Lusin’s Theorem and Continuous Extension

Here we give proofs for two versions of Lusin’s Theorem, one from Exercise 44, Ch2 in Folland’s Real Analysis and the other from the textbook used for my first year undergraduate mathematical analysis course in Beijing.  The latter version is a stronger result which in addition discusses the condition for a real-valued function defined on a subset of \mathbb{R}^n to be extended to the whole of \mathbb{R}^n. A more general result in topology is the Tietze Extension Theorem. 

See the full post here: Lusin’s Theorem and Continuous Extension

Here we let \mu denote the Lebesgue measure on \mathbb{R}.

Lusin’s Theorem (Version 1)[Exercise 2.44, Folland]. Suppose E\subset \mathbb{R}^n is Lebesgue  measurable, f: E\to \mathbb{R} is Lebesgue measurable and \epsilon> 0, there is a compact set F\subset E such that \mu(F^c)<\epsilon and f|_F is continuous.

Lusin’s Theorem(Version 2)[Huan]. Suppose E\subset \mathbb{R}^n is Lebesgue measurable and f: E\to \bar{\mathbb{R}} is a Lebesgue measurable extended real valued function with \mu(|f|=+ \infty)=0, then  \forall \epsilon >0, \exists g\in C(E) such that \mu(f\neq g)<\epsilon, where C(E) denotes the space of continuous function on E

Continuous Extension Theorem[Huan]. Suppose E\subset \mathbb{R}^n, then f can be extended to a continuous function on \mathbb{R}^n if and only if f can be extended to a continuous function on the closure \bar{E} of E.

Tietze Extension Theorem. Let X be normal and F \subset X be closed and let f: F \to R be continuous. Then there is a map g: X \to R such that
g(x) = f(x) for all x\in F. (Note that in topology, by a map we mean a continuous function. )


The “Dunce Cap” Space Is Contractible

Here is the exercise 6 on P. 50 in the book Topology and Geometry by Glen Bredon. I put it here because I found the drawing of this cap very lovely. Indeed I like that most of the pictures in this book are lovely sketches.

Question. The “dunce cap” space is the quotient of a triangle (and interior) obtained by identifying all three edges in an inconsistent manner. That is, if the vertices of the triangle are p, q, r then we identify the line segment (p, q) with (q, r) and with (p, r) in the
orientation indicated by the order given of the vertices. (See Figure 1-6.) Show that
the dunce cap is contractible.the dunce cap

Following the development in the book, I will use the following theorem that the homotopy type of a mapping cylinder or cone
depends only on the homotopy class of the map [Theorem 14.18, Topology and Geometry by Glen Bredon]. The idea is to identify the dunce cap as a mapping cone.

Theorem 14.18. If f_0\simeq f_1:X\to Y \text{ then } M_{f_0 } \simeq M_{f_1}\text { rel } X+Y \text{ and }C_{f_0}\simeq C_{f_1}\text{ rel } Y+\mathrm{vertex}.

Proof of the Qustion. Suppose f: S^1\to S^1 is a map from S^1 to itself. The cone C_f=M_f/S^1 \times\{1\} for f is obtained by pinching the top of the mapping cylinder to a point. As M_f  is the cylinder S_1\times [0,1] with the bottom pasted to S_1 by the map f, C_f is D_2 with \partial D_2 pasted to S_1 by the map f. So the dunce cap is just C_f \text{ with } f: S^1\to S^1 defined as

f(e^{2\pi i t})= \begin{cases} e^{2\pi i (3t)}, 0\leq t\leq 2/3\\ e^{2\pi i(2- 3t)}, 2/3\leq t\leq 1. \end{cases}

which is homotopic to the identity by a linear homotopy (note that we make the choice of f for an easy definition of the homotopy)

H(e^{2\pi i t},s)= \begin{cases} e^{2\pi i (3t(1-s)+st}, 0\leq t\leq 2/3\\ e^{2\pi i[(2- 3t)(1-s)+st]}, 2/3\leq t\leq 1. \end{cases}.

