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1. If two functions \( f, g : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) are both integrable, then \( f + g \) is also integrable, and\n\n\[ \n{\int }_{{\mathbb{R}}^{n}}\left( {f + g}\right) \left| {{d}^{n}\mathbf{x}}\right| = {\int }_{{\mathbb{R}}^{n}}f\left| {{d}^{n}\mathbf{x}}\right| + {\int }_{{\mathbb{R}}^{n}}g\le... | 1. For any subset \( A \subset {\mathbb{R}}^{n} \), we have\n\n4.1.35\n\nApplying this to each cube \( C \in {\mathcal{D}}_{N}\left( {\mathbb{R}}^{n}\right) \), we get\n\n\[ \n{U}_{N}\left( f\right) + {U}_{N}\left( g\right) \geq {U}_{N}\left( {f + g}\right) \; \geq \;{L}_{N}\left( {f + g}\right) \geq {L}_{N}\left( f\ri... | Yes |
1. A bounded function \( f \) with bounded support is integrable if and only if both \( {f}^{ + } \) and \( {f}^{ - } \) are integrable. | 1. If \( f \) is integrable, then by part 4 of Proposition 4.1.14, \( \left| f\right| \) is integrable, and so are \( {f}^{ + } = \frac{1}{2}\left( {\left| f\right| + f}\right) \) and \( {f}^{ - } = \frac{1}{2}\left( {\left| f\right| - f}\right) \), by parts 1 and 2. Now assume that \( {f}^{ + } \) and \( {f}^{ - } \) ... | Yes |
Proposition 4.1.16. If \( {f}_{1}\left( \mathbf{x}\right) \) is integrable on \( {\mathbb{R}}^{n} \) and \( {f}_{2}\left( \mathbf{y}\right) \) is integrable on \( {\mathbb{R}}^{m} \), then the function\n\n\( g\left( {\mathbf{x},\mathbf{y}}\right) = {f}_{1}\left( \mathbf{x}\right) {f}_{2}\left( \mathbf{y}\right) \; \) o... | Proof. First suppose \( {f}_{1} \) and \( {f}_{2} \) are nonnegative. For any \( {A}_{1} \subset {\mathbb{R}}^{n} \) and \( {A}_{2} \subset {\mathbb{R}}^{m} \), we have\n\n\[{M}_{{A}_{1} \times {A}_{2}}\left( g\right) = {M}_{{A}_{1}}\left( {f}_{1}\right) {M}_{{A}_{2}}\left( {f}_{2}\right) ;\;{m}_{{A}_{1} \times {A}_{2}... | Yes |
Lemma 4.1.19. An interval \( I = \left\lbrack {a, b}\right\rbrack \) has length \( \left| {b - a}\right| \) . | Proof. Of the cubes (i.e., intervals) \( C \in {\mathcal{D}}_{N}\left( \mathbb{R}\right) \), at most two contain an endpoint \( a \) or \( b \) . The others are either entirely in \( I \) or entirely outside; on those\n\n\[ \n{M}_{C}\left( {\mathbf{1}}_{I}\right) = {m}_{C}\left( {\mathbf{1}}_{I}\right) = \left\{ \begin... | No |
Proposition 4.1.20 (Volume of \( n \) -dimensional parallelogram). The \( n \) -dimensional parallelogram\n\n\[ P\overset{\text{ def }}{ = }{I}_{1} \times \cdots \times {I}_{n} \subset {\mathbb{R}}^{n} \]\n\nformed by the product of intervals \( {I}_{i} = \left\lbrack {{a}_{i},{b}_{i}}\right\rbrack \) has volume\n\n\[ ... | Proof. This follows immediately from Proposition 4.1.16, applied to\n\n\[ {\mathbf{1}}_{P}\left( \mathbf{x}\right) = {\mathbf{1}}_{{I}_{1}}\left( {x}_{1}\right) {\mathbf{1}}_{{I}_{2}}\left( {x}_{2}\right) \ldots {\mathbf{1}}_{{I}_{n}}\left( {x}_{n}\right) \] | Yes |
Theorem 4.1.21 (Sum of volumes). If two disjoint sets \( A, B \) in \( {\mathbb{R}}^{n} \) are pavable, then so is their union, and the volume of the union is the sum of the volumes: | Proof. Since \( {\mathbf{1}}_{A \cup B} = {\mathbf{1}}_{A} + {\mathbf{1}}_{B} \) if \( A \) and \( B \) are disjoint, the result follows from part 1 of Proposition 4.1.14. | Yes |
Proposition 4.1.22 (Volume invariant under translation). Let \( A \) be any pavable subset of \( {\mathbb{R}}^{n} \) and \( \overrightarrow{\mathbf{v}} \in {\mathbb{R}}^{n} \) any vector. Denote by \( A + \overrightarrow{\mathbf{v}} \) the set \( A \) translated by \( \overrightarrow{\mathbf{v}} \). Then \( A + \overri... | Proof. Let \( {K}_{N} \) be the set of cubes \( C \in {\mathcal{D}}_{N}\left( {\mathbb{R}}^{n}\right) \) with \( C \subset A \) (i.e., the set of cubes at the \( N \) th level that are entirely inside \( A \) ), and let \( {H}_{N} \) be the set of cubes \( C \in {\mathcal{D}}_{N}\left( {\mathbb{R}}^{n}\right) \) such t... | Yes |
Proposition 4.1.23 (Set with volume 0). A bounded set \( X \subset {\mathbb{R}}^{n} \) has volume 0 if and only if for every \( \epsilon > 0 \) there exists \( N \) such that | \[ \begin{array}{l} \mathop{\sum }\limits_{{C \in {\mathcal{D}}_{N}\left( {\mathbb{R}}^{n}\right) }}{\operatorname{vol}}_{n}\left( C\right) \leq \epsilon . \\ C \cap X \neq \varnothing \end{array} \] | No |
Proposition 4.1.24 (Scaling volume). If \( A \subset {\mathbb{R}}^{n} \) has volume and \( t \in \mathbb{R} \), then \( {tA} \) has volume, and\n\n\[ \n{\operatorname{vol}}_{n}\left( {tA}\right) = {\left| t\right| }^{n}{\operatorname{vol}}_{n}\left( A\right) \n\] | Proof. By Proposition 4.1.20, this is true if \( A \) is a parallelogram, in particular, a cube \( C \in {\mathcal{D}}_{N} \) . Assume \( A \) is a subset of \( {\mathbb{R}}^{n} \) whose volume is well defined. This means that \( {\mathbf{1}}_{A} \) is integrable, or, equivalently, that\n\n\[ \n\mathop{\lim }\limits_{{... | Yes |
What is the center of gravity of the right triangle \( T \) shown in Figure 4.2.1? | The area of \( T \), i.e., \( {\int }_{T}{dxdy} \), is \( {ab}/2 \). Using the techniques of Section 4.5, we can easily compute\n\n\[ \n{\int }_{T}{xdxdy} = \frac{{a}^{2}b}{6}\;\text{ and }\;{\int }_{T}{ydxdy} = \frac{a{b}^{2}}{6} \]\n\nso by equation 4.2.1, the center of gravity of \( T \) is the point \( \overline{\m... | Yes |
Example 4.2.4 (Computing \( \pi \) using Buffon’s needle). Toss a needle of length 1 on a piece of lined paper, with lines parallel to the \( x \) -axis spaced 1 apart. The sample space should be the space of positions of the needle. We will be interested in the probability that the needle intersects a line, so we will... | So the probability that the needle will intersect a line is \[ \mathbf{P}\left( A\right) = \frac{2}{\pi }{\int }_{A}\left| {d\theta ds}\right| = \frac{2}{\pi }{\int }_{0}^{\pi }\frac{1}{2}\sin {\theta d\theta } = \frac{2}{\pi }. \] Thus the probability of intersecting the line is \( 2/\pi \) . This provides a (cumberso... | Yes |
Theorem 4.2.7 (The central limit theorem). If an experiment and a random variable have expectation \( E \) and standard deviation \( \sigma \), then if the experiment is repeated \( n \) times, with average result \( \bar{x} \), the probability that \( \overline{x}\; \) is between \( \;E + \frac{\sigma }{\sqrt{n}}a\; \... | \[ \frac{1}{\sqrt{2\pi }}{\int }_{a}^{b}{e}^{-{y}^{2}/2}{dy} \] | Yes |
