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Let \( F : {R}^{2} \rightarrow {R}^{2} \) be given by\n\n\[ F\left( {x, y}\right) = \left( {{x}^{2} - {y}^{2},{2xy}}\right) ,\;\left( {x, y}\right) \in {R}^{2}. \]\n\n\( F \) is easily seen to be differentiable, and its differential \( d{F}_{p} \) at \( p = \left( {x, y}\right) \) is\n\n\[ d{F}_{p} = \left( \begin{matr... | For instance, \( d{F}_{\left( 1,1\right) }\left( {2,3}\right) = \left( {-2,{10}}\right) \). | Yes |
Let \( F : {R}^{2} \rightarrow {R}^{2} \) be given by\n\n\[ F\left( {x, y}\right) = \left( {{e}^{x}\cos y,{e}^{x}\sin y}\right) ,\;\left( {x, y}\right) \in {R}^{2}. \] | The component functions of \( F \), namely, \( u\left( {x, y}\right) = {e}^{x}\cos y, v\left( {x, y}\right) = {e}^{x}\sin y \), have continuous partial derivatives of all orders. Thus, \( F \) is differentiable.\n\nIt is instructive to see, geometrically, how \( F \) transforms curves of the \( {xy} \) plane. For insta... | Yes |
If \( \alpha \left( t\right) = \left( {x\left( t\right), y\left( t\right), z\left( t\right) }\right) \) is a parametrized curve in \( {S}^{2} \), then \( {2x}{x}^{\prime } + {2y}{y}^{\prime } + {2z}{z}^{\prime } = 0 \) | which shows that the vector \( \left( {x, y, z}\right) \) is normal to the sphere at the point \( \left( {x, y, z}\right) \) . Thus, \( \bar{N} = \left( {x, y, z}\right) \) and \( N = \left( {-x, - y, - z}\right) \) are fields of unit normal vectors in \( {S}^{2} \) . We fix an orientation in \( {S}^{2} \) by choosing ... | Yes |
Consider the cylinder \( \left\{ {\left( {x, y, z}\right) \in {R}^{3};{x}^{2} + {y}^{2} = 1}\right\} \). By an argument similar to that of the previous example, we see that \( \bar{N} = \left( {x, y,0}\right) \) and \( N = \left( {-x, - y,0}\right) \) are unit normal vectors at \( \left( {x, y, z}\right) \). We fix an ... | By considering a curve \( \left( {x\left( t\right), y\left( t\right), z\left( t\right) }\right) \) contained in the cylinder, that is, with \( {\left( x\left( t\right) \right) }^{2} + {\left( y\left( t\right) \right) }^{2} = 1 \), we are able to see that, along this curve, \( N\left( t\right) = \) \( \left( {-x\left( t... | Yes |
Let us analyze the point \( p = \left( {0,0,0}\right) \) of the hyperbolic paraboloid \( z = {y}^{2} - {x}^{2} \) . For this, we consider a parametrization \( \mathbf{x}\left( {u, v}\right) \) given by\n\n\[ \mathbf{x}\left( {u, v}\right) = \left( {u, v,{v}^{2} - {u}^{2}}\right) ,\] | and compute the normal vector \( N\left( {u, v}\right) \) . We obtain successively\n\n\[ {\mathbf{x}}_{u} = \left( {1,0, - {2u}}\right) \]\n\n\[ {\mathbf{x}}_{v} = \left( {0,1,{2v}}\right) \]\n\n\[ N = \left( {\frac{u}{\sqrt{{u}^{2} + {v}^{2} + \frac{1}{4}}},\frac{-v}{\sqrt{{u}^{2} + {v}^{2} + \frac{1}{4}}},\frac{1}{2\... | Yes |
The differential \( {\mathrm{{dN}}}_{\mathrm{p}} : {\mathrm{T}}_{\mathrm{p}}\left( \mathrm{S}\right) \rightarrow {\mathrm{T}}_{\mathrm{p}}\left( \mathrm{S}\right) \) of the Gauss map is a self-adjoint linear map (cf. the appendix to Chap. 3). | Since \( d{N}_{p} \) is linear, it suffices to verify that \( \left\langle {d{N}_{p}\left( {w}_{1}\right) ,{w}_{2}}\right\rangle = \) \( \left\langle {{w}_{1}, d{N}_{p}\left( {w}_{2}\right) }\right\rangle \) for a basis \( \left\{ {{w}_{1},{w}_{2}}\right\} \) of \( {T}_{p}\left( S\right) \) . Let \( \mathbf{x}\left( {u... | Yes |
Consider the surface of revolution obtained by rotating the curve \( z = {y}^{4} \) about the \( z \) axis. We shall show that at \( p = \left( {0,0,0}\right) \) the differential \( d{N}_{p} = 0 \) . | To see this, we observe that the curvature of the curve \( z = {y}^{4} \) at \( p \) is equal to zero. Moreover, since the \( {xy} \) plane is a tangent plane to the surface at \( p \), the normal vector \( N\left( p\right) \) is parallel to the \( z \) axis. Therefore, any normal section at \( p \) is obtained from th... | Yes |
In the plane of Example 1, all normal sections are straight lines; hence, all normal curvatures are zero. Thus, the second fundamental form is identically zero at all points. | This agrees with the fact that \( {dN} \equiv 0 \) . | Yes |
We shall compute the Gaussian curvature of the points of the torus covered by the parametrization (cf. Example 6 of Sec. 2-2) | \[ \mathbf{x}\left( {u, v}\right) = \left( {\left( {a + r\cos u}\right) \cos v,\left( {a + r\cos u}\right) \sin v, r\sin u}\right) ,\] \[ 0 < u < {2\pi },\;0 < v < {2\pi }.\]\n\nFor the computation of the coefficients \( e, f, g \), we need to know \( N \) (and thus \( {\mathbf{x}}_{u} \) and \( {\mathbf{x}}_{v} \) ), ... | Yes |
Consider the surface obtained by rotating the curve \( z = {y}^{3} \), \( - 1 < z < 1 \), about the line \( z = 1 \) (see Fig. 3-18). A simple computation shows that the points generated by the rotation of the origin \( O \) are parabolic points. | We shall omit this computation, because we shall prove shortly (Example 4) that the parallels and the meridians of a surface of revolution are lines of curvature; this, together with the fact that, for the points in question, the meridians (curves of the form \( y = {x}^{3} \)) have zero curvature and the parallel is a... | No |
Very often a surface is given as the graph of a differentiable function (cf. Prop. 1, Sec. 2-2) \( z = h\left( {x, y}\right) \), where \( \left( {x, y}\right) \) belong to an open set \( U \subset {R}^{2} \). It is, therefore, convenient to have at hand formulas for the relevant concepts in this case. | To obtain such formulas let us parametrize the surface by\n\n\[ \mathbf{x}\left( {u, v}\right) = \left( {u, v, h\left( {u, v}\right) }\right) ,\;\left( {u, v}\right) \in U, \]\n\nwhere \( u = x, v = y \). A simple computation shows that\n\n\[ {\mathbf{x}}_{u} = \left( {1,0,{h}_{u}}\right) ,\;{\mathbf{x}}_{v} = \left( {... | Yes |
A vector field in the usual torus \( T \) is obtained by parametrizing the meridians of \( T \) by arc length and defining \( w\left( p\right) \) as the velocity vector of the meridian through \( p \) (Fig. 3-30). Notice that \( \left| {w\left( p\right) }\right| = 1 \) for all \( p \in T \) . | It is left as an exercise (Exercise 2) to verify that \( w \) is differentiable. | No |
A similar procedure, this time on the sphere \( {S}^{2} \) and using the semimeridians of \( {S}^{2} \), yields a vector field \( w \) defined in the sphere minus the two poles \( N \) and \( S \) . To obtain a vector field defined in the whole sphere, | reparametrize all the semimeridians by the same parameter \( t, - 1 < t < 1 \), and define \( v\left( p\right) = \left( {1 - {t}^{2}}\right) w\left( p\right) \) for \( p \in {S}^{2} - \{ N\} \cup \{ S\} \) and \( v\left( N\right) = v\left( S\right) = 0 \) (Fig. 3-31). | Yes |
