_2\left(D_1 f\right) \] on \(G\).

Mn-proofProof

Fix \((a, b) \in G\). For \(\delta >0\) let \(S_\delta =[a, a+\delta ] \times [b, b+\delta ]\). If \(S_\delta \subseteq G\), then

\[ \begin{aligned} \int _{S_\delta } D_1\left(D_2 f\right) d \lambda _2 & =\int _b^{b+\delta } \int _a^{a+\delta }\left(D_1\left(D_2 f\right)\right)(x, y) d x d y \\ & =\int _b^{b+\delta }\left[\left(D_2 f\right)(a+\delta , y)-\left(D_2 f\right)(a, y)\right] d y \\ & =f(a+\delta , b+\delta )-f(a+\delta , b)-f(a, b+\delta )+f(a, b) \end{aligned} \]

where the first equality follows from Fubini’s theorem, and the second and third equalities follow from the fundamental theorem of calculus.

An analogous computation of \(\int_{S_\delta } D_2\left(D_1 f\right) d \lambda_2\) gives the same result. Thus,

\[ \int _{S_\delta }\left[D_1\left(D_2 f\right)-D_2\left(D_1 f\right)\right] d \lambda _2=0 \]

for all \(\delta\) such that \(S_\delta \subseteq G\). If \(\left(D_1\left(D_2 f\right)\right)(a, b)>\left(D_2\left(D_1 f\right)\right)(a, b)\), then by continuity of \(D_1\left(D_2 f\right)\) and \(D_2\left(D_1 f\right)\) the integrand in the equation above is positive on \(S_\delta\) for sufficiently small \(\delta\), contradicting the fact that the integral above equals 0. Similarly, the inequality \(\left(D_1\left(D_2 f\right)\right)(a, b)<\left(D_2\left(D_1 f\right)\right)(a, b)\) also contradicts the equation above for small \(\delta\). We thus conclude that

\[ \left(D_1\left(D_2 f\right)\right)(a, b)=\left(D_2\left(D_1 f\right)\right)(a, b), \]

as required.

ProblemProblem 11

Define \(f: \mathbb {R}^{2} \rightarrow \mathbb {R}\) by the formula \[ f(x, y) = \begin{cases} \frac{x y\left(x^{2}-y^{2}\right)}{x^{2}+y^{2}} & \text{ if }(x, y) \neq (0,0) \\ 0 & \text{ if }(x, y) = (0,0)\end{cases} \]

1. Prove that \(D_{1}\left(D_{2} f\right)\) and \(D_{2}\left(D_{1} f\right)\) exist everywhere on \(\mathbb {R}^{2}\).

2. Show that \(\left(D_{1}\left(D_{2} f\right)\right)(0,0) \neq \left(D_{2}\left(D_{1} f\right)\right)(0,0)\).

3. Explain why the previous part does not contradict the theorem on equality of partial derivatives.

ProblemProblem 12

Let \(\mu_1\), \(\mu_2\) be measures on \((\mathbb {R}, \mathscr {B}\left(\mathbb {R}\right))\), where \(\mu_1\) is absolutely continuous with respect to Lebesgue measure with density \(f_1(x) = e^{-2x} \cdot I_{[0,+\infty )}(x)\), and \(\mu_2 = \sum_{k=1}^{+\infty }\frac{1}{3^k}\delta_k\). Let \(\mu = \mu_1 \otimes \mu_2\). Find \(\int_{\mathbb {R}^{2}} g \; d\mu\) for the following functions:

1. \(g(x,y) = x^2 + y-3\)

2. \(g(x,y) = e^{x + y}\)

Do not forget to state which theorem you are using, Fubini’s or Tonelli’s.

HintHint

Enum-item(1)

\(\sum_{k=1}^{+\infty }k \cdot p^k = \frac{p}{(1-p)^2}\) for \(\left|p\right| < 1\)

Enum-item(2)