What is the probability that a fair coin tossed 1000 times will come up heads between 510 and 520 times? | In principle, this is straightforward: just compute the sum\n\n\[ \frac{1}{{2}^{1000}}\mathop{\sum }\limits_{{k = {510}}}^{{520}}\left( \begin{matrix} {1000} \\ k \end{matrix}\right) \]\n\nIn practice, computing these numbers would be extremely cumbersome; it is much easier to use the central limit theorem. Our individ... | Yes |
Theorem 4.3.1 (Criterion for integrability). A function \( f : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) is integrable if and only if it is bounded with bounded support, and for all \( \epsilon > 0 \), there exists \( N \) such that  sufficiently large) the sum of their volumes is less than epsilon, then the function is integrable: we can make the difference between the upper ... | No |
Theorem 4.3.6. Any continuous function \( {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) with bounded support is integrable. | Proof. Theorem 4.3.6 follows almost immediately from Theorem 1.6.11. Let \( f \) be continuous with bounded (hence compact) support. Since \( \sup \left| f\right| \) is realized on the support, \( f \) is bounded. By Theorem 1.6.11, \( f \) is uniformly continuous: choose \( \epsilon \) and find \( \delta > 0 \) such t... | Yes |
Corollary 4.3.7. Let \( X \subset {\mathbb{R}}^{n} \) be compact and let \( f : X \rightarrow \mathbb{R} \) be continuous. Then the graph \( {\Gamma }_{f} \subset {\mathbb{R}}^{n + 1} \) has volume 0. | Proof. Since \( X \) is compact, it is bounded, and there is a number \( A \) such that for all \( N \), the number of cubes \( C \in {\mathcal{D}}_{N}\left( {\mathbb{R}}^{n}\right) \) such that \( X \cap C \neq \varnothing \) is at most \( A{2}^{nN} \) . Choose \( \epsilon > 0 \), and use Theorem 1.6.11 to find \( \de... | Yes |
Corollary 4.3.8. Let \( U \subset {\mathbb{R}}^{n} \) be open and let \( f : U \rightarrow \mathbb{R} \) be a continuous function. Then any compact part \( Y \) of the graph of \( f \) has \( \left( {n + 1}\right) \) - dimensional volume 0. | Proof. The projection of \( Y \) into \( {\mathbb{R}}^{n} \) is compact, and the restriction of \( f \) to that projection satisfies the hypotheses of Corollary 4.3.7. | No |
Theorem 4.3.9. A function \( f : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \), bounded with bounded support, is integrable if it is continuous except on a set of volume 0 . | Proof. Denote by \( \Delta \) (\ | No |
Corollary 4.3.10. Let \( f : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) be integrable, and let \( g : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) be a bounded function. If \( f = g \) except on a set of volume 0, then \( g \) is integrable, and \[ {\int }_{{\mathbb{R}}^{n}}f\left| {{d}^{n}\mathbf{x}}\right| = {\int }_{{... | Proof. The support of \( g \) is bounded, since the support of \( f \) is bounded and so is the support of \( g - f \) . For any \( N \), a cube \( C \in {\mathcal{D}}_{N}\left( {\mathbb{R}}^{n}\right) \) where \( {\operatorname{osc}}_{C}\left( g\right) > \epsilon \) is either one where \( {\operatorname{osc}}_{C}\left... | Yes |
Corollary 4.3.12. Any polynomial function \( p \) can be integrated over any set \( A \) of finite volume; that is, \( p \cdot {\mathbf{1}}_{A} \) is integrable. | Proof. The function \( p \cdot {\mathbf{1}}_{A} \) meets the conditions of Theorem 4.3.9: it is bounded with bounded support and is continuous except on the boundary of \( A \), which has volume 0 . | Yes |
Example 4.4.3 (A set with measure 0, undefined volume). The set of rational numbers in the interval \( \left\lbrack {0,1}\right\rbrack \) has measure 0. | You can list them in order, for instance, as \( 1,1/2,1/3,2/3,1/4,2/4,3/4,1/5,\ldots \) (The list is infinite and includes some numbers more than once.) Center an open interval of length \( \epsilon /2 \) at 1, an open interval of length \( \epsilon /4 \) at \( 1/2 \), an open interval of length \( \epsilon /8 \) at \(... | Yes |
Theorem 4.4.4 (A countable union of sets of measure 0 has measure 0). Let \( i \mapsto {X}_{i} \) be a sequence of sets of measure 0 . Then\n\n\( {X}_{1} \cup {X}_{2} \cup \ldots \; \) is a set of measure 0 . | Proof of Theorem 4.4.4: We can turn the sequences\n\n\[ \n{B}_{1, i},\ldots ,{B}_{j, i},\ldots \n\]\n\ninto a single sequence by listing the boxes in some order, just as we did for the rational numbers in Example 4.4.3.\n\nFor instance, one could list first the boxes where \( i + j = 2 \), then those where \( i + j = 3... | No |
Example 4.4.3 (and, more generally, any countable set) corresponds to the case of Theorem 4.4.4 where each \( {X}_{i} \) is a single point. | Proof of Theorem 4.4.4. Since there exist infinite sequences \( i \mapsto {B}_{j, i} \) of boxes (one sequence for each \( j \) ) such that\n\n\[ \n{X}_{1} \subset {B}_{1,1} \cup {B}_{1,2} \cup \ldots ,\;\text{ and }\;\sum \operatorname{vol}{B}_{1, i} \leq \frac{\epsilon }{2} \n\]\n\n\[ \n{X}_{2} \subset {B}_{2,1} \cup... | Yes |
Corollary 4.4.5. Let \( {B}_{R}\left( \mathbf{0}\right) \) be the ball of radius \( R \) centered at \( \mathbf{0} \) . If for all \( R \geq 0 \) the subset \( X \subset {\mathbb{R}}^{n} \) satisfies \( {\operatorname{vol}}_{n}\left( {X \cap {B}_{R}\left( \mathbf{0}\right) }\right) = 0 \), then \( X \) has measure 0. | Proof. Since \( X = { \cup }_{m = 1}^{\infty }\left( {X \cap {B}_{m}\left( \mathbf{0}\right) }\right) \), it is a countable union of sets of volume 0 , hence measure 0 . | Yes |
Proposition 4.4.6. Let \( X \) be a subspace of \( {\mathbb{R}}^{n} \) of dimension \( k < n \) . Then \( X \) has measure 0, and any translate of \( X \) has measure 0 . | Proof. The first statement follows from Proposition 4.3.5 and Corollary 4.4.5; the second from Proposition 4.1.22. | No |
Consider the function\n\n\[ f\left( x\right) = \left\{ \begin{array}{ll} \frac{1}{q} & \text{ if }x = \frac{p}{q}\text{ is rational, written in lowest terms with }q > 0\text{ and }\left| x\right| \leq 1 \\ 0 & \text{ if }x\text{ is irrational or }\left| x\right| > 1. \end{array}\right. \]\n\nThis function is integrable... | For instance, \( f\left( {3/4}\right) = 1/4 \), but arbitrarily close to \( 3/4 \) we have irrational numbers, giving \( f\left( x\right) = 0 \). But such values form a set of measure 0. The function is continuous at the irrationals: arbitrarily close to any irrational number \( x \) you will find rational numbers \( p... | No |