So the dunce cap is homotopic to C_{id}\simeq D^2 which is contractible.

[Solutions] Ch 2 Manifolds in “Spacetime and geometry”

These are parts of the solutions to exercises from:

Carroll, S. M. (2005). Spacetime and geometry: An introduction to general relativity. Addison Wesley.



  1. First consider mapping the infinite cylinder to a semi-infinite cylinder, then consider projecting points on it to a plane.Capture

Let L=1,

(acos\theta,asin\theta,z)\mapsto ^f (acos\theta,asin\theta,e^z)\mapsto ^g (acos\theta/e^z,asin\theta/e^z,1).

g\circ f is the map desired whose image is an open set R^2\backslash (0,0).

2. No. \mathbb{R}^k must be k-dimension manifolds. Furthermore, the dimension of a manifold is unique.

Suppose that U is a subset of a manifold M, if there are two charts with different dimension m and $latex n$, then there exists \Phi_1, \Phi_2\in C^{\infty}, \Phi_1:U\to \mathbb{R}^m, \Phi_2:U\to \mathbb{R}^n. Then \Phi_2\circ \Phi_1 is a diffeomorphism from \Phi_1(U) to \Phi_2(U). Hence m must be equal to n.

Note: Here I use such an assertion( J. Milnor, Topology from Differentiable Viewpoint, P4).

Assertion. If f is a diffeomorphism between open sets U\subset R^k and V\subset R^l, then k must equal l, and the linear mapping

df_x:R^k\to R^l

must be nonsingular.

Proof. The composition f^{-1}\circ f is the identity map of U; hence d(f^{-1})_v\circ df_x is the identity map of R^k. Similarly df_x\circ d(f^{-1})_v is the identity map of R^l. Thus df_x has a two-sided inverse, and it follows that k=l.

By the way, k=l can also be seen from the max\{rank(BA), rank(AB)\}=min\{rank(A),rank(B)\}, A,B denote the linear map d(f^{-1})_v and df_x respectively.

3. Trivial.

4. First two can be verified directly.

Composition formula:


[X,Y]^{\mu}=[X,Y](x^{\mu})=X(Y(x^{\mu}))-Y(X(x^{\mu}))=X(Y^{\mu})-Y(X^{\mu})\\ =X^{\mu}\partial_{\lambda}Y^{\mu}-Y^{\lambda}\partial_{\lambda}X^{\mu}.


[X,Y]^{\mu'}=X^{\mu'}\partial_{\lambda'}Y^{\mu'}-Y^{\lambda'}\partial_{\lambda'}X^{\mu'}\\  =X^{\mu}\partial_{\lambda}(\frac{\partial x^{\mu'}}{\partial x^{\mu}}Y^{\mu})-Y^{\lambda}\partial_{\lambda}(\frac{\partial x^{\mu'}}{\partial x^{\mu}}X^{\mu})\\ =X^{\lambda}Y^{\mu}\partial_{\lambda}(\frac{\partial x^{\mu'}}{\partial x^{\mu}})+\frac{\partial x^{\mu'}}{\partial x^{\mu}}x^{\lambda}\partial_{\lambda}Y^{\mu}-X^{\lambda}Y^{\mu}\partial_{\lambda}(\frac{\partial x^{\mu'}}{\partial x^{\mu}})-\frac{\partial x^{\mu'}}{\partial x^{\mu}}Y^{\lambda}\partial_{\lambda}X^{\mu}  \\ = \frac{\partial x^{\mu'}}{\partial x^{\mu}}(X^{\mu}\partial_{\lambda}Y^{\mu}-Y^{\lambda}\partial_{\lambda}X^{\mu})= \frac{\partial x^{\mu'}}{\partial x^{\mu}}([X,Y]^{\mu}).