We want to find a field of directions \( {r}^{\prime } \) on \( S - \{ \left( {0,0,0}\right) \} \) that is orthogonal to \( r \) at each point and to determine the integral curves of \( {r}^{\prime }. | We begin by parametrizing \( S \) by\n\n\[ \mathbf{x}\left( {u, v}\right) = \left( {u, v,{u}^{2} - {v}^{2}}\right) ,\;u = x,\;v = y. \]\n\nThe family \( \left\{ {C}_{\alpha }\right\} \) is given by \( {u}^{2} - {v}^{2} = \) const. \( \neq 0 \) (or rather by the image under \( \mathbf{x} \) of this set). If \( {u}^{\pri... | Yes |
Example 4. An almost trivial example, but one which illustrates the mechanism of the above method, is given by the hyperbolic paraboloid \( z = {x}^{2} - {y}^{2} \) . As usual we parametrize the entire surface by\n\n\[ \n\\mathbf{x}\\left( {u, v}\\right) = \\left( {u, v,{u}^{2} - {v}^{2}}\\right) .\n\] | A simple computation shows that\n\n\[ \ne = \\frac{2}{{\\left( 1 + 4{u}^{2} + 4{v}^{2}\\right) }^{1/2}},\\;f = 0,\\;g = - \\frac{2}{{\\left( 1 + 4{u}^{2} + 4{v}^{2}\\right) }^{1/2}}.\n\]\n\nThus, the equation of the asymptotic curves can be written as\n\n\[ \n\\frac{2}{{\\left( 1 + 4{u}^{2} + 4{v}^{2}\\right) }^{1/2}}\... | Yes |
The simplest examples of ruled surfaces are the tangent surfaces to a regular curve (cf. Example 4, Sec. 2-3), the cylinders and the cones. | A cylinder is a ruled surface generated by a one-parameter family of lines \( \{ \alpha \left( t\right), w\left( t\right) \}, t \in I \), where \( \alpha \left( I\right) \) is contained in a plane \( P \) and \( w\left( t\right) \) is parallel to a fixed direction in \( {R}^{3} \) (Fig. 3-33(a)). A cone is a ruled surf... | Yes |
Example 2. Let \( {S}^{1} \) be the unit circle \( {x}^{2} + {y}^{2} = 1 \) in the \( {xy} \) plane, and let \( \alpha \left( s\right) \) be a parametrization of \( {S}^{1} \) by arc length. For each \( s \), let \( w\left( s\right) = {\alpha }^{\prime }\left( s\right) + {e}_{3} \) , where \( {e}_{3} \) is the unit vec... | \[ \mathbf{x}\left( {s, v}\right) = \alpha \left( s\right) + v\left( {{\alpha }^{\prime }\left( s\right) + {e}_{3}}\right) \] is a ruled surface. It can be put into a more familiar form if we write \[ \mathbf{x}\left( {s, v}\right) = \left( {\cos s - v\sin s,\sin s + v\cos s, v}\right) \] and notice that \( {x}^{2} + {... | Yes |
Let \( S \) be the hyperbolic paraboloid \( z = {kxy}, k \neq 0 \). To show that \( S \) is a ruled surface, we observe that the lines \( y = z/{tk}, x = t \), for each \( t \neq 0 \) belong to \( S \). If we take the intersection of this family of lines with the plane \( z = 0 \), we obtain the curve \( x = t, y = 0, ... | \[ \alpha \left( t\right) = \left( {t,0,0}\right) ,\;w\left( t\right) = \frac{\left( 0,1, kt\right) }{\sqrt{1 + {k}^{2}{t}^{2}}}. \] This gives a ruled surface (Fig. 3-35) \[ \mathbf{x}\left( {t, v}\right) = \alpha \left( t\right) + {vw}\left( t\right) = \left( {t,\frac{v}{\sqrt{1 + {k}^{2}{t}^{2}}},\frac{vkt}{\sqrt{1 ... | Yes |
Let \( S \) be a regular surface and \( \alpha = \alpha \left( s\right) \) a curve on \( S \) parametrized by arc length. Assume that \( \alpha \) is nowhere tangent to an asymptotic direction. Consider the ruled surface \[ \mathbf{x}\left( {s, v}\right) = \alpha \left( s\right) + v\frac{N\left( s\right) \land {N}^{\pr... | To show that \( \mathbf{x} \) is a developable surface, we shall check that condition (9) holds for \( \mathbf{x} \) . In fact, by a straightforward computation, we obtain \[ \left\langle {\frac{N \land {N}^{\prime }}{\left| {N}^{\prime }\right| } \land {\left( \frac{N \land {N}^{\prime }}{\left| {N}^{\prime }\right| }... | Yes |
It is easily checked that \( E = G = {a}^{2}{\cosh }^{2}v, F = 0 \), and \( {\mathbf{x}}_{uu} + {\mathbf{x}}_{vv} = 0 \) . Thus, the helicoid is a minimal surface. It has the additional property that it is the only minimal surface, other than the plane, which is also a ruled surface. | We can give a proof of the last assertion if we assume that the zeros of the Gaussian curvature of a minimal surface are isolated (for a proof, see, for instance, the survey of Osserman quoted at the end of this section, p. 76). Granted this, we shall proceed as follows.\n\nAssume that the surface is not a plane. Then ... | No |
Enneper's surface is the parametrized surface\n\n\\[ \n\\mathbf{x}\\left( {u, v}\\right) = \\left( {u - \\frac{{u}^{3}}{3} + u{v}^{2}, v - \\frac{{v}^{3}}{3} + v{u}^{2},{u}^{2} - {v}^{2}}\\right) ,\\;\\left( {u, v}\\right) \\in {R}^{2},\n\\] | An interesting feature of Enneper's surface is that it has self-intersections. This can be shown by setting \\( u = \\rho \\cos \\theta, v = \\rho \\sin \\theta \\) and writing\n\n\\[ \n\\mathbf{x}\\left( {\\rho ,\\theta }\\right) = \\left( {\\rho \\cos \\theta - \\frac{{\\rho }^{3}}{3}\\cos {3\\theta },\\rho \\sin \\t... | Yes |
Example 8 (Scherk's Minimal Surface). This is given by\n\n\[ \mathbf{x}\left( {u, v}\right) = \left( {\arg \frac{\zeta + i}{\zeta - i},\arg \frac{\zeta + 1}{\zeta - 1},\log \left| \frac{{\zeta }^{2} + 1}{{\zeta }^{2} - 1}\right| }\right) ,\n\]\n\n\( \zeta \neq \pm 1,\zeta \neq \pm i, \)\n\nwhere \( \zeta = u + {iv} \),... | We easily compute that\n\n\[ \arg \frac{\zeta + i}{\zeta - i} = {\tan }^{-1}\frac{2u}{{u}^{2} + {v}^{2} - 1} \]\n\n\[ \arg \frac{\zeta + 1}{\zeta - 1} = {\tan }^{-1}\frac{-{2v}}{{u}^{2} + {v}^{2} - 1} \]\n\n\[ \log \left| \frac{{\zeta }^{2} + 1}{{\zeta }^{2} - 1}\right| = \frac{1}{2}\log \frac{{\left( {u}^{2} - {v}^{2}... | Yes |
Let \( \varphi \) be a map of the coordinate neighborhood \( \overline{\mathbf{x}}\left( U\right) \) of the cylinder given in Example 2 of Sec. 2-5 into the plane \( \mathbf{x}\left( {R}^{2}\right) \) of Example 1 of Sec. 2-5, defined by \( \varphi = \mathbf{x} \circ {\overline{\mathbf{x}}}^{-1} \) (we have changed \( ... | In fact, each vector \( w \), tangent to the cylinder at a point \( p \in \overline{\mathbf{x}}\left( U\right) \), is tangent to a curve \( \overline{\mathbf{x}}\left( {u\left( t\right), v\left( t\right) }\right) \), where \( \left( {u\left( t\right), v\left( t\right) }\right) \) is a curve in \( U \subset {R}^{2} \) .... | Yes |
We shall show that the catenoid is locally isometric to the helicoid of Example 3, Sec. 2-5. | Let us make the following change of parameters:\n\n\[ \bar{u} = u,\;\bar{v} = a\sinh v,\;0 < u < {2\pi }, - \infty < v < \infty ,\]\n\nwhich is possible since the map is clearly one-to-one, and the Jacobian\n\n\[ \frac{\partial \left( {\bar{u},\bar{v}}\right) }{\partial \left( {u, v}\right) } = a\cosh v \]\n\nis nonzer... | Yes |
We shall prove that the one-sheeted cone (minus the vertex)\n\n\\[ \nz = + k\\sqrt{{x}^{2} + {y}^{2}},\\;\\left( {x, y}\\right) \\neq \\left( {0,0}\\right) ,\n\\]\n\nis locally isometric to a plane. | The idea is to show that a cone minus a generator can be \ | No |