Theorem 4.4.8 (What functions are integrable). Let \( f : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) be bounded with bounded support. Then \( f \) is integrable if and only if it is continuous except on a set of measure 0. | Proof. We will start with the harder direction: if \( f : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \), bounded with bounded support, is continuous almost everywhere, then it is integrable. We will use the criterion for integrability given by Theorem 4.3.1; thus we want to prove that for all \( \epsilon > 0 \) there exis... | Yes |
Lemma 4.4.9. Let \( i \mapsto {B}_{i} \) be a sequence satisfying \( \Delta \subset \cup {B}_{i} \) and \( \sum {\operatorname{vol}}_{n}{B}_{i} < \epsilon \) . Then \( {\operatorname{osc}}_{{B}_{i}}\left( f\right) > \epsilon \) on only finitely many boxes \( {B}_{i} \) . | Proof of Lemma 4.4.9. Assume the lemma is false. Then there exist an infinite subsequence of boxes \( j \mapsto {B}_{{i}_{j}} \) and two infinite sequences of points, \( j \mapsto {\mathbf{x}}_{j}, j \mapsto {\mathbf{y}}_{j} \) in \( {B}_{{i}_{j}} \), such that \( \left| {f\left( {\mathbf{x}}_{j}\right) - f\left( {\mat... | Yes |
Corollary 4.4.11. Let \( f \) and \( g \) be integrable functions on \( {\mathbb{R}}^{n} \) such that\n\n\( f \geq g \) and \( \int f\left( \mathbf{x}\right) \left| {{d}^{n}\mathbf{x}}\right| = \int g\left( \mathbf{x}\right) \left| {{d}^{n}\mathbf{x}}\right| \) . Then\n\n\[ \n\{ \mathbf{x} \mid f\left( \mathbf{x}\right... | Proof. The function \( f - g \) is integrable, hence continuous almost everywhere. Thus if \( f > g \) on a set not of measure 0, there exists \( {\mathbf{x}}_{0} \) such that \( f\left( {\mathbf{x}}_{0}\right) > g\left( {\mathbf{x}}_{0}\right) \) and \( f - g \) is continuous at \( {\mathbf{x}}_{0} \) . Then there exi... | Yes |
Corollary 4.4.12 (Product of integrable functions is integrable). If two functions \( f, g : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) are both integrable, then \( {fg} \) is also integrable. | Proof. If \( f \) is continuous except on a set \( {X}_{f} \) of measure 0, and \( g \) is continuous except on a set \( {X}_{g} \) of measure 0, then (Proposition 1.5.29) \( {fg} \) is continuous except on a subset of \( {X}_{f} \cup {X}_{g} \) . By Theorem 4.4.4, this subset has measure 0 . Thus by Theorem 4.4.8, \( ... | Yes |
Corollary 4.4.13 (Integration is translation invariant). If a function \( f : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) is integrable, then for any \( \overrightarrow{\mathbf{v}} \in {\mathbb{R}}^{n} \), the function \( \mathbf{x} \mapsto f\left( {\mathbf{x} - \overrightarrow{\mathbf{v}}}\right) \) is integrable. | Corollary 4.4.13 follows from Theorem 4.4.8. | Yes |
Now let’s integrate an unspecified function \( f : {\mathbb{R}}^{2} \rightarrow \mathbb{R} \) over the area bordered on the top by the parabolas \( y = {x}^{2} \) and \( y = {\left( x - 2\right) }^{2} \) and on the bottom by the straight lines \( y = - x \) and \( y = x - 2 \), as shown in Figure 4.5.4. | Let's start again by sweeping our pencil from left to right, which corresponds to the outer integral being with respect to \( x \) . The limits for the outer integral are clearly \( x = 0 \) and \( x = 2 \), giving\n\n\[ \n{\int }_{0}^{2}\left( {\int {fdy}}\right) {dx} \n\]\n\nAs we sweep our pencil from left to right,... | No |
Example 4.5.3 (Setting up a multiple integral in \( {\mathbb{R}}^{3} \) ). Already in \( {\mathbb{R}}^{3} \) this kind of visualization becomes much harder. Suppose we want to integrate a function over the pyramid \( P \) shown in Figure 4.5.5 and given by the formula\n\n\[ P = \left\{ {\left. {\left( \begin{array}{l} ... | There are six ways to apply Fubini's theorem, which in this case because of symmetries will result in the same expressions with the variables permuted. Let us vary \( z \) first. For instance, we can lift a piece of paper and see how it intersects the pyramid at various heights. Clearly the paper will only intersect th... | No |
In the preceding examples, we could deduce the upper and lower limits from pictures of the region over which we want to integrate. Here is a case where the picture doesn't give the answer. Let us set up a multiple integral over the ellipse \( E \) defined by \( E = \\left\\{ {\\left. \\left( \\begin{array}{l} x \\\\ y ... | The boundary of \( E \) is defined by \( {y}^{2} + {xy} + {x}^{2} - 1 = 0 \), so on the boundary, \( y = \\frac{-x \\pm \\sqrt{{x}^{2} - 4\\left( {{x}^{2} - 1}\\right) }}{2} = \\frac{-x \\pm \\sqrt{4 - 3{x}^{2}}}{2}. Such values exist only if \( 4 - 3{x}^{2} \\geq 0 \), i.e., \( \\left| x\\right| \\leq 2\\sqrt{3}/3 \).... | Yes |
Let \( f : {\mathbb{R}}^{2} \rightarrow \mathbb{R} \) be the function \( f\left( \begin{array}{l} x \\ y \end{array}\right) = {xy}{\mathbf{1}}_{S}\left( \begin{array}{l} x \\ y \end{array}\right) \), where \( S \) is the unit square, as shown in Figure 4.5.7. Then \[ {\int }_{{\mathbb{R}}^{2}}f\left( \begin{array}{l} x... | \[ = {\int }_{0}^{1}{\left\lbrack \frac{{x}^{2}y}{2}\right\rbrack }_{x = 0}^{x = 1}{dy} = {\int }_{0}^{1}\frac{y}{2}{dy} = \frac{1}{4}.\;\bigtriangleup \] | Yes |
Let us integrate the function \( {e}^{-{y}^{2}} \) over the triangle shown in Figure 4.5.8:\n\n\[ T = \left\{ {\left. {\left( \begin{array}{l} x \\ y \end{array}\right) \in {\mathbb{R}}^{2}}\right| \;0 \leq x \leq y \leq 1}\right\} . | Fubini's theorem gives us two ways to write this integral as an iterated one-dimensional integral:\n\n\[ {\int }_{0}^{1}\left( {{\int }_{x}^{1}{e}^{-{y}^{2}}{dy}}\right) {dx}\;\text{ and }\;{\int }_{0}^{1}\left( {{\int }_{0}^{y}{e}^{-{y}^{2}}{dx}}\right) {dy}. \]\n\nThe first cannot be computed in elementary terms, sin... | Yes |
Example 4.5.7 (Volume of a ball in \( {\mathbb{R}}^{n} \) ). Let \( {B}_{R}^{n}\left( \mathbf{0}\right) \) be the ball of radius \( R \) in \( {\mathbb{R}}^{n} \), centered at \( \mathbf{0} \), and let \( {b}_{n}\left( R\right) \) be its volume. By Proposition 4.1.24, \( {b}_{n}\left( R\right) = {R}^{n}{b}_{n}\left( 1\... | By Fubini's theorem,\n\n\[ \underset{\begin{matrix} \text{vol. of} \\ \text{unit ball in }{\mathbb{R}}^{n} \end{matrix}}{\underbrace{{\beta }_{n}}} = {\int }_{{B}_{1}^{n}\left( \mathbf{0}\right) }\left| {{d}^{n}\mathbf{x}}\right| = {\int }_{-1}^{1}\overset{1}{\overbrace{\left( {\int }_{{B}_{\sqrt{1 - {x}_{n}^{2}}}^{n -... | Yes |