5. Set  V_{1}=\partial _{x}, V_{2} = A(x, y)\partial _{x} + B(x, y)\partial _{y}. The commutator will be given by

[V_{1} , V_{2} ] = [\partial _{x} A(x, y)]\partial _{x} + [\partial _{x} B(x, y)]\partial _{y}.

Set V_{2} =x \partial _{x} + \partial _{y}. Then V_1, V_2 are nowhere vanishing, and their commutator is nowhere vanishing as well.




\theta=arccos \frac{z}{\sqrt{x^2+y^2+z^2}}=arccos\frac{\lambda}{\sqrt{1+\lambda^2}}


(\lambda\in [0,2\pi))


Cartesian: (-sin\lambda, cos\lambda,1),



(a) \frac{\partial x^{\mu}}{\partial x^{\nu'}}=\frac{\partial(x,z)}{\partial(\chi,\theta)}=\left(\begin{array}{cc} {cosh\chi sin\theta}&{sinh\chi cos\theta}\\{sinh\chi cos\theta}&{-cosh\chi sin\theta} \end{array}\right)
(b)dx=cosh\chi sin\theta dx+sinh\chi cos\theta d\theta,
dz=sinh\chi cos\theta dx-cosh\chi sin\theta d\theta,



(a)d*F=3\partial_{[\mu}F_{\nu\rho ]}=1/2[\partial_{\mu}(\epsilon^{\nu_1\nu_2}_{\nu\rho}F_{\nu_1\nu_2})+\partial_{\nu}(\epsilon^{\nu_1\nu_2}_{\rho\mu}F_{\nu_1\nu_2})+\partial_{\rho}(\epsilon^{\nu_1\nu_2}_{\mu\nu}F_{\nu_1\nu_2})]=\epsilon^{\sigma}_{\mu\nu\rho}J_{\rho}=*J
*F=\left( \begin{array}{cccc} 0&0 &0 &0 \\0 &0 &0 &0 \\0  &0 &0 &qsin\theta\\ 0&0 &-qsin\theta&0 \end{array}\right)
F=(- **F)=\left( \begin{array}{cccc} 0&-q/r^2 &0 &0 \\q/r^2 &0 &0 &0 \\0  &0 &0 &0\\ 0&0 &0&0 \end{array}\right)
B_{\mu}=0, \mu=r,\theta,\phi




[Solutions]Ch 1 Special Relativity and Flat Spacetime


9. For a system of discrete point particles the energy-momentum tensor take the form

T^{\mu \nu}=\Sigma_a \frac{p_{\mu}^{(a)} p_{\nu}^{(a)} }{p^{(a)}}\delta^{(3)}(x-x^{(a)}),

where the index a labels the different particles. Show that, for a dense collection of particles with isotopically distributed velocities, we can smooth over the individual particle worldlines to obtain the perfect-fluid energy-momentum tensor

T^{\mu\nu}=(\rho +p)U^{\mu}U^{\nu} + p\eta^{\mu\nu}.


Doing average over 4-volume,

{\bf T}^{\mu\nu}=\dfrac{1}{\Delta V_4}\int_{\Delta V_4}T^{\mu\nu}d V=\dfrac{1}{\sqrt{-g} d^3x^i dx^0}\int_{\Delta V_4}T^{\mu\nu} \sqrt{-g} d^3x^i dx^0 .

Then a) in {\bf T}^{\mu\nu} only delta-functions depend on x, b) metric determinant g is a macroscopic quantity, is constant over selected volume and can also be taken away from the integral.

{\bf T}^{\mu\nu}=\dfrac{1}{d^3x^i dx^0}\sum_a\dfrac{p_a^\mu p_a^\nu}{p_a^0}\int_{\Delta V_4} \delta^{(3)}({\bf x}-{\bf x}^{(a)}) d^3 x^i dx^0 = \dfrac{1}{d^3x^i}\sum_a\dfrac{p_a^\mu p_a^\nu}{p_a^0}

In the last expression the sum is taken over the particles which have worldlines passing through \Delta V_4.