We shall compute the Christoffel symbols for a surface of revolution parametrized by (cf. Example 4, Sec. 2-3)\n\n\\[ \n\\mathbf{x}\\left( {u, v}\\right) = \\{ f\\left( v\\right) \\cos u, f\\left( v\\right) \\sin u, g\\left( v\\right) ),\\;f\\left( v\\right) \\neq 0.\n\\] | Since\n\n\\[ \nE = {\\left( f\\left( v\\right) \\right) }^{2},\\;F = 0,\\;G = {\\left\\{ {f}^{\\prime }\\left( v\\right) \\right\\) }^{2} + {\\left( {g}^{\\prime }\\left( v\\right) \\right) }^{2},\n\\]\n\nwe obtain\n\n\\[ \n{E}_{u} = 0,\\;{E}_{v} = {2f}{f}^{\\prime },\n\\]\n\n\\[ \n{F}_{u} = {F}_{v} = 0,\\;{G}_{u} = 0,... | Yes |
Let \( C \) be a parallel of colatitude \( \varphi \) (see Fig. 4-12) of an oriented unit sphere and let \( {w}_{0} \) be a unit vector, tangent to \( C \) at some point \( p \) of \( C \) . Let us determine the parallel transport of \( {w}_{0} \) along \( C \), parametrized by arc length \( s \), with \( s = 0 \) at \... | Consider the cone which is tangent to the sphere along \( C \) . The angle \( \psi \) at the vertex of this cone is given by \( \psi = \left( {\pi /2}\right) - \varphi \) . By the above property, the problem reduces to the determination of the parallel transport of \( {w}_{0} \), along \( C \) , relative to the tangent... | Yes |
The previous example is a particular case of an interesting geometric construction of the parallel transport. Let \( C \) be a regular curve on a surface \( S \) and assume that \( C \) is nowhere tangent to an asymptotic direction. Consider the envelope of the family of tangent planes of \( S \) along \( C \) (cf. Exa... | Now, we shall prove later in this book (Sec. 4-6, theorem of Minding) that a surface of zero Gaussian curvature is locally isometric to a plane. Thus, we can map a neighborhood \( V \subset \sum \) of \( p \) into a plane \( P \) by an isometry \( \varphi : V \rightarrow P \) . To obtain the parallel transport of \( w ... | No |
The great circles of a sphere \( {S}^{2} \) are geodesics. | Indeed, the great circles \( C \) are obtained by intersecting the sphere with a plane that passes through the center \( O \) of the sphere. The principal normal at a point \( p \in C \) lies in the direction of the line that connects \( p \) to \( O \) because \( C \) is a circle of center \( O \) . Since \( {S}^{2} \... | Yes |
We are going to obtain some conclusions from these equations. First, as expected, the meridians \( u = \) const. and \( v = v\left( s\right) \), parametrized by arc length \( s \), are geodesics. Indeed, the first equation of (4a) is trivially satisfied by \( u = \) const. The second equation becomes\n\n\[ {v}^{\prime ... | Since the first fundamental form along the meridian \( u = \) const. \( v = v\left( s\right) \) yields\n\n\[ \left( {{\left( {f}^{\prime }\right) }^{2} + {\left( {g}^{\prime }\right) }^{2}}\right) {\left( {v}^{\prime }\right) }^{2} = 1, \]\n\nwe conclude that\n\n\[ {\left( {v}^{\prime }\right) }^{2} = \frac{1}{{\left( ... | Yes |
Example 1. Let \( S \) be a surface of revolution given by (cf. Sec. 3-3, Example 4)\n\n\[ x = \varphi \left( v\right) \cos u,\;y = \varphi \left( v\right) \sin u,\;z = \psi \left( v\right) ,\;0 < u < {2\pi }, \]\n\nwhere\n\n\[ \varphi \left( v\right) = C\cos v,\;C > 1, \]\n\n\[ \psi \left( v\right) = \int \sqrt{1 - {C... | By using expressions already known (Sec. 3-3, Example 4), we obtain\n\n\[ E = {C}^{2}{\cos }^{2}v \]\n\n\[ F = 0, \]\n\n\[ G = 1\text{,} \]\n\n\[ e = - C\cos v\left( \sqrt{1 - {C}^{2}{\sin }^{2}v}\right) \]\n\n\[ f = 0, \]\n\n\[ g = - \frac{C\cos v}{\sqrt{1 - {C}^{2}{\sin }^{2}v}} \]\n\nhence,\n\n\[ {k}_{1} = \frac{e}{... | Yes |
We shall prove that \( \pi \) is a covering map. | We first observe that when \( \left( {{u}_{0},{v}_{0}}\right) \in P \), the mapping \( \pi \) restricted to the band\n\n\[ R = \left\{ {\left( {u, v}\right) \in P;{u}_{0} - \pi \leq u \leq {u}_{0} + \pi }\right\} \]\n\ncovers \( S \) entirely. Actually, \( \pi \) restricted to the interior of \( R \) is a parametrizati... | Yes |
We shall prove that \( \pi \) is a covering map (see Fig. 5-23). | It is clear that \( \pi \) is continuous and that \( \pi \left( H\right) = {S}^{1} \) . This verifies condition 1 . To verify condition 2, let \( p \in {S}^{1} \) . We shall prove that \( U = {S}^{1} - \{ q\} \), where \( q \in {S}^{1} \) is the point symmetric to \( p \), is a distinguished neighborhood of \( p \) . I... | Yes |
Under what conditions is a local homeomorphism a global homeomorphism? | The notion of covering space allows us to break up this question into two questions as follows:\n\n1. Under what conditions is a local homeomorphism a covering map?\n\n2. Under what conditions is a covering map a global homeomorphism?\n\nA simple answer to question 1 is given by the following proposition.\n\nPROPOSITIO... | Yes |
Example 4. Let\n\n\[ \n{S}^{1} = \left\{ {\left( {x, y}\right) \in {R}^{2};x = \cos t, y = \sin t, t \in R}\right\} \n\] \n\nbe the unit circle and define a map \( \pi : {S}^{1} \rightarrow {S}^{1} \) by \n\n\[ \n\pi \left( {\cos t,\sin t}\right) - \left( {\cos {kt},\sin {kt}}\right) , \n\] \n\nwhere \( k \) is a posit... | Geometrically, \( \pi \) wraps the first \( {S}^{1}k \) times onto the second \( {S}^{1} \) . Notice that the inverse image of a point \( p \in {S}^{1} \) contains exactly \( k \) points. Thus, \( \pi \) is a \( k \) -sheeted covering of \( {S}^{1} \) . | Yes |
Let \( \alpha : \left\lbrack {0, l}\right\rbrack \rightarrow {R}^{2} \) be a plane, continuous closed curve. Choose a point \( {p}_{0} \in {R}^{2},{p}_{0} \notin \alpha \left( \left\lbrack {0, l}\right\rbrack \right) \), and let \( \varphi : \left\lbrack {0, l}\right\rbrack \rightarrow {S}^{1} \) be given by\n\n\[ \var... | Notice that by moving \( {p}_{0} \) along an arc \( \beta \) which does not meet \( \alpha \left( \left\lbrack {0, l}\right\rbrack \right) \) the winding number remains unchanged. Indeed, the position maps of \( \alpha \) relative to any two points of \( \beta \) can clearly be joined by a homotopy. It follows that the... | Yes |
THEOREM 1 (Differentiable Jordan Curve Theorem). Let \( \alpha : \left\lbrack {0, l}\right\rbrack \) \( \rightarrow {\mathrm{R}}^{2} \) be a plane, regular, closed, simple curve. Then \( {\mathrm{R}}^{2} \rightarrow \alpha \left( \left\lbrack {0, l}\right\rbrack \right) \) has exactly two connected components, and \( \... | Proof. Let \( {N}_{\epsilon }\left( \alpha \right) \) be a tubular neighborhood of \( \alpha \left( \left\lbrack {0, l}\right\rbrack \right) \) . This is constructed in the same way as that used for the tubular neighborhood of a compact surface (cf. Sec. 2-7). We recall that \( {N}_{\epsilon }\left( \alpha \right) \) i... | Yes |