Exercise 4.5.10 asks you to show that if two points \( \\mathbf{x},\\mathbf{y} \) are chosen in the unit square, without privileging any part of the square, then \( E\\left( {\\left| \\mathbf{x} - \\mathbf{y}\\right| }^{2}\\right) = 1/3 \) | \[ E\\left( {\\left( x - y\\right) }^{2}\\right) = {\\int }_{0}^{1}{\\int }_{0}^{1}{\\left( x - y\\right) }^{2}{dxdy} = {\\int }_{0}^{1}\\left( {{\\int }_{0}^{1}{x}^{2} - {2xy} + {y}^{2}{dx}}\\right) {dy} \]\n\n\[ = {\\int }_{0}^{1}{\\left\\lbrack \\frac{{x}^{3}}{3} - {x}^{2}y + {y}^{2}x\\right\\rbrack }_{0}^{1}{dy} = ... | No |
Theorem 4.5.10 (Fubini’s theorem). Let \( f : {\mathbb{R}}^{n} \times {\mathbb{R}}^{m} \rightarrow \mathbb{R} \) be an integrable function, and suppose that for each \( \mathbf{x} \in {\mathbb{R}}^{n} \), the function \( \mathbf{y} \mapsto f\left( {\mathbf{x},\mathbf{y}}\right) \) is integrable. Then the function | \[ \mathbf{x} \mapsto {\int }_{{\mathbb{R}}^{m}}f\left( {\mathbf{x},\mathbf{y}}\right) \left| {{d}^{m}\mathbf{y}}\right| \] is integrable, and \[ {\int }_{{\mathbb{R}}^{n + m}}f\left( {\mathbf{x},\mathbf{y}}\right) \left| {{d}^{n}\mathbf{x}}\right| \left| {{d}^{m}\mathbf{y}}\right| = {\int }_{{\mathbb{R}}^{n}}\left( {{... | Yes |
Proposition 4.6.4 (Product rules). If \( {f}_{1},\ldots ,{f}_{n} \) are functions that are integrated exactly by an integration rule, i.e., \[ {\int }_{a}^{b}{f}_{j}\left( x\right) {dx} = \mathop{\sum }\limits_{i}{w}_{i}{f}_{j}\left( {p}_{i}\right) \;\text{ for }j = 1,\ldots, n \] then the product \[ f\left( \mathbf{x}... | Proof. This follows immediately from Proposition 4.1.16. Indeed, \[ {\int }_{{\left\lbrack a, b\right\rbrack }^{n}}f\left( \mathbf{x}\right) \left| {{d}^{n}\mathbf{x}}\right| = \left( {{\int }_{a}^{b}{f}_{1}\left( {x}_{1}\right) d{x}_{1}}\right) \ldots \left( {{\int }_{a}^{b}{f}_{n}\left( {x}_{n}\right) d{x}_{n}}\right... | Yes |
Each weight for the two-dimensional Simpson's rule is the product of two one-dimensional Simpson weights from Definition 4.6.1. In the very simple case where we wish to integrate over a square, dividing it into only four subsquares, and sampling the function at each vertex, we have nine samples in all. | Let us do this with the square of sidelength \( b - a = 2 \) . Since \( n = 1 \) (this square corresponds, in the one-dimensional case, to the first piece in Figure 4.6.2), the one-dimensional weights are\n\n\[ \n{w}_{1} = {w}_{3} = \frac{b - a}{6n} \cdot 1 = \frac{1}{3}\;\text{ and }\;{w}_{2} = \frac{b - a}{6n} \cdot ... | Yes |
In Example 4.5.9 we computed the expected value for the determinant of a \( 2 \times 2 \) matrix. Now let try the same for \( 3 \times 3 \) matrices, using MATHEMATICA’s Montecarlo program to approximate\n\n\[{\int }_{C}\left| {\det A}\right| \left| {{d}^{9}\mathbf{x}}\right|\]\n\n4.6.25\n\ni.e., to evaluate the averag... | Four runs of length 50000 gave the following estimates of the integral:\n\n\[{.13453},{.13519},{.13409},{.13456}.\]\n\n4.6.26\n\nHow accurate are these guesses? We might expect the first two digits to be correct, and the third to be off by 1 or 2 . What does the theory say?\n\nThe same program calculates the standard d... | Yes |
Example 4.8.2 (The function \( {\Delta }_{3} \) ). If\n\n\[ A = \left\lbrack \begin{array}{lll} 1 & 3 & 4 \\ 0 & 1 & 1 \\ 1 & 2 & 0 \end{array}\right\rbrack \text{, then}{A}_{\left\lbrack 2,1\right\rbrack } = \left\lbrack \begin{array}{lll} 1 & 3 & 4 \\ 0 & 1 & 1 \\ 1 & 2 & 0 \end{array}\right\rbrack = \left\lbrack \be... | Equation 4.8.10: The first line is equation 4.8.5 ; the second is the inductive assumption. and equation 4.8.5 corresponds to\n\n\[ {\Delta }_{3}\left( A\right) = 1\underset{i = 1}{\underbrace{{\Delta }_{2}\left( \left\lbrack \begin{array}{ll} 1 & 1 \\ 2 & 0 \end{array}\right\rbrack \right) }} - 0\underset{i = 2}{\unde... | Yes |
Theorem 4.8.4. If \( A \) and \( B \) are \( n \times n \) matrices, then | Proof. The serious case is the one in which \( A \) is invertible. If \( A \) is invertible, consider the function\n\n\[ f\left( B\right) = \frac{\det \left( {AB}\right) }{\det A}.\n\nAs you can readily check (Exercise 4.8.8), it has properties 1, 2, and 3, which characterize the determinant function. Since the determi... | No |
Corollary 4.8.5. If a matrix \( A \) is invertible, then\n\n\[ \det {A}^{-1} = \frac{1}{\det A} \] | Proof. Just compute: \( \det A\det {A}^{-1} = \det \left( {A{A}^{-1}}\right) = \det I = 1 \) . | Yes |
Theorem 4.8.8. For any \( n \times n \) matrix \( A \) , \[ \det A = \det {A}^{\top }\text{.} \] | Proof. First we will see that the theorem is true for elementary matrices. \[ \det {E}_{3} = - \det I = - 1 \] The equation \( D\left( {A}_{2}\right) = {\mu D}\left( {A}_{1}\right) \) in the proof of uniqueness can be rewritten All type 1 matrices are diagonal as \( \det E = \mu \det I = \mu \), where \( E \) is an ele... | Yes |
Theorem 4.8.9 (Determinant of triangular matrix). If a matrix is triangular, then its determinant is the product of the entries along the diagonal. | Proof. We will prove the result for upper triangular matrices; the result for lower triangular matrices then follows from Theorem 4.8.8. The proof is by induction. Theorem 4.8.9 is clearly true for a \( 1 \times 1 \) triangular matrix (note that any \( 1 \times 1 \) matrix is triangular). If \( A \) is triangular of si... | Yes |
Example 4.8.13 (Computing the determinant by permutations). Let \( n = 3 \), and let \( A \) be the matrix \( A = \left\lbrack \begin{array}{lll} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array}\right\rbrack \) . There are six possible permutations of \( n = 3 \) numbers, so we have the following, where the labeling of ... | So \( \det A = {45} + {84} + {96} - {48} - {72} - {105} = 0 \) . Can you see why this determinant had to be \( 0{?}^{10}\bigtriangleup \) | Yes |
Example 4.8.16 (Derivative of the determinant of \( 2 \times 2 \) matrices). The determinant function of \( 2 \times 2 \) matrices can be considered as the function det : \( {\mathbb{R}}^{4} \rightarrow \mathbb{R} \) given by\n\n\( \det \left( \begin{array}{l} a \\ b \\ c \\ d \end{array}\right) = {ad} - {bc},\; \) wit... | Proof. 1. By Theorem 4.8.12, the determinant is a polynomial in the entries of the matrix, hence certainly differentiable. (For instance, the formula \( {ad} - {bc} \) is a polynomial in the variables \( a, b, c, d \) .)\n\n2. Since (Proposition 1.7.14) \( \left\lbrack {\mathbf{D}\det \left( I\right) }\right\rbrack B \... | No |
Corollary 4.8.17. If \( P \) is invertible, then for any matrix \( A \) we have\n\n\[ \operatorname{tr}\left( {{P}^{-1}{AP}}\right) = \operatorname{tr}A \] | Proof. This uses Theorem 4.8.15 and Theorem 4.8.6 (basis independence of the determinant): | No |