Consider symmetry,

{\bf T}^{0 0} = \dfrac{1}{d^3x^i}\sum_a p_a^0 \equiv \rho

{\bf T}^{i 0} = \dfrac{1}{d^3x^i}\sum_a p_a^i. Because of isotropy, {\bf T}^{i 0}= 0.

{\bf T}^{i j} = \dfrac{1}{d^3x^i}\sum_a\dfrac{p_a^i p_a^j}{p_a^0}. The sum should produce a symmetric macroscopic 3-tensor of second order. But all 3-tensors are defined by 3 eigenvectors. Because no preferred direction exits and no preferred directions correspond to the case when the matrix has all eigenvalues equal, that is when the matrix is proportional to kronecker delta. The coefficient of proportionality is the pressure: {\bf T}^{i j} \equiv P \delta^{ij}.

Replace :

\delta ^{ij}\to \eta^{ij}+U^i U^j

T^{00}\to \rho U^0U^0,

then we get the result.

10. Using the tensor transformation law applied to F_{\mu\nu}, show how the electric and magnetic field 3-vectors E and B transform under

(a) a rotation about the y-axis;

(b) a boost along the z-axis.


(a) \Lambda^{\mu'}_{\nu}=\left(\begin{array}{cccc} 1&0&0&0 \\ 0&cos\theta&0&sin\theta\\ 0&0&1&0\\ 0&-sin\theta&0&cos\theta \end{array}\right)

(b)\Lambda^{\mu'}_{\nu}=\left(\begin{array}{cccc} cosh\theta&0&0&-sinh\theta \\ 0&1&0&0\\ 0&0&1&0\\ -sinh\theta&0&0&cosh\theta \end{array}\right)

\overline{F}^{\mu\nu}=\Lambda F \Lambda^T=\left(\begin{array}{cccc}                            0& - E1*cosh(x) - B2*sinh(x)& B1*sinh(x) - E2*cosh(x)& E3*sinh(x)^2 - E3*cosh(x)^2\\ E1*cosh(x) + B2*sinh(x)&                     0&                      B3&- B2*cosh(x) - E1*sinh(x)\\ E2*cosh(x) - B1*sinh(x)&                       -B3&                       0&     B1*cosh(x) - E2*sinh(x)\\ E3*cosh(x)^2 - E3*sinh(x)^2&   B2*cosh(x) + E1*sinh(x)&E2*sinh(x) - B1*cosh(x)&                           0 \end{array}\right).

13. Consider adding to the Lagrangian for electromagnetism an additional term of the form \mathcal{L}'=\tilde{\epsilon}_{\mu\nu\rho\sigma}F^{\mu\nu}F{\rho\sigma}

(a) Express \mathcal{L}' in terms of E and B.

(b) Show that including \mathcal{L}'  does not affect Maxwell’s equations. Can you think of a deep reason for this?


(a) \tilde{\epsilon}_{\mu\nu\rho\sigma}F^{\mu\nu}F{\rho\sigma}=\tilde{\epsilon}_{ijk}(F^i F^{jk}-F^{i0}F^{jk}+F^{ij}F^{0k}-F^{ij}F^{k0})=4F^{0i}F^{jk}\tilde{\epsilon}^{ijk}=-8\vec{E}\vec{B}

(b) The expression in (a) can be expressed by 4-dimension divergence:

\tilde{\epsilon}_{\mu\nu\rho\sigma}F^{\mu\nu}F{\rho\sigma}=4\partial_i[\tilde{\epsilon}^{ijlm}A_k (\partial_l A_m)]

Integrate \mathcal{L}' over 4-dimension space, according to Stokes theorem, we can get the additional term in Lagrangian L. This term vanishes when varying action S