The unique asymptotic line that passes through a paraholic point \( \mathrm{p} \in \mathrm{U} \subset \mathrm{S} \) of a surface \( \mathrm{S} \) of curvature \( \mathrm{K} \equiv 0 \) is an (open) segment of a (straight) line in \( \mathrm{S} \) . | Proof. Since \( p \) is not umbilical, it is possible to parametrize a neighborhood \( V \subset U \) of \( p \) by \( \mathbf{x}\left( {u, v}\right) = \mathbf{x} \) in such a way that the coordinate curves are lines of curvature. Suppose that \( v = \) const. is an asymptotic curve; that is, it has zero normal curvatu... | Yes |
Let \( {S}^{2} = \left\{ {\left( {x, y, z}\right) \in {R}^{3};{x}^{2} + {y}^{2} + {z}^{2} = 1}\right\} \) be the unit sphere and let \( A : {S}^{2} \rightarrow {S}^{2} \) be the antipodal map; i.e., \( A\left( {x, y, z}\right) = \) \( \left( {-x, - y, - z}\right) \) . Let \( {P}^{2} \) be the set obtained from \( {S}^{... | From the fact that \( {S}^{2} \) is a regular surface and \( A \) is a diffeomorphism, it follows that \( {P}^{2} \) together with the family \( \left\{ {{U}_{\alpha },\pi \circ {\mathbf{x}}_{\alpha }}\right\} \) is an abstract surface, to be denoted again by \( {P}^{2}.{P}^{2} \) is called the real projective plane. | Yes |
Example 3. Let \( S = {R}^{2} \) be a plane with coordinates \( \left( {u, v}\right) \) and define an inner product at each point \( q = \left( {u, v}\right) \in {R}^{2} \) by setting\n\n\[ \n{\left\langle \frac{\partial }{\partial u},\frac{\partial }{\partial u}\right\rangle }_{q} = E = 1,\;{\left\langle \frac{\partia... | Actually the geometry of \( H \) is an exact model for the non-Euclidean geometry of Lobachewski, in which all the axioms of Euclid, except the axiom of parallels, are assumed (cf. Sec. 4-5). To make this point clear, we shall compute the geodesics of \( H \) .\n\nIf we look at the differential equations for the geodes... | No |
Let \( {R}^{2} \) be a plane with coordinates \( \left( {x, y}\right) \) and \( {T}_{m, n} : {R}^{2} \rightarrow \) \( {R}^{2} \) be the map (translation) \( {T}_{m, n}\left( {x, y}\right) = \left( {x + m, y + n}\right) \), where \( m \) and \( n \) are integers. Define an equivalence relation in \( {R}^{2} \) by \( \l... | Let \( {i}_{\alpha } : {U}_{\alpha } \subset {R}^{2} \rightarrow {R}^{2} \) be a family of parametrizations of \( {R}^{2} \), where \( {i}_{\alpha } \) is the identity map, such that \( {U}_{\alpha } \cap {T}_{m, n}\left( {U}_{\alpha }\right) = \phi \) for all \( m, n \) . Since \( {T}_{m, n} \) is a diffeomorphism, it... | Yes |
We claim that the real projective plane \( {P}^{2} \) is nonorientable. | To prove this, we first make the following general observation. Whenever an abstract surface \( S \) contains an open set \( M \) diffeomorphic to a Möbius strip (Sec. 2-6, Example 3), it is nonorientable. Otherwise, there exists a family of parametrizations covering \( S \) with the property that all coordinate change... | Yes |
For instance, let \( f : {R}^{2} \rightarrow R \) be given by \( f\left( {x, y}\right) = \left( {{x}^{2}/{a}^{2}}\right) - \left( {{y}^{2}/{b}^{2}}\right) - 1 \) . Observe that \( f \) is continuous, \( 0 \in R \) is a closed set in \( R \), and \( \left( {0, + \infty }\right) \) is an open set in \( R \) . Thus, the s... | On the other hand, the set\n\n\[ \nA = \left\{ {\left( {x, y}\right) \in {R}^{2},{x}^{2} + {y}^{2} < 1}\right\} \n\]\n\n\[ \n\cup \left\{ {\left( {x, y}\right) \in {R}^{2};{x}^{2} + {y}^{2} = 1, x > 0, y > 0}\right\} \n\]\n\nis neither open nor closed (Fig. A5-2). | No |
LEMMA 1. Call a sequence \( \left\{ {\mathrm{x}}_{\mathrm{i}}\right\} \) of real numbers a Cauchy sequence if given \( \epsilon > 0 \), there exists \( {\mathrm{i}}_{0} \) such that \( \left| {{\mathrm{x}}_{\mathrm{i}},{\mathrm{x}}_{\mathrm{j}}}\right| < \epsilon \) for all \( \mathrm{i},\mathrm{j} > {\mathrm{i}}_{0} \... | Proof. Let \( \left\{ {x}_{i}\right\} \rightarrow {x}_{0} \) . Then, if \( \epsilon > 0 \) is given, there exists \( {i}_{0} \) such that \( \left| {{x}_{i} - {x}_{0}}\right| < \epsilon /2 \) for \( i > {i}_{0} \) . Thus, for \( i, j > {i}_{0} \), we have\n\n\[ \left| {{x}_{i} - {x}_{j}}\right| \leq \left| {{x}_{i} - {... | Yes |
Theorem 1.1.1. Let \( \mu \) be a measure on \( \left( {\Omega ,\mathcal{F}}\right) \n\n(i) monotonicity. If \( A \subset B \) then \( \mu \left( A\right) \leq \mu \left( B\right) \). | Proof. (i) Let \( B - A = B \cap {A}^{c} \) be the difference of the two sets. Using + to denote disjoint union, \( B = A + \left( {B - A}\right) \) so\n\n\[ \n\mu \left( B\right) = \mu \left( A\right) + \mu \left( {B - A}\right) \geq \mu \left( A\right) .\n\] | Yes |
Theorem 1.1.2. Associated with each Stieltjes measure function \( F \) there is a unique measure \( \mu \) on \( \left( {\mathbf{R},\mathcal{R}}\right) \) with \( \mu (\left( {a, b\rbrack }\right) = F\left( b\right) - F\left( a\right) \) | The proof of Theorem 1.1.2 is a long and winding road, so we will content ourselves to describe the main ideas involved in this section and to hide the remaining details in the appendix in Section A.1. The choice of \ | No |
Lemma 1.1.3. If \( \mathcal{S} \) is a semialgebra then \( \overline{\mathcal{S}} = \{ \) finite disjoint unions of sets in \( \mathcal{S}\} \) is an algebra, called the algebra generated by \( \mathcal{S} \) . | Proof. Suppose \( A = { + }_{i}{S}_{i} \) and \( B = { + }_{j}{T}_{j} \), where + denotes disjoint union and we assume the index sets are finite. Then \( A \cap B = { + }_{i, j}{S}_{i} \cap {T}_{j} \in \overline{\mathcal{S}} \) . As for complements, if \( A = { + }_{i}{S}_{i} \) then \( {A}^{c} = { \cap }_{i}{S}_{i}^{c... | Yes |
Lemma 1.1.5. Suppose only that (i) holds.\n\n(a) If \( A,{B}_{i} \in \overline{\mathcal{S}} \) with \( A = { + }_{i = 1}^{n}{B}_{i} \) then \( \bar{\mu }\left( A\right) = \mathop{\sum }\limits_{i}\bar{\mu }\left( {B}_{i}\right) \) .\n\n(b) If \( A,{B}_{i} \in \overline{\mathcal{S}} \) with \( A \subset { \cup }_{i = 1}... | Proof. Observe that it follows from the definition that if \( A = { + }_{i}{B}_{i} \) is a finite disjoint union of sets in \( \overline{\mathcal{S}} \) and \( {B}_{i} = { + }_{j}{S}_{i, j} \), then\n\n\[ \bar{\mu }\left( A\right) = \mathop{\sum }\limits_{{i, j}}\mu \left( {S}_{i, j}\right) = \mathop{\sum }\limits_{i}\... | Yes |
Theorem 1.2.1. Any distribution function \( F \) has the following properties:\n\n(i) \( F \) is nondecreasing.\n\n(ii) \( \mathop{\lim }\limits_{{x \rightarrow \infty }}F\left( x\right) = 1,\mathop{\lim }\limits_{{x \rightarrow - \infty }}F\left( x\right) = 0 \) .\n\n(iii) \( F \) is right continuous, i.e. \( \mathop{... | Proof. To prove (i), note that if \( x \leq y \) then \( \{ X \leq x\} \subset \{ X \leq y\} \), and then use (i) in Theorem 1.1.1 to conclude that \( P\left( {X \leq x}\right) \leq P\left( {X \leq y}\right) \) .\n\nTo prove (ii), we observe that if \( x \uparrow \infty \), then \( \{ X \leq x\} \uparrow \Omega \), whi... | Yes |