Theorem 4.8.15\n\n\\[ \n\\operatorname{tr}\\left( {{P}^{-1}{AP}}\\right) \\;\\text{ 和 }\\; \\triangleq \\;\\left\\lbrack {\\mathbf{D}\\det \\left( I\\right) }\\right\\rbrack \\left( {{P}^{-1}{AP}}\\right) \n\\]\n | \n\\[ \n= \\;\\mathop{\\lim }\\limits_{{h \\rightarrow 0}}\\frac{\\det \\left( {I + h{P}^{-1}{AP}}\\right) - \\det I}{h} \n\\]\n\n\\[ \n= \\;\\mathop{\\lim }\\limits_{{h \\rightarrow 0}}\\frac{\\det \\left( {{P}^{-1}\\left( {P + {hAP}}\\right) }\\right) - \\det I}{h} \n\\]\n\n\\[ \n= \\;\\mathop{\\lim }\\limits_{{h \\r... | Yes |
If \( A = \left\lbrack \begin{array}{ll} 0 & 1 \\ 1 & 1 \end{array}\right\rbrack \), then | \n\[
{\chi }_{A}\left( t\right) = \det \left( {\left\lbrack \begin{array}{ll} t & 0 \\ 0 & t \end{array}\right\rbrack - \left\lbrack \begin{array}{ll} 0 & 1 \\ 1 & 1 \end{array}\right\rbrack }\right) = \det \left\lbrack \begin{matrix} t & - 1 \\ - 1 & t - 1 \end{matrix}\right\rbrack = {t}^{2} - t - 1.\;\bigtriangleup
\... | Yes |
Theorem 4.8.20. Let \( A \) be a square matrix. The eigenvalues of \( A \) are the roots of \( {\chi }_{A} \) . | Proof. If \( \lambda \) is a root of \( {\chi }_{A} \), then \( \det \left( {{\lambda I} - A}\right) = 0 \), so (by Theorem 4.8.3 and the dimension formula) \( \ker \left( {{\lambda I} - A}\right) \neq \{ \overline{\mathbf{0}}\} \) . If \( \overline{\mathbf{v}} \in \ker \left( {{\lambda I} - A}\right) \) is a nonzero v... | Yes |
Corollary 4.8.21. If \( A \) is a triangular matrix, the diagonal entries of \( A \) are the eigenvalues of \( A \), each appearing as many times as its multiplicity as a root of \( {\chi }_{A} \) . | The proof is the object of Exercise 4.8.12. | No |
Corollary 4.8.22. If the roots of \( {\chi }_{A} \) are simple, then \( {\mathbb{C}}^{n} \) admits an eigenbasis for \( A \) . | Proof. If the roots are simple, there are \( n \) of them, and the corresponding eigenvectors are linearly independent by Theorem 2.7.7, providing an eigenbasis. | Yes |
Theorem 4.8.23. For any vector \( \overrightarrow{\mathbf{w}} \), the polynomial \( p \) divides \( {\chi }_{A} \) . | The roots of \( p \) are eigenvalues, so they are also roots of \( {\chi }_{A} \) . The problem is that we need to show that a root of \( p \) cannot have higher multiplicity than it does as a root of \( {\chi }_{A} \) . This requires three intermediate statements, all of great interest in their own right. | No |
Proposition 4.8.24. If \( A \) is an \( n \times n \) complex matrix, there exists an invertible matrix \( P \) such that \( {P}^{-1}{AP} \) is upper triangular. Equivalently, there is a basis \( {\overrightarrow{\mathbf{v}}}_{1},\ldots ,{\overrightarrow{\mathbf{v}}}_{n} \) such that the matrix of \( A \) in that basis... | Proof. The proof is by induction on \( n \) . It is obvious if \( n = 1 \), so suppose \( n \geq 2 \) and assume the result for all \( \left( {n - 1}\right) \times \left( {n - 1}\right) \) matrices.\n\nFind an eigenvector \( {\overrightarrow{\mathbf{v}}}_{1} \) with eigenvalue \( {\lambda }_{1} \) (which exists by the ... | Yes |
Theorem 4.8.26. All square matrices are in the closure of the diagonalizable ones: for every complex square matrix \( A \) there is a sequence of complex diagonalizable matrices \( {A}_{i} \) that converges to \( A \) . | Proof. Suppose that \( B\overset{\text{ def }}{ = }{P}^{-1}{AP} \) is upper triangular, with entries \( {\lambda }_{1},\ldots ,{\lambda }_{n} \) on the diagonal. Choose sequences \( {\lambda }_{i, m} \) such that for all \( m \) the numbers \( {\lambda }_{1, m},\ldots ,{\lambda }_{n, m} \) are distinct, and such that \... | Yes |
Theorem 4.8.27 (The Cayley-Hamilton theorem). If \( A \) is any square matrix, then \( {\chi }_{A}\left( A\right) = \left\lbrack 0\right\rbrack \) . | Algebraic proofs of this theorem are quite difficult, but with Theorem 4.8.26 it is easy, and its proof (with hints) is the object of Exercise 4.8.14. | No |
Corollary 4.9.2. If \( S \) is an orthogonal matrix, then\n\n\[{\operatorname{vol}}_{n}S\left( A\right) = {\operatorname{vol}}_{n}A\] | Proof. Since \( {S}^{\top }S = I \) (margin note next to Definition 2.4.15) we have \( \left( {\det S}\right) \left( {\det {S}^{\top }}\right) = {\left( \det S\right) }^{2} = 1 \), so \( \left| {\det S}\right| = 1 \) . | Yes |
Lemma 4.9.6. The sequence of pavings \( N \mapsto T\left( {{\mathcal{D}}_{N}\left( {\mathbb{R}}^{n}\right) }\right) \) is a nested partition. | Proof of Lemma 4.9.6. We must check the two conditions of Definition 4.7.3 of a nested partition. The first condition is that small paving pieces must fit inside big paving pieces: if we pave \( {\mathbb{R}}^{n} \) with blocks \( T\left( C\right) \), then if\n\n\[ {C}_{1} \in {\mathcal{D}}_{{N}_{1}}\left( {\mathbb{R}}^... | Yes |
Lemma 4.9.7. If \( S, T : {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{n} \) are linear transformations, then\n\n\[ \n{\operatorname{vol}}_{n}\left( {S \circ T}\right) \left( Q\right) = {\operatorname{vol}}_{n}S\left( Q\right) {\operatorname{vol}}_{n}T\left( Q\right) \n\] | Proof of Lemma 4.9.7. This follows from equation 4.9.20, substituting \( S \) for \( T \) and \( T\left( Q\right) \) for \( A \) :\n\n\[ \n{\operatorname{vol}}_{n}\left( {S \circ T}\right) \left( Q\right) = {\operatorname{vol}}_{n}S\left( {T\left( Q\right) }\right) = {\operatorname{vol}}_{n}S\left( Q\right) {\operatorn... | Yes |
Theorem 4.9.8 (Linear change of variables). Let \( T : {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{n} \) be an invertible linear transformation, and \( f : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) an integrable function. Then \( f \circ T \) is integrable, and | \[ {\int }_{{\mathbb{R}}^{n}}f\left( \mathbf{y}\right) \left| {{d}^{n}\mathbf{y}}\right| = \underset{\begin{matrix} \text{ corrects for } \\ \text{ stretching by }T \end{matrix}}{\underbrace{\left| \det T\right| }}{\int }_{{\mathbb{R}}^{n}}\underset{f\left( \mathbf{y}\right) }{\underbrace{f\left( {T\left( \mathbf{x}\ri... | Yes |
Example 4.9.9 (Linear change of variables). The linear transformation given by \( T = \left\lbrack \begin{array}{ll} a & 0 \\ 0 & b \end{array}\right\rbrack \) transforms the unit disc into an ellipse, as shown in Figure 4.9.6. The area of the ellipse is then given by | Area of ellipse \( = {\int }_{\text{ellipse }}\left| {{d}^{2}\mathbf{y}}\right| = \underset{ab}{\underbrace{\left| \det \left\lbrack \begin{array}{ll} a & 0 \\ 0 & b \end{array}\right\rbrack \right| }}\underset{\pi = \text{area of unit disc }}{\underbrace{{\int }_{\text{disc }}\left| {{d}^{2}\mathbf{x}}\right| }} = \le... | Yes |
To compute\n\n\[ \n{\int }_{0}^{\pi }\sin x{e}^{\cos x}{dx} \n\] | traditionally, one sets \( u = \cos x \), so that \( {du} = - \sin {xdx} \) . Then for \( x = 0 \) , we have \( u = \cos 0 = 1 \), and for \( x = \pi \), we have \( u = \cos \pi = - 1 \), so\n\n\[ \n{\int }_{0}^{\pi }\sin x{e}^{\cos x}{dx} = {\int }_{1}^{-1} - {e}^{u}{du} = {\int }_{-1}^{1}{e}^{u}{du} = e - \frac{1}{e}... | Yes |
Consider the paraboloid of Figure 4.10.2, given by\n\n\\[ \nz = f\\left( \\begin{array}{l} x \\\\ y \\end{array}\\right) = \\left\\{ \\begin{array}{ll} {x}^{2} + {y}^{2} & \\text{ if }{x}^{2} + {y}^{2} \\leq {R}^{2} \\\\ 0 & \\text{ if }{x}^{2} + {y}^{2} > {R}^{2} \\end{array}\\right.\n\\]\n | Usually one would write the integral\n\n\\[ \n{\\int }_{{\\mathbb{R}}^{2}}f\\left( \\begin{array}{l} x \\\\ y \\end{array}\\right) \\left| {dxdy}\\right| \\;\\text{ as }\\;{\\int }_{{D}_{R}}\\left( {{x}^{2} + {y}^{2}}\\right) \\left| {dxdy}\\right|\n\\]\n\nwhere\n\n\\[ \n{D}_{R} = \\left\\{ {\\left. {\\left( \\begin{ar... | No |
The lemniscate looks like a figure eight; the name comes from the Latin word for ribbon. We will compute the area of the right lobe \( A \) of the lemniscate given by the equation | \[ = {\int }_{-\pi /4}^{\pi /4}\frac{\cos {2\theta }}{2}{d\theta } = {\left\lbrack \frac{\sin {2\theta }}{4}\right\rbrack }_{-\pi /4}^{\pi /4} = \frac{1}{2}. \] | Yes |
Proposition 4.10.7 (Change of variables for spherical coordinates). Let \( f \) be an integrable function defined on \( {\mathbb{R}}^{3} \), and let the spherical coordinate map \( S \) map a region \( B \) of \( \left( {r,\theta ,\varphi }\right) \) -space to a region \( A \) in \( \left( {x, y, z}\right) \) -space. S... | The \( {r}^{2}\cos \varphi \) corrects for distortion induced by the spherical coordinates map. Again, we postpone the justification for this formula. | No |
Let's integrate the function \( z \) over the upper half of the unit ball, denoted \( A \): | \[ {\int }_{A}z\left| {dxdydz}\right| \text{becomes} \]\n\[ {\int }_{B}\underset{z}{\underbrace{\left( r\sin \varphi \right) }}\left( {{r}^{2}\cos \varphi }\right) \left| {drd\theta d\varphi }\right| = {\int }_{0}^{1}\left( {{\int }_{0}^{\pi /2}\left( {{\int }_{0}^{2\pi }{r}^{3}\sin \varphi \cos {\varphi d\theta }}\rig... | Yes |
Proposition 4.10.10 (Change of variables for cylindrical coordinates). Let \( f \) be an integrable function defined on \( {\mathbb{R}}^{3} \), and suppose that the cylindrical coordinate map \( C \) maps a region \( B \subset \left( {0,\infty }\right) \times \lbrack 0,{2\pi }) \times \mathbb{R} \) of \( \left( {r,\the... | \[ {\int }_{A}f\left( \begin{array}{l} x \\ y \\ z \end{array}\right) \left| {dxdydz}\right| = {\int }_{B}f\left( \begin{matrix} r\cos \theta \\ r\sin \theta \\ z \end{matrix}\right) r\left| {drd\theta dz}\right| \] | Yes |
Let us integrate \( \left( {{x}^{2} + {y}^{2}}\right) z \) over the region \( A \subset {\mathbb{R}}^{3} \) that is the part of the cone \( {z}^{2} \geq {x}^{2} + {y}^{2} \) where \( 0 \leq z \leq 1 \) (see Figure 4.10.6). | \[ {\int }_{A}({x}^{2} + {y}^{2})z\;|{dx}\;{dy}\;{dz}| = {\int }_{B}{r}^{2}z(\underset{ = 1}{\underbrace{{\cos }^{2}\theta + {\sin }^{2}\theta }})\;r\;|{dr}\;{d\theta }\;{dz}| = {\int }_{B}({r}^{2}z)\;r\;|{dr}\;{d\theta }\;{dz}| \] \[ = {\int }_{0}^{2\pi }\left( {{\int }_{0}^{1}\left( {{\int }_{r}^{1}{r}^{3}{zdz}}\righ... | Yes |
Theorem 4.10.12 (Change of variables formula). Let \( X \) be a compact subset of \( {\mathbb{R}}^{n} \) with boundary \( \partial X \) of volume 0 ; let \( U \subset {\mathbb{R}}^{n} \) be an open set containing \( X \) . Let \( \Phi : U \rightarrow {\mathbb{R}}^{n} \) be a \( {C}^{1} \) mapping that is injective on \... | The theorem is proved in Appendix A19. | Yes |
Consider the ratio of equation 4.10.29 in the case of polar coordinates, when \( \Phi = P \) . If a rectangle \( C \) in the \( \left( {r,\theta }\right) \) plane, containing the point \( \left( \begin{matrix} {r}_{0} \\ {\theta }_{0} \end{matrix}\right) \), has sides of length \( {\Delta r} \) and \( {\Delta \theta } ... | \[ {\int }_{Y}f\left| {{d}^{n}\mathbf{y}}\right| = {\int }_{X}\left( {f \circ P}\right) r\left| {drd\theta }\right| \] where \( r \) is the ratio of the volumes of infinitesimal paving blocks. | Yes |
Suppose you wish to find the area of the region \( X \subset {\mathbb{R}}^{2} \) given by\n\n\[ 1 \leq {xy} \leq 2\text{ and }{x}^{2} \leq y \leq 2{x}^{2}. \]\n\nWhat change of variables is appropriate? | First we draw the hyperbolas given by the equalities \( 1 = {xy} \) and \( {xy} = 2 \), and the parabolas given by \( {x}^{2} = y \) and \( y = 2{x}^{2} \), as shown in Figure 4.10.9. This figure suggests that setting \( u = {xy} \) would be one good choice; the equation \( {x}^{2} \leq y \leq 2{x}^{2} \) suggests that... | No |
Example 4.10.19 (A less standard change of variables). The region \( T \) defined by\n\n\[ \n{\\left( \\frac{x}{1 - z}\\right) }^{2} + {\\left( \\frac{y}{1 + z}\\right) }^{2} \\leq 1,\\; - 1 < z < 1 \n\] | looks like the curvy-sided tetrahedron pictured in Figure 4.10.10; we will compute its volume. Notice that horizontal slices of \( T \) are ellipses, so we will use \ | No |
Theorem 4.11.2 (Convergence for Riemann integrals). Let \( k \mapsto {f}_{k} \) be a sequence of integrable functions \( {\mathbb{R}}^{n} \rightarrow \mathbb{R} \), all with support in a fixed ball \( B \subset {\mathbb{R}}^{n} \), and converging uniformly to a function \( f \) . Then \( f \) is integrable, and \[ \mat... | Proof. Choose \( \epsilon > 0 \) and \( K \) so large that \( \mathop{\sup }\limits_{{\mathbf{x} \in {\mathbb{R}}^{n}}}\left| {f\left( \mathbf{x}\right) - {f}_{k}\left( \mathbf{x}\right) }\right| < \epsilon \) when \( k > K \) . Then when \( k > K \), we have, for any \( N \), \[ {L}_{N}\left( f\right) > {L}_{N}\left( ... | Yes |
the mass of the integral is contained in a square 1 high and 1 wide. As \( k \rightarrow \infty \) this mass drifts off to infinity and gets lost: | \n\[ \mathop{\lim }\limits_{{k \rightarrow \infty }}{\int }_{0}^{\infty }{f}_{k}\left( x\right) {dx} = 1,\;\text{ but }{\int }_{0}^{\infty }\mathop{\lim }\limits_{{k \rightarrow \infty }}{f}_{k}\left( x\right) {dx} = {\int }_{0}^{\infty }{0dx} = 0. \] | Yes |