Theorem 1.2.2. If \( F \) satisfies (i),(ii), and (iii) in Theroem 1.2.1, then it is the distribution function of some random variable. | Proof. Let \( \Omega = \left( {0,1}\right) ,\mathcal{F} = \) the Borel sets, and \( P = \) Lebesgue measure. If \( \omega \in \left( {0,1}\right) \) , let\n\n\[ X\left( \omega \right) = \sup \{ y : F\left( y\right) < \omega \} \]\n\nOnce we show that\n\n\( \left( \star \right) \)\n\n\[ \{ \omega : X\left( \omega \right... | Yes |
Uniform distribution on \( \left( {0,1}\right) .f\left( x\right) = 1 \) for \( x \in \left( {0,1}\right) \) and 0 otherwise. Distribution function: | \[ F\left( x\right) = \left\{ \begin{array}{ll} 0 & x \leq 0 \\ x & 0 \leq x \leq 1 \\ 1 & x > 1 \end{array}\right. \] | Yes |
Exponential distribution with rate \( \lambda .f\left( x\right) = \lambda {e}^{-{\lambda x}} \) for \( x \geq 0 \) and 0 otherwise. Distribution function: | \[ F\left( x\right) = \left\{ \begin{array}{ll} 0 & x \leq 0 \\ 1 - {e}^{-x} & x \geq 0 \end{array}\right. \] | Yes |
Theorem 1.2.3. For \( x > 0 \) ,\n\n\[ \left( {{x}^{-1} - {x}^{-3}}\right) \exp \left( {-{x}^{2}/2}\right) \leq {\int }_{x}^{\infty }\exp \left( {-{y}^{2}/2}\right) {dy} \leq {x}^{-1}\exp \left( {-{x}^{2}/2}\right) \] | Proof. Changing variables \( y = x + z \) and using \( \exp \left( {-{z}^{2}/2}\right) \leq 1 \) gives\n\n\[ {\int }_{x}^{\infty }\exp \left( {-{y}^{2}/2}\right) {dy} \leq \exp \left( {-{x}^{2}/2}\right) {\int }_{0}^{\infty }\exp \left( {-{xz}}\right) {dz} = {x}^{-1}\exp \left( {-{x}^{2}/2}\right) \]\n\nFor the other d... | Yes |
Uniform distribution on the Cantor set. The Cantor set \( C \) is defined by removing \( \left( {1/3,2/3}\right) \) from \( \left\lbrack {0,1}\right\rbrack \) and then removing the middle third of each interval that remains. We define an associated distribution function by setting \( F\left( x\right) = 0 \) for \( x \l... | There is no \( f \) for which (1.2.1) holds because such an \( f \) would be equal to 0 on a set of measure 1 . From the definition, it is immediate that the corresponding measure has \( \mu \left( {C}^{c}\right) = 0 \) . | Yes |
Theorem 1.3.1. If \( \{ \omega : X\left( \omega \right) \in A\} \in \mathcal{F} \) for all \( A \in \mathcal{A} \) and \( \mathcal{A} \) generates \( \mathcal{S} \) (i.e., \( \mathcal{S} \) is the smallest \( \sigma \) -field that contains \( \mathcal{A} \) ), then \( X \) is measurable. | Proof. Writing \( \{ X \in B\} \) as shorthand for \( \{ \omega : X\left( \omega \right) \in B\} \), we have\n\n\[ \left\{ {X \in { \cup }_{i}{B}_{i}}\right\} = { \cup }_{i}\left\{ {X \in {B}_{i}}\right\} \]\n\n\[ \left\{ {X \in {B}^{c}}\right\} = \{ X \in B{\} }^{c} \]\n\nSo the class of sets \( \mathcal{B} = \{ B : \... | Yes |
Theorem 1.3.2. If \( X : \left( {\Omega ,\mathcal{F}}\right) \rightarrow \left( {S,\mathcal{S}}\right) \) and \( f : \left( {S,\mathcal{S}}\right) \rightarrow \left( {T,\mathcal{T}}\right) \) are measurable maps, then \( f\left( X\right) \) is a measurable map from \( \left( {\Omega ,\mathcal{F}}\right) \) to \( \left(... | Proof. Let \( B \in \mathcal{T}.\;\{ \omega : f\left( {X\left( \omega \right) }\right) \in B\} = \left\{ {\omega : X\left( \omega \right) \in {f}^{-1}\left( B\right) }\right\} \in \mathcal{F} \), since by assumption \( {f}^{-1}\left( B\right) \in \mathcal{S} \) | Yes |
Theorem 1.3.3. If \( {X}_{1},\ldots {X}_{n} \) are random variables and \( f : \left( {{\mathbf{R}}^{n},{\mathcal{R}}^{n}}\right) \rightarrow \left( {\mathbf{R},\mathcal{R}}\right) \) is measurable, then \( f\left( {{X}_{1},\ldots ,{X}_{n}}\right) \) is a random variable. | Proof. In view of Theorem 1.3.2, it suffices to show that \( \left( {{X}_{1},\ldots ,{X}_{n}}\right) \) is a random vector. To do this, we observe that if \( {A}_{1},\ldots ,{A}_{n} \) are Borel sets then\n\n\[ \left\{ {\left( {{X}_{1},\ldots ,{X}_{n}}\right) \in {A}_{1} \times \cdots \times {A}_{n}}\right\} = { \cap }... | Yes |
Theorem 1.3.4. If \( {X}_{1},\ldots ,{X}_{n} \) are random variables then \( {X}_{1} + \ldots + {X}_{n} \) is a random variable. | Proof. In view of Theorem 1.3.3 it suffices to show that \( f\left( {{x}_{1},\ldots ,{x}_{n}}\right) = {x}_{1} + \ldots + {x}_{n} \) is measurable. To do this, we use Example 1.3.1 and note that \( \left\{ {x : {x}_{1} + \ldots + {x}_{n} < a}\right\} \) is an open set and hence is in \( {\mathcal{R}}^{n} \) . | Yes |
Theorem 1.3.5. If \( {X}_{1},{X}_{2},\ldots \) are random variables then so are\n\n\[ \mathop{\inf }\limits_{n}{X}_{n}\;\mathop{\sup }\limits_{n}{X}_{n}\;\mathop{\limsup }\limits_{n}{X}_{n}\;\mathop{\liminf }\limits_{n}{X}_{n} \] | Proof. Since the infimum of a sequence is \( < a \) if and only if some term is \( < a \) (if all terms are \( \geq a \) then the infimum is), we have\n\n\[ \left\{ {\mathop{\inf }\limits_{n}{X}_{n} < a}\right\} = { \cup }_{n}\left\{ {{X}_{n} < a}\right\} \in \mathcal{F} \]\n\nA similar argument shows \( \left\{ {\math... | Yes |
Lemma 1.4.1. Let \( \varphi \) and \( \psi \) be simple functions.\n\n(iii) \( \int \varphi + {\psi d\mu } = \int {\varphi d\mu } + \int {\psi d\mu } \) . | Proof. (i) and (ii) are immediate consequences of the definition. To prove (iii), suppose\n\n\[ \varphi = \mathop{\sum }\limits_{{i = 1}}^{m}{a}_{i}{1}_{{A}_{i}}\;\text{ and }\;\psi = \mathop{\sum }\limits_{{j = 1}}^{n}{b}_{j}{1}_{{B}_{j}} \]\n\nTo make the supports of the two functions the same, we let \( {A}_{0} = { ... | Yes |
Lemma 1.4.2. If (i) and (iii) hold then we have:\n\n(iv) If \( \varphi \leq \psi \) a.e. then \( \int {\varphi d\mu } \leq \int {\psi d\mu } \).\n\n(v) If \( \varphi = \psi \) a.e. then \( \int {\varphi d\mu } = \int {\psi d\mu } \). | Proof. By (iii), \( \int {\psi d\mu } = \int {\varphi d\mu } + \int \left( {\psi - \varphi }\right) {d\mu } \) and the second integral is \( \geq 0 \) by (i), so (iv) holds. \( \varphi = \psi \) a.e. implies \( \varphi \leq \psi \) a.e. and \( \psi \leq \varphi \) a.e. so (v) follows from two applications of (iv). | Yes |
Lemma 1.4.3. Let \( E \) be a set with \( \mu \left( E\right) < \infty \) . If \( f \) and \( g \) are bounded functions that vanish on \( {E}^{c} \) then:\n\n(i) If \( f \geq 0 \) a.e. then \( \int {fd\mu } \geq 0 \) .\n\n(ii) For any \( a \in \mathbf{R},\int {afd\mu } = a\int {fd\mu } \) .\n\n(iii) \( \int f + {gd\mu... | Proof. Since we can take \( \varphi \equiv 0 \) ,(i) is clear from the definition. To prove (ii), we observe that if \( a > 0 \), then \( {a\varphi } \leq {af} \) if and only if \( \varphi \leq f \), so\n\n\[ \int {afd\mu } = \mathop{\sup }\limits_{{\varphi \leq f}}\int {a\varphi d\mu } = \mathop{\sup }\limits_{{\varph... | Yes |