Theorem 4.11.4 (Dominated convergence for Riemann integrals). Let \( {f}_{k} : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) be a sequence of \( R \) -integrable functions. Suppose there exists \( R \) such that all \( {f}_{k} \) have their support in \( {B}_{R}\left( \mathbf{0}\right) \) and satisfy \( \left| {f}_{k}\rig... | \[ \mathop{\lim }\limits_{{k \rightarrow \infty }}{\int }_{{\mathbb{R}}^{n}}{f}_{k}\left( \mathbf{x}\right) \left| {{d}^{n}\mathbf{x}}\right| = {\int }_{{\mathbb{R}}^{n}}f\left( \mathbf{x}\right) \left| {{d}^{n}\mathbf{x}}\right| \] | Yes |
Theorem 4.11.7. Let \( k \mapsto {f}_{k}, k \mapsto {g}_{k} \) be two sequences of \( R \) -integrable functions such that\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{\infty }{\int }_{{\mathbb{R}}^{n}}\left| {{f}_{k}\left( \mathbf{x}\right) }\right| \left| {{d}^{n}\mathbf{x}}\right| < \infty ,\;\mathop{\sum }\limits_{{k = 1... | Proof of Theorem 4.11.7. Set \( {h}_{k} = {f}_{k} - {g}_{k} \), and \( {H}_{l} = \mathop{\sum }\limits_{{k = 1}}^{l}{h}_{k} \) . To prove equation 4.11.18 we need to show that\n\n\[ \mathop{\lim }\limits_{{l \rightarrow \infty }}{\int }_{{\mathbb{R}}^{n}}{H}_{l}\left( \mathbf{x}\right) \left| {{d}^{n}\mathbf{x}}\right|... | Yes |
Proposition 4.11.9. If \( f \) is R-integrable, then it is L-integrable, and its Lebesgue integral equals its Riemann integral. | Proof. Just take \( {f}_{1} = f \), and set \( {f}_{k} = 0 \) for \( k = 2,3,\ldots \) Clearly \( \sum {f}_{k} = f \) everywhere, and inequality 4.11.19 is satisfied:\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{\infty }\int \left| {{f}_{k}\left( \mathbf{x}\right) }\right| \left| {{d}^{n}\mathbf{x}}\right| = \int \left| {{f}... | Yes |
Example 4.11.10 (L-integrable function that is not R-integrable). Let \( f \) be the indicator function of the rationals, i.e., \[ f\left( x\right) = \left\{ \begin{array}{ll} 1 & \text{ if }x \in \mathbb{Q} \\ 0 & \text{ otherwise. } \end{array}\right. \] | This function equals 0 almost everywhere, so it is Lebesgue integrable with integral \( 0.\;\bigtriangleup \) | Yes |
Example 4.11.12 (An unbounded, integrable function). The function \( f\left( x\right) = {\mathbf{1}}_{\left\lbrack 0,1\right\rbrack }\left( x\right) \ln x \) is L-integrable, even though it isn’t bounded. | As shown in Figure 4.11.4, it can be written as a sum of bounded functions:\n\n\[ f\left( x\right) \underset{L}{ = }\mathop{\sum }\limits_{{i = 0}}^{\infty }{f}_{i}\left( x\right) ,\;\text{ where }{f}_{i} = \left( {{\mathbf{1}}_{\left( {2}^{-\left( {i + 1}\right) },{2}^{-i}\right\rbrack }\left( x\right) }\right) \ln x.... | Yes |
Example 4.11.13 (A function that is not Lebesgue integrable). Some improper one-dimensional integrals do not correspond to Lebesgue-integrable functions: integrals whose existence depends on cancellations, like\n\n\[ \n{\int }_{0}^{\infty }\frac{\sin x}{x}{dx} \n\] | As you may recall from one-variable calculus, this improper integral is defined to be\n\n\[ \n\mathop{\lim }\limits_{{A \rightarrow \infty }}{\int }_{0}^{A}\frac{\sin x}{x}{dx} \n\]\n\nand we can show that the limit exists, for instance, by saying that the series\n\n\[ \n\mathop{\sum }\limits_{{k = 0}}^{\infty }{\int }... | Yes |
Proposition 4.11.14 (The Lebesgue integral is linear). If \( f \) and \( g \) are L-integrable and \( a, b \) are constants, then \( {af} + {bg} \) is L-integrable and\n\n\[{\int }_{{\mathbb{R}}^{n}}\left( {{af} + {bg}}\right) \left( \mathbf{x}\right) \left| {{d}^{n}\mathbf{x}}\right| = a{\int }_{{\mathbb{R}}^{n}}f\lef... | Proof. If \( f = \sum {f}_{k} \) and \( g = \sum {g}_{k} \), then \( {af} + {bg} = \sum \left( {a{f}_{k} + b{g}_{k}}\right) \) . Indeed, the series \( \sum \left( {a{f}_{k} + b{g}_{k}}\right) \) will converge except on the union of the sets where the series \( \mathop{\sum }\limits_{k}\left| {f\left( \mathbf{x}\right) ... | Yes |
Proposition 4.11.15. If \( f \) is L-integrable on \( {\mathbb{R}}^{n} \), and \( g \) is \( R \) -integrable on \( {\mathbb{R}}^{n} \), then \( {fg} \) is L-integrable. | Proof. Since \( f \) is L-integrable, we can set \( f = \mathop{\sum }\limits_{k}{f}_{k} \), where the functions \( {f}_{k} \) are R-integrable and\n\n\[ \mathop{\sum }\limits_{k}{\int }_{{\mathbb{R}}^{n}}\left| {{f}_{k}\left( \mathbf{x}\right) }\right| \begin{Vmatrix}{{d}^{n}\mathbf{x}}\end{Vmatrix} < \infty \]\n\nWe ... | Yes |
Proposition 4.11.16. If \( f \) and \( g \) are \( L \) -integrable and \( f\underset{L}{ \leq }g \), then \[ {\int }_{{\mathbb{R}}^{n}}f\left( \mathbf{x}\right) \left| {{d}^{n}\mathbf{x}}\right| \leq {\int }_{{\mathbb{R}}^{n}}g\left( \mathbf{x}\right) \left| {{d}^{n}\mathbf{x}}\right| \] | Proof. By Proposition 4.11.14, the statement is equivalent to saying that \( 0 \leq g - f \) implies \( 0 \leq {\int }_{{\mathbb{R}}^{n}}\left( {g - f}\right) \left( \mathbf{x}\right) \left| {{d}^{n}\mathbf{x}}\right| \) . Replace \( g - f \) by \( f \) ; the statement becomes \( 0 \leq f \Rightarrow 0 \leq {\int }_{{\... | Yes |
Theorem 4.11.18. (Monotone convergence theorem: Lebesgue integrals). 1. Let \( 0 \leq {f}_{1}\underset{L}{ \leq }{f}_{2}\underset{L}{ \leq }\cdots \) be a sequence of L-integrable nonnegative functions. If \[ \mathop{\sup }\limits_{k}{\int }_{{\mathbb{R}}^{n}}{f}_{k}\left( \mathbf{x}\right) \left| {{d}^{n}\mathbf{x}}\r... | Proof. 1. Apply Theorem 4.11.17 to the sup, rewritten as the series \[ f = {f}_{1} + \left( {{f}_{2} - {f}_{1}}\right) + \cdots + \left( {{f}_{k} - {f}_{k - 1}}\right) + \cdots = {g}_{1} + {g}_{2} + \cdots + {g}_{k} + \cdots . \] 2. If \( f \) were L-integrable, then by Proposition 4.11.16 and \( f \geq {f}_{k} \) , \[... | Yes |
Theorem 4.11.19 (Dominated convergence theorem: Lebesgue integrals). Let \( k \mapsto {f}_{k} \) be a sequence of L-integrable functions that converges pointwise almost everywhere to some function \( f \) . Suppose there is an L-integrable function \( F : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) such that \( \left| {... | \[ {\int }_{{\mathbb{R}}^{n}}f\left( \mathbf{x}\right) \left| {{d}^{n}\mathbf{x}}\right| = \mathop{\lim }\limits_{{k \rightarrow \infty }}{\int }_{{\mathbb{R}}^{n}}{f}_{k}\left( \mathbf{x}\right) \left| {{d}^{n}\mathbf{x}}\right| \] | Yes |