Lemma 1.4.4. Let \( {E}_{n} \uparrow \Omega \) have \( \mu \left( {E}_{n}\right) < \infty \) and let \( a \land b = \min \left( {a, b}\right) \). Then \[ {\int }_{{E}_{n}}f \land {nd\mu } \uparrow \int {fd\mu }\;\text{ as }n \uparrow \infty \] | Proof. It is clear that from (iv) in Lemma 1.4.3 that the left-hand side increases as \( n \) does. Since \( h = \left( {f \land n}\right) {1}_{{E}_{n}} \) is a possibility in the sup, each term is smaller than the integral on the right. To prove that the limit is \( \int {fd\mu } \), observe that if \( 0 \leq h \leq f... | Yes |
Lemma 1.4.5. Suppose \( f, g \geq 0 \) .\n\n(i) \( \int {fd\mu } \geq 0 \)\n\n(ii) If \( a > 0 \) then \( \int {afd\mu } = a\int {fd\mu } \)\n\n(iii) \( \int f + {gd\mu } = \int {fd\mu } + \int {gd\mu } \)\n\n(iv) If \( 0 \leq g \leq f \) a.e. then \( \int {gd\mu } \leq \int {fd\mu } \)\n\n(v) If \( 0 \leq g = f \) a.e... | Proof. (i) is trivial from the definition. (ii) is clear, since when \( a > 0,{ah} \leq {af} \) if and only if \( h \leq f \) and we have \( \int {ahd\mu } = a\int {hdu} \) for \( h \) in the defining class. For (iii), we observe that if \( f \geq h \) and \( g \geq k \), then \( f + g \geq h + k \) so taking the sup o... | No |
Lemma 1.4.6. If \( f = {f}_{1} - {f}_{2} \) where \( {f}_{1},{f}_{2} \geq 0 \) and \( \int {f}_{i}{d\mu } < \infty \) then\n\n\[ \int {fd\mu } = \int {f}_{1}{d\mu } - \int {f}_{2}{d\mu } \] | Proof. \( {f}_{1} + {f}^{ - } = {f}_{2} + {f}^{ + } \) and all four functions are \( \geq 0 \), so by (iii) of Lemma 1.4.5,\n\n\[ \int {f}_{1}{d\mu } + \int {f}^{ - }{d\mu } = \int {f}_{1} + {f}^{ - }{d\mu } = \int {f}_{2} + {f}^{ + }{d\mu } = \int {f}_{2}{d\mu } + \int {f}^{ + }{d\mu } \]\n\nRearranging gives the desi... | Yes |
Theorem 1.4.7. Suppose \( f \) and \( g \) are integrable.\n\n(i) If \( f \geq 0 \) a.e. then \( \int {fd\mu } \geq 0 \) . | Proof. (i) is trivial. | No |
Theorem 1.5.1. Jensen’s inequality. Suppose \( \varphi \) is convex, that is,\n\n\[ \n{\lambda \varphi }\left( x\right) + \left( {1 - \lambda }\right) \varphi \left( y\right) \geq \varphi \left( {{\lambda x} + \left( {1 - \lambda }\right) y}\right) \n\]\n\nfor all \( \lambda \in \left( {0,1}\right) \) and \( x, y \in \... | Proof. Let \( c = \int {fd\mu } \) and let \( \ell \left( x\right) = {ax} + b \) be a linear function that has \( \ell \left( c\right) = \varphi \left( c\right) \) and \( \varphi \left( x\right) \geq \ell \left( x\right) \) . To see that such a function exists, recall that convexity implies\n\n\[ \n\mathop{\lim }\limit... | Yes |
Theorem 1.5.2. Hölder’s inequality. If \( p, q \in \left( {1,\infty }\right) \) with \( 1/p + 1/q = 1 \) then\n\n\[ \int \left| {fg}\right| {d\mu } \leq \parallel f{\parallel }_{p}\parallel g{\parallel }_{q} \] | Proof. If \( \parallel f{\parallel }_{p} \) or \( \parallel g{\parallel }_{q} = 0 \) then \( \left| {fg}\right| = 0 \) a.e., so it suffices to prove the result when \( \parallel f{\parallel }_{p} \) and \( \parallel g{\parallel }_{q} > 0 \) or by dividing both sides by \( \parallel f{\parallel }_{p}\parallel g{\paralle... | Yes |
Theorem 1.5.3. Bounded convergence theorem. Let \( E \) be a set with \( \mu \left( E\right) < \infty \) . Suppose \( {f}_{n} \) vanishes on \( {E}^{c},\left| {{f}_{n}\left( x\right) }\right| \leq M \), and \( {f}_{n} \rightarrow f \) in measure. Then | \[ \int {fd\mu } = \mathop{\lim }\limits_{{n \rightarrow \infty }}\int {f}_{n}{d\mu } \] | No |
Consider the real line \( \mathbf{R} \) equipped with the Borel sets \( \mathcal{R} \) and Lebesgue measure \( \lambda \) . The functions \( {f}_{n}\left( x\right) = 1/n \) on \( \left\lbrack {0, n}\right\rbrack \) and 0 otherwise on show that the conclusion of Theorem 1.5.3 does not hold when \( \mu \left( E\right) = ... | Proof. Let \( \epsilon > 0,{G}_{n} = \left\{ {x : \left| {{f}_{n}\left( x\right) - f\left( x\right) }\right| < \epsilon }\right\} \) and \( {B}_{n} = E - {G}_{n} \) . Using (iii) and (vi) from Theorem 1.4.7,\n\n\[ \left| {\int {fd\mu }-\int {f}_{n}{d\mu }}\right| = \left| {\int \left( {f - {f}_{n}}\right) {d\mu }}\righ... | No |
Example 1.5.1 shows that we may have strict inequality in Theorem 1.5.4. The functions \( {f}_{n}\left( x\right) = n{1}_{(0,1/n\rbrack }\left( x\right) \) on \( \left( {0,1}\right) \) equipped with the Borel sets and Lebesgue measure show that this can happen on a space of finite measure. | Proof. Let \( {g}_{n}\left( x\right) = \mathop{\inf }\limits_{{m \geq n}}{f}_{m}\left( x\right) .{f}_{n}\left( x\right) \geq {g}_{n}\left( x\right) \) and as \( n \uparrow \infty \) ,\n\n\[ \n{g}_{n}\left( x\right) \uparrow g\left( x\right) = \mathop{\liminf }\limits_{{n \rightarrow \infty }}{f}_{n}\left( x\right) \n\]... | Yes |
Theorem 1.5.5. Monotone convergence theorem. If \( {f}_{n} \geq 0 \) and \( {f}_{n} \uparrow f \) then\n\n\[ \int {f}_{n}{d\mu } \uparrow \int {fd\mu } \] | Proof. Fatou’s lemma, Theorem 1.5.4, implies liminf \( \int {f}_{n}{d\mu } \geq \int {fd\mu } \) . On the other hand, \( {f}_{n} \leq f \) implies \( \lim \sup \int {f}_{n}{d\mu } \leq \int {fd\mu } \) . | Yes |
Theorem 1.5.6. Dominated convergence theorem. If \( {f}_{n} \rightarrow f \) a.e., \( \left| {f}_{n}\right| \leq g \) for all \( n \), and \( g \) is integrable, then \( \int {f}_{n}{d\mu } \rightarrow \int {fd\mu } \) . | Proof. \( {f}_{n} + g \geq 0 \) so Fatou’s lemma implies\n\n\[ \mathop{\liminf }\limits_{{n \rightarrow \infty }}\int {f}_{n} + {gd\mu } \geq \int f + {gd\mu } \]\n\nSubtracting \( \int {gd\mu } \) from both sides gives\n\n\[ \mathop{\liminf }\limits_{{n \rightarrow \infty }}\int {f}_{n}{d\mu } \geq \int {fd\mu } \]\n\... | Yes |
Theorem 1.6.2. Jensen’s inequality. Suppose \( \varphi \) is convex, that is,\n\n\[ \n{\lambda \varphi }\left( x\right) + \left( {1 - \lambda }\right) \varphi \left( y\right) \geq \varphi \left( {{\lambda x} + \left( {1 - \lambda }\right) y}\right) \n\] \n\nfor all \( \lambda \in \left( {0,1}\right) \) and \( x, y \in ... | To recall the direction in which the inequality goes note that if \( P\left( {X = x}\right) = \lambda \) and \( P\left( {X = y}\right) = 1 - \lambda \) then \n\n\[ \n{E\varphi }\left( X\right) = {\lambda \varphi }\left( x\right) + \left( {1 - \lambda }\right) \varphi \left( y\right) \geq \varphi \left( {{\lambda x} + \... | Yes |
Theorem 1.6.4. Chebyshev’s inequality. Suppose \( \varphi : \mathbf{R} \rightarrow \mathbf{R} \) has \( \varphi \geq 0 \), let \( A \in \mathcal{R} \) and let \( {i}_{A} = \inf \{ \varphi \left( y\right) : y \in A\} \) . \[ {i}_{A}P\left( {X \in A}\right) \leq E\left( {\varphi \left( X\right) ;X \in A}\right) \leq {E\v... | Proof. The definition of \( {i}_{A} \) and the fact that \( \varphi \geq 0 \) imply that \[ {i}_{A}{1}_{\left( X \in A\right) } \leq \varphi \left( X\right) {1}_{\left( X \in A\right) } \leq \varphi \left( X\right) \] So taking expected values and using part (c) of Theorem 1.6.1 gives the desired result. | Yes |
Theorem 1.6.8. Suppose \( {X}_{n} \rightarrow X \) a.s. Let \( g, h \) be continuous functions with\n\n(i) \( g \geq 0 \) and \( g\left( x\right) \rightarrow \infty \) as \( \left| x\right| \rightarrow \infty \) ,\n\n(ii) \( \left| {h\left( x\right) }\right| /g\left( x\right) \rightarrow 0 \) as \( \left| x\right| \rig... | Proof. By subtracting a constant from \( h \), we can suppose without loss of generality that \( h\left( 0\right) = 0 \) . Pick \( M \) large so that \( P\left( {\left| X\right| = M}\right) = 0 \) and \( g\left( x\right) > 0 \) when \( \left| x\right| \geq M \) . Given a random variable \( Y \), let \( \bar{Y} = Y{1}_{... | Yes |
Theorem 1.6.9. Change of variables formula. Let \( X \) be a random element of \( \left( {S,\mathcal{S}}\right) \) with distribution \( \mu \), i.e., \( \mu \left( A\right) = P\left( {X \in A}\right) \). If \( f \) is a measurable function from \( \left( {S,\mathcal{S}}\right) \) to \( \left( {\mathbf{R},\mathcal{R}}\r... | Proof. We will prove this result by verifying it in four increasingly more general special cases that parallel the way that the integral was defined in Section 1.4. The reader should note the method employed, since it will be used several times below.\n\nCASE 1: INDICATOR FUNCTIONS. If \( B \in \mathcal{S} \) and \( f ... | Yes |
If \( X \) has an exponential distribution with rate 1 then | \[ E{X}^{k} = {\int }_{0}^{\infty }{x}^{k}{e}^{-x}{dx} = k! \] So the mean of \( X \) is 1 and variance is \( E{X}^{2} - {\left( EX\right) }^{2} = 2 - {1}^{2} = 1 \) . If we let \( Y = X/\lambda \) , then by Exercise 1.2.5, \( Y \) has density \( \lambda {e}^{-{\lambda y}} \) for \( y \geq 0 \), the exponential density... | No |
If \( X \) has a standard normal distribution, | \[ {EX} = \int x{\left( 2\pi \right) }^{-1/2}\exp \left( {-{x}^{2}/2}\right) {dx} = 0\;\text{ (by symmetry) } \] \[ \operatorname{var}\left( X\right) = E{X}^{2} = \int {x}^{2}{\left( 2\pi \right) }^{-1/2}\exp \left( {-{x}^{2}/2}\right) {dx} = 1 \] | Yes |
We say that \( X \) has a Bernoulli distribution with parameter \( p \) if \( P\left( {X = 1}\right) = p \) and \( P\left( {X = 0}\right) = 1 - p \). | Clearly, \[ {EX} = p \cdot 1 + \left( {1 - p}\right) \cdot 0 = p \] Since \( {X}^{2} = X \), we have \( E{X}^{2} = {EX} = p \) and \[ \operatorname{var}\left( X\right) = E{X}^{2} - {\left( EX\right) }^{2} = p - {p}^{2} = p\left( {1 - p}\right) \] | Yes |
We say that \( X \) has a Poisson distribution with parameter \( \lambda \) if\n\n\[ P\left( {X = k}\right) = {e}^{-\lambda }{\lambda }^{k}/k!\text{ for }k = 0,1,2,\ldots \] | To evaluate the moments of the Poisson random variable, we use a little inspiration to observe that for \( k \geq 1 \)\n\n\[ E\left( {X\left( {X - 1}\right) \cdots \left( {X - k + 1}\right) }\right) = \mathop{\sum }\limits_{{j = k}}^{\infty }j\left( {j - 1}\right) \cdots \left( {j - k + 1}\right) {e}^{-\lambda }\frac{{... | Yes |
Theorem 1.7.1. There is a unique measure \( \mu \) on \( \mathcal{F} \) with\n\n\[ \mu \left( {A \times B}\right) = {\mu }_{1}\left( A\right) {\mu }_{2}\left( B\right) \] | Proof. By Theorem 1.1.4 it is enough to show that if \( A \times B = { + }_{i}\left( {{A}_{i} \times {B}_{i}}\right) \) is a finite or countable disjoint union then\n\n\[ \mu \left( {A \times B}\right) = \mathop{\sum }\limits_{i}\mu \left( {{A}_{i} \times {B}_{i}}\right) \]\n\nFor each \( x \in A \), let \( I\left( x\r... | No |
Theorem 1.7.2. Fubini’s theorem. If \( f \geq 0 \) or \( \int \left| f\right| {d\mu } < \infty \) then\n\n\[{\int }_{X}{\int }_{Y}f\left( {x, y}\right) {\mu }_{2}\left( {dy}\right) {\mu }_{1}\left( {dx}\right) = {\int }_{X \times Y}{fd\mu } = {\int }_{Y}{\int }_{X}f\left( {x, y}\right) {\mu }_{1}\left( {dx}\right) {\mu... | Proof. We will prove only the first equality, since the second follows by symmetry. Two technical things that need to be proved before we can assert that the first integral makes sense are:\n\nWhen \( x \) is fixed, \( y \rightarrow f\left( {x, y}\right) \) is \( \mathcal{B} \) measurable.\n\n\( x \rightarrow {\int }_{... | Yes |
Lemma 1.7.3. If \( E \in \mathcal{F} \) then \( {E}_{x} \in \mathcal{B} \) . | Proof. \( {\left( {E}^{c}\right) }_{x} = {\left( {E}_{x}\right) }^{c} \) and \( {\left( { \cup }_{i}{E}_{i}\right) }_{x} = { \cup }_{i}{\left( {E}_{i}\right) }_{x} \), so if \( \mathcal{E} \) is the collection of sets \( E \) for which \( {E}_{x} \in \mathcal{B} \), then \( \mathcal{E} \) is a \( \sigma \) -algebra. Si... | Yes |
Lemma 1.7.4. If \( E \in \mathcal{F} \) then \( g\left( x\right) \equiv {\mu }_{2}\left( {E}_{x}\right) \) is \( \mathcal{A} \) measurable and\n\n\[{\int }_{X}{gd}{\mu }_{1} = \mu \left( E\right)\] | Proof. If conclusions hold for \( {E}_{n} \) and \( {E}_{n} \uparrow E \), then Theorem 1.3.5 and the monotone convergence theorem imply that they hold for \( E \) . Since \( {\mu }_{1} \) and \( {\mu }_{2} \) are \( \sigma \) -finite, it is enough then to prove the result for \( E \subset F \times G \) with \( {\mu }_... | Yes |
Let \( X = Y = \{ 1,2,\ldots \} \) with \( \mathcal{A} = \mathcal{B} = \) all subsets and \( {\mu }_{1} = {\mu }_{2} = \) counting measure. For \( m \geq 1 \), let \( f\left( {m, m}\right) = 1 \) and \( f\left( {m + 1, m}\right) = - 1 \), and let \( f\left( {m, n}\right) = 0 \) otherwise. We claim that\n\n\[ \mathop{\s... | In words, if we sum the columns first, the first one gives us a 1 and the others 0 , while if we sum the rows each one gives us a 0 . | Yes |
Let \( X = \left( {0,1}\right), Y = \left( {1,\infty }\right) \), both equipped with the Borel sets and Lebesgue measure. Let \( f\left( {x, y}\right) = {e}^{-{xy}} - 2{e}^{-{2xy}} \). | \[ {\int }_{0}^{1}{\int }_{1}^{\infty }f\left( {x, y}\right) {dydx} = {\int }_{0}^{1}{x}^{-1}\left( {{e}^{-x} - {e}^{-{2x}}}\right) {dx} > 0 \] \[ {\int }_{1}^{\infty }{\int }_{0}^{1}f\left( {x, y}\right) {dxdy} = {\int }_{1}^{\infty }{y}^{-1}\left( {{e}^{-{2y}} - {e}^{-y}}\right) {dy} < 0 \] | Yes |
Example 1.7.3. Let \( X = \left( {0,1}\right) \) with \( \mathcal{A} = \) the Borel sets and \( {\mu }_{1} = \) Lebesgue measure. Let \( Y = \left( {0,1}\right) \) with \( \mathcal{B} = \) all subsets and \( {\mu }_{2} = \) counting measure. Let \( f\left( {x, y}\right) = 1 \) if \( x = y \) and 0 otherwise | \[
{\int }_{Y}f\left( {x, y}\right) {\mu }_{2}\left( {dy}\right) = 1\;\text{ for all }x\text{ so }\;{\int }_{X}{\int }_{Y}f\left( {x, y}\right) {\mu }_{2}\left( {dy}\right) {\mu }_{1}\left( {dx}\right) = 1
\]
\[
{\int }_{X}f\left( {x, y}\right) {\mu }_{1}\left( {dx}\right) = 0\;\text{ for all }y\text{ so }\;{\int }_{Y... | Yes |
Example 1.7.4. By the axiom of choice and the continuum hypothesis one can define an order relation \( { < }^{\prime } \) on \( \left( {0,1}\right) \) so that \( \left\{ {x : x{ < }^{\prime }y}\right\} \) is countable for each \( y \) . Let \( X = Y = \left( {0,1}\right) \), let \( \mathcal{A} = \mathcal{B} = \) the Bo... | \[ {\int }_{X}f\left( {x, y}\right) {\mu }_{1}\left( {dx}\right) = 0\;\text{ for all }y \] \[ {\int }_{Y}f\left( {x, y}\right) {\mu }_{2}\left( {dy}\right) = 1\;\text{ for all }x \] | Yes |