Theorem 4.11.22 (Differentiating under the integral sign). Let \( f\left( {t,\mathbf{x}}\right) : {\mathbb{R}}^{n + 1} \rightarrow \mathbb{R} \) be a function such that for each fixed \( t \), the integral\n\n\[ F\left( t\right) = {\int }_{{\mathbb{R}}^{n}}f\left( {t,\mathbf{x}}\right) \left| {{d}^{n}\mathbf{x}}\right|... | Proof. Just compute\n\n\[ {F}^{\prime }\left( t\right) = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{F\left( {t + h}\right) - F\left( t\right) }{h} = \mathop{\lim }\limits_{{h \rightarrow 0}}{\int }_{{\mathbb{R}}^{n}}\frac{f\left( {t + h,\mathbf{x}}\right) - f\left( {t,\mathbf{x}}\right) }{h}\left| {{d}^{n}\mathbf{x... | Yes |
Let\n\n\\[ \n{f}_{a}\left( x\right) = {e}^{-a{x}^{2}},\;\text{ so that }\;{\widehat{f}}_{a}\left( \xi \right) = {\int }_{\mathbb{R}}{e}^{-a{x}^{2}}{e}^{2\pi i\xi x}{dx}. \n\\] | We can't compute this Fourier transform directly, but equation 4.11.85 (or Theorem 4.11.22) gives\n\n\\[ \n{\widehat{{f}_{a}}}^{\prime }\left( \xi \right) = {\pi i}{\int }_{-\infty }^{\infty }{e}^{2\pi ix\xi }\left( {{2x}{e}^{-a{x}^{2}}}\right) {dx}. \n\\]\n\nThis can be integrated by parts (justified by Exercise 4.11.... | Yes |
The Fourier transform of both sides of the differential equation\n\n\\[ \n{a}_{p}{D}^{p}f + \cdots + {a}_{0}f = g \n\\] | is\n\n\\[ \n\\underset{\\text{product of }\\widehat{f}\\text{ and a polynomial }}{\\underbrace{\\left( {{a}_{p}{\\left( -2\\pi i\\xi \\right) }^{p} + {a}_{p - 1}{\\left( -2\\pi i\\xi \\right) }^{p - 1} + \\cdots + {a}_{0}}\\right) \\widehat{f}}} = \\widehat{g}, \n\\]\n\nwhich gives\n\n\\[ \n\\widehat{f} = \\frac{\\wide... | Yes |
Proposition 5.1.1 (Volume of a \( k \) -parallelogram in \( {\mathbb{R}}^{k} \) ). Let \( {\overrightarrow{\mathbf{v}}}_{1},\ldots ,{\overrightarrow{\mathbf{v}}}_{k} \) be \( k \) vectors in \( {\mathbb{R}}^{k} \), so that \( T = \left\lbrack {{\overrightarrow{\mathbf{v}}}_{1},\ldots ,{\overrightarrow{\mathbf{v}}}_{k}}... | Proof of Proposition 5.1.1:\n\nRecall that if \( A \) and \( B \) are \( n \times n \) Proof. \( \sqrt{\det \left( {{T}^{\top }T}\right) } = \sqrt{\left( {\det {T}^{\top }}\right) \left( {\det T}\right) } = \sqrt{{\left( \det T\right) }^{2}} = \left| {\det T}\right| \) matrices, then\n\n\[ \n\det A\det B = \det \left( ... | Yes |
Example 5.1.2 (Volume of two-dimensional and three-dimensional parallelograms). When \( k = 2 \), we have | \[ \det \left( {{T}^{\top }T}\right) = \det \left( {\left\lbrack \begin{matrix} {\overrightarrow{\mathbf{v}}}_{1}^{\top } \\ {\overrightarrow{\mathbf{v}}}_{2}^{\top } \end{matrix}\right\rbrack \left\lbrack \begin{matrix} {\overrightarrow{\mathbf{v}}}_{1} & {\overrightarrow{\mathbf{v}}}_{2} \end{matrix}\right\rbrack }\r... | Yes |
Example 5.1.4 (Volume of a 3-parallelogram in \( {\mathbb{R}}^{4} \) ). Let \( P \) be the 3-parallelogram \( P \) in \( {\mathbb{R}}^{4} \) spanned by \( {\overrightarrow{\mathbf{v}}}_{1} = \left\lbrack \begin{array}{l} 1 \\ 0 \\ 0 \\ 1 \end{array}\right\rbrack ,{\overrightarrow{\mathbf{v}}}_{2} = \left\lbrack \begin{... | Set \( T = \left\lbrack {{\overrightarrow{\mathbf{v}}}_{1},{\overrightarrow{\mathbf{v}}}_{2},{\overrightarrow{\mathbf{v}}}_{3}}\right\rbrack \) ; then \[ {T}^{\top }T = \left\lbrack \begin{array}{lll} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{array}\right\rbrack \;\text{ and }\;\det \left( {{T}^{\top }T}\right) = 4,\;\t... | Yes |
Example 5.2.4 (Parametrization of a cone). The subset of \( {\mathbb{R}}^{3} \) of equation \( {x}^{2} + {y}^{2} - {z}^{2} = 0 \), shown in Figure 5.2.1, is not a manifold in the neighborhood of the vertex, which is at the origin. However, the subset\n\n\[ M = \\left\\{ {\\left. \\left( \\begin{array}{l} x \\\\ y \\\\ ... | In the language of Definition 5.2.3 we have \( U = \\left\\lbrack {0,1}\\right\\rbrack \\times \\left\\lbrack {0,{2\\pi }}\\right\\rbrack \) . We will set \( X = \\partial U \), so that \( U - X = \\left( {0,1}\\right) \\times \\left( {0,{2\\pi }}\\right) \), and \( X \) consists of the four line segments \( \\left( \\... | No |
Consider the surface in Figure 5.2.3, which is obtained by rotating the curve of equation \( {\left( 1 - x\right) }^{3} = {z}^{2} \) in the \( \left( {x, z}\right) \) -plane around the \( z \) -axis. This surface has the equation \( {\left( 1 - \sqrt{{x}^{2} + {y}^{2}}\right) }^{3} = {z}^{2} \) . The curve can be param... | Figure 5.2.3 shows the image of this parametrization for \( 0 \leq \theta \leq {3\pi }/2 \) and \( \left| t\right| \leq 1 \) . It can be guessed from the picture, and proved from the formula, that the points in \( \left\lbrack {-1,1}\right\rbrack \times \left\lbrack {0,{3\pi }/2}\right\rbrack \) where \( t = \pm 1 \) a... | Yes |
Let \( {\gamma }_{1} \) and \( {\gamma }_{2} \) be two parametrizations of \( {S}^{2} \) by spherical coordinates, but with different poles. Call \( {P}_{1},{P}_{1}^{\prime } \) the poles for \( {\gamma }_{1} \) and \( {P}_{2},{P}_{2}^{\prime } \) the poles for \( {\gamma }_{2} \) . Then \( {\gamma }_{2}^{-1} \circ {\g... | Indeed, some single point in the domain of \( {\gamma }_{1} \) maps to \( {P}_{2} \cdot {}^{2} \) But as shown in Figure 5.2.4, \( {\gamma }_{2} \) maps a whole segment to \( {P}_{2} \), so that \( {\gamma }_{2}^{-1} \circ {\gamma }_{1} \) \ | No |
Theorem 5.2.11. Both \( {U}_{1}^{\mathrm{{OK}}} = {U}_{1} - \left( {{X}_{1} \cup {Y}_{2}}\right) \) and \( {U}_{2}^{\mathrm{{OK}}} = {U}_{2} - \left( {{X}_{2} \cup {Y}_{1}}\right) \) are open subsets of \( {\mathbb{R}}^{k} \) with boundaries of \( k \) -dimensional volume 0, and\n\n\[ \Phi \overset{\text{ def }}{ = }{\... | Proof. By Proposition 3.2.11 we have\n\n\[ \left\lbrack {\mathbf{D}\Phi \left( \mathbf{x}\right) }\right\rbrack = \left\lbrack {\mathbf{D}{\gamma }_{2}^{-1}\left( {{\gamma }_{1}\left( \mathbf{x}\right) }\right) }\right\rbrack \left\lbrack {\mathbf{D}{\gamma }_{1}\left( \mathbf{x}\right) }\right\rbrack \] \nsince both \... | Yes |
Proposition 5.3.3 (Integral independent of parametrization). Let \( M \) be a \( k \) -dimensional manifold in \( {\mathbb{R}}^{n} \) and \( f : M \rightarrow \mathbb{R} \) a function. If \( U \) and \( V \) are subsets of \( {\mathbb{R}}^{k} \) and \( {\gamma }_{1} : U \rightarrow M,{\gamma }_{2} : V \rightarrow M \) ... | Proof. Define \( \Phi = {\gamma }_{2}^{-1} \circ {\gamma }_{1} : {U}^{\mathrm{{OK}}} \rightarrow {V}^{\mathrm{{OK}}} \) to be the \ | No |
Corollary 5.3.4. Every point of a \( k \) -dimensional manifold \( M \subset {\mathbb{R}}^{n} \) has a neighborhood with finite \( k \) -dimensional volume. | Proof. This follows from Theorem 5.2.6 and Proposition 5.3.3 (and Theorem 4.3.6, which says that the integral, hence volume, exists). | Yes |
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