Lemma 2.1.1. Without loss of generality we can suppose each \( {\mathcal{A}}_{i} \) contains \( \Omega \) . In this case the condition is equivalent to\n\n\[ P\left( {{ \cap }_{i = 1}^{n}{A}_{i}}\right) = \mathop{\prod }\limits_{{i = 1}}^{n}P\left( {A}_{i}\right) \;\text{ whenever }{A}_{i} \in {\mathcal{A}}_{i} \]\n\ns... | Proof. If \( {\mathcal{A}}_{1},{\mathcal{A}}_{2},\ldots ,{\mathcal{A}}_{n} \) are independent and \( {\overline{\mathcal{A}}}_{i} = {\mathcal{A}}_{i} \cup \{ \Omega \} \) then \( {\overline{\mathcal{A}}}_{1},{\overline{\mathcal{A}}}_{2},\ldots ,{\overline{\mathcal{A}}}_{n} \) are independent, since if \( {A}_{i} \in {\... | No |
Theorem 2.1.2. \( \pi - \lambda \) Theorem. If \( \mathcal{P} \) is a \( \pi \) -system and \( \mathcal{L} \) is a \( \lambda \) -system that contains \( \mathcal{P} \) then \( \sigma \left( \mathcal{P}\right) \subset \mathcal{L} \) . | The proof is hidden away in Section A. 1 of the Appendix. | No |
Theorem 2.1.3. Suppose \( {\mathcal{A}}_{1},{\mathcal{A}}_{2},\ldots ,{\mathcal{A}}_{n} \) are independent and each \( {\mathcal{A}}_{i} \) is a \( \pi \) -system. Then \( \sigma \left( {\mathcal{A}}_{1}\right) ,\sigma \left( {\mathcal{A}}_{2}\right) ,\ldots ,\sigma \left( {\mathcal{A}}_{n}\right) \) are independent. | Proof. Let \( {A}_{2},\ldots ,{A}_{n} \) be sets with \( {A}_{i} \in {\mathcal{A}}_{i} \), let \( F = {A}_{2} \cap \cdots \cap {A}_{n} \) and let \( \mathcal{L} = \) \( \{ A : P\left( {A \cap F}\right) = P\left( A\right) P\left( F\right) \} \) . Since \( P\left( {\Omega \cap F}\right) = P\left( \Omega \right) P\left( F... | Yes |
Theorem 2.1.4. In order for \( {X}_{1},\ldots ,{X}_{n} \) to be independent, it is sufficient that for all \( {x}_{1},\ldots ,{x}_{n} \in ( - \infty ,\infty \rbrack \)\n\n\[ P\left( {{X}_{1} \leq {x}_{1},\ldots ,{X}_{n} \leq {x}_{n}}\right) = \mathop{\prod }\limits_{{i = 1}}^{n}P\left( {{X}_{i} \leq {x}_{i}}\right) \] | Proof. Let \( {\mathcal{A}}_{i} = \) the sets of the form \( \left\{ {{X}_{i} \leq {x}_{i}}\right\} \) . Since\n\n\[ \left\{ {{X}_{i} \leq x}\right\} \cap \left\{ {{X}_{i} \leq y}\right\} = \left\{ {{X}_{i} \leq x \land y}\right\} \]\n\nwhere \( {\left( x \land y\right) }_{i} = {x}_{i} \land {y}_{i} = \min \left\{ {{x}... | No |
Theorem 2.1.5. Suppose \( {\mathcal{F}}_{i, j},1 \leq i \leq n,1 \leq j \leq m\left( i\right) \) are independent and let \( {\mathcal{G}}_{i} = \sigma \left( {{ \cup }_{j}{\mathcal{F}}_{i, j}}\right) \) . Then \( {\mathcal{G}}_{1},\ldots ,{\mathcal{G}}_{n} \) are independent. | Proof. Let \( {\mathcal{A}}_{i} \) be the collection of sets of the form \( { \cap }_{j}{A}_{i, j} \) where \( {A}_{i, j} \in {\mathcal{F}}_{i, j}.{\mathcal{A}}_{i} \) is a \( \pi \) -system that contains \( \Omega \) and contains \( { \cup }_{j}{\mathcal{F}}_{i, j} \) so Theorem 2.1.3 implies \( \sigma \left( {\mathca... | Yes |
Theorem 2.1.6. If for \( 1 \leq i \leq n,1 \leq j \leq m\left( i\right) ,{X}_{i, j} \) are independent and \( {f}_{i} \) : \( {\mathbf{R}}^{m\left( i\right) } \rightarrow \mathbf{R} \) are measurable then \( {f}_{i}\left( {{X}_{i,1},\ldots ,{X}_{i, m\left( i\right) }}\right) \) are independent. | Proof. Let \( {\mathcal{F}}_{i, j} = \sigma \left( {X}_{i, j}\right) \) and \( {\mathcal{G}}_{i} = \sigma \left( {{ \cup }_{j}{\mathcal{F}}_{i, j}}\right) \) . Since \( {f}_{i}\left( {{X}_{i,1},\ldots ,{X}_{i, m\left( i\right) }}\right) \in {\mathcal{G}}_{i} \), the desired result follows from Theorem 2.1.5 and Exercis... | No |
Theorem 2.1.7. Suppose \( {X}_{1},\ldots ,{X}_{n} \) are independent random variables and \( {X}_{i} \) has distribution \( {\mu }_{i} \), then \( \left( {{X}_{1},\ldots ,{X}_{n}}\right) \) has distribution \( {\mu }_{1} \times \cdots \times {\mu }_{n} \) . | Proof. Using the definitions of (i) \( {A}_{1} \times \cdots \times {A}_{n} \) ,(ii) independence,(iii) \( {\mu }_{i} \), and (iv) \( {\mu }_{1} \times \cdots \times {\mu }_{n} \)\n\n\[ P\left( {\left( {{X}_{1},\ldots ,{X}_{n}}\right) \in {A}_{1} \times \cdots \times {A}_{n}}\right) = P\left( {{X}_{1} \in {A}_{1},\ldot... | Yes |
Theorem 2.1.8. Suppose \( X \) and \( Y \) are independent and have distributions \( \mu \) and \( \nu \) . If \( h : {\mathbf{R}}^{2} \rightarrow \mathbf{R} \) is a measurable function with \( h \geq 0 \) or \( E\left| {h\left( {X, Y}\right) }\right| < \infty \) then\n\n\[ \n{Eh}\left( {X, Y}\right) = \iint h\left( {x... | Proof. Using Theorem 1.6.9 and then Fubini's theorem (Theorem 1.7.2) we have\n\n\[ \n{Eh}\left( {X, Y}\right) = {\int }_{{\mathbf{R}}^{2}}{hd}\left( {\mu \times \nu }\right) = \iint h\left( {x, y}\right) \mu \left( {dx}\right) \nu \left( {dy}\right) \n\]\n\nTo prove the second result, we start with the result when \( f... | Yes |
Theorem 2.1.9. If \( {X}_{1},\ldots ,{X}_{n} \) are independent and have (a) \( {X}_{i} \geq 0 \) for all \( i \), or (b) \( E\left| {X}_{i}\right| < \infty \) for all \( i \) then \[ E\left( {\mathop{\prod }\limits_{{i = 1}}^{n}{X}_{i}}\right) = \mathop{\prod }\limits_{{i = 1}}^{n}E{X}_{i} \] i.e., the expectation on ... | Proof. \( X = {X}_{1} \) and \( Y = {X}_{2}\cdots {X}_{n} \) are independent by Theorem 2.1.6 so taking \( f\left( x\right) = \left| x\right| \) and \( g\left( y\right) = \left| y\right| \) we have \( E\left| {{X}_{1}\cdots {X}_{n}}\right| = E\left| {X}_{1}\right| E\left| {{X}_{2}\cdots {X}_{n}}\right| \), and it follo... | Yes |
It can happen that \( E\left( {XY}\right) = {EX} \cdot {EY} \) without the variables being independent. Suppose the joint distribution of \( X \) and \( Y \) is given by the following table\n\n\[\n\begin{matrix} & & & & Y & \\ & & & 1 & 0 & - 1 \\ & & 1 & 0 & a & 0 \\ X & 0 & b & c & b & \\ & - 1 & 0 & a & 0 & \end{mat... | Things are arranged so that \( {XY} \equiv 0 \) . Symmetry implies \( {EX} = 0 \) and \( {EY} = 0 \), so \( E\left( {XY}\right) = 0 = {EXEY} \) . The random variables are not independent since\n\n\[\nP\left( {X = 1, Y = 1}\right) = 0 < {ab} = P\left( {X = 1}\right) P\left( {Y = 1}\right)\n\] | Yes |
Theorem 2.1.10. If \( X \) and \( Y \) are independent, \( F\left( x\right) = P\left( {X \leq x}\right) \), and \( G\left( y\right) = \) \( P\left( {Y \leq y}\right) \), then\n\n\[ P\left( {X + Y \leq z}\right) = \int F\left( {z - y}\right) {dG}\left( y}\right) \] | Proof. Let \( h\left( {x, y}\right) = {1}_{\left( x + y \leq z\right) } \) . Let \( \mu \) and \( \nu \) be the probability measures with distribution functions \( F \) and \( G \) . Since for fixed \( y \)\n\n\[ \int h\left( {x, y}\right) \mu \left( {dx}\right) = \int {1}_{( - \infty, z - y\rbrack }\left( x\right) \mu... | Yes